logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proofs

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?

1. A proof is a logically consistent argument that leads from ahypothesis p to a conclusion q.

2. Whether p and q are true in a larger sense is immaterial tothe internal consistency of the proof itself: In some proofswe deliberately start with what we believe to be a falsehypothesis and lead it to a contradiction.

3. So a proof must be a logical construct that connects p to qand which is so that, independent of the truth values of pand q, the construct remains valid. That is, the logicalstructure must be a compound logical statement that isalways true.

4. Compound logical statements that are always true arecalled tautologies.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?1. A proof is a logically consistent argument that leads from a

hypothesis p to a conclusion q.

2. Whether p and q are true in a larger sense is immaterial tothe internal consistency of the proof itself: In some proofswe deliberately start with what we believe to be a falsehypothesis and lead it to a contradiction.

3. So a proof must be a logical construct that connects p to qand which is so that, independent of the truth values of pand q, the construct remains valid. That is, the logicalstructure must be a compound logical statement that isalways true.

4. Compound logical statements that are always true arecalled tautologies.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?1. A proof is a logically consistent argument that leads from a

hypothesis p to a conclusion q.2. Whether p and q are true in a larger sense is immaterial to

the internal consistency of the proof itself

: In some proofswe deliberately start with what we believe to be a falsehypothesis and lead it to a contradiction.

3. So a proof must be a logical construct that connects p to qand which is so that, independent of the truth values of pand q, the construct remains valid. That is, the logicalstructure must be a compound logical statement that isalways true.

4. Compound logical statements that are always true arecalled tautologies.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?1. A proof is a logically consistent argument that leads from a

hypothesis p to a conclusion q.2. Whether p and q are true in a larger sense is immaterial to

the internal consistency of the proof itself: In some proofswe deliberately start with what we believe to be a falsehypothesis and lead it to a contradiction.

3. So a proof must be a logical construct that connects p to qand which is so that, independent of the truth values of pand q, the construct remains valid. That is, the logicalstructure must be a compound logical statement that isalways true.

4. Compound logical statements that are always true arecalled tautologies.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?1. A proof is a logically consistent argument that leads from a

hypothesis p to a conclusion q.2. Whether p and q are true in a larger sense is immaterial to

the internal consistency of the proof itself: In some proofswe deliberately start with what we believe to be a falsehypothesis and lead it to a contradiction.

3. So a proof must be a logical construct that connects p to qand which is so that, independent of the truth values of pand q, the construct remains valid.

That is, the logicalstructure must be a compound logical statement that isalways true.

4. Compound logical statements that are always true arecalled tautologies.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?1. A proof is a logically consistent argument that leads from a

hypothesis p to a conclusion q.2. Whether p and q are true in a larger sense is immaterial to

the internal consistency of the proof itself: In some proofswe deliberately start with what we believe to be a falsehypothesis and lead it to a contradiction.

3. So a proof must be a logical construct that connects p to qand which is so that, independent of the truth values of pand q, the construct remains valid. That is, the logicalstructure must be a compound logical statement that isalways true.

4. Compound logical statements that are always true arecalled tautologies.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?1. A proof is a logically consistent argument that leads from a

hypothesis p to a conclusion q.2. Whether p and q are true in a larger sense is immaterial to

the internal consistency of the proof itself: In some proofswe deliberately start with what we believe to be a falsehypothesis and lead it to a contradiction.

3. So a proof must be a logical construct that connects p to qand which is so that, independent of the truth values of pand q, the construct remains valid. That is, the logicalstructure must be a compound logical statement that isalways true.

4. Compound logical statements that are always true arecalled tautologies.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?

5. So to understand proofs, we need to collect tautologies thatcan serve as templates for proofs.

6. Thankfully there is a fairly small number of such“templates”.

7. But we must be careful, because the use of these templatesin actual proofs is neither prescriptive, nor schematic. Sothey are another set of reasoning structures that we mustautomatize.

8. We present them all at once to provide all possible tools.Facility with the tools will come with practice.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?5. So to understand proofs, we need to collect tautologies that

can serve as templates for proofs.

6. Thankfully there is a fairly small number of such“templates”.

7. But we must be careful, because the use of these templatesin actual proofs is neither prescriptive, nor schematic. Sothey are another set of reasoning structures that we mustautomatize.

8. We present them all at once to provide all possible tools.Facility with the tools will come with practice.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?5. So to understand proofs, we need to collect tautologies that

can serve as templates for proofs.6. Thankfully there is a fairly small number of such

“templates”.

7. But we must be careful, because the use of these templatesin actual proofs is neither prescriptive, nor schematic. Sothey are another set of reasoning structures that we mustautomatize.

8. We present them all at once to provide all possible tools.Facility with the tools will come with practice.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?5. So to understand proofs, we need to collect tautologies that

can serve as templates for proofs.6. Thankfully there is a fairly small number of such

“templates”.7. But we must be careful, because the use of these templates

in actual proofs is neither prescriptive, nor schematic.

Sothey are another set of reasoning structures that we mustautomatize.

8. We present them all at once to provide all possible tools.Facility with the tools will come with practice.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?5. So to understand proofs, we need to collect tautologies that

can serve as templates for proofs.6. Thankfully there is a fairly small number of such

“templates”.7. But we must be careful, because the use of these templates

in actual proofs is neither prescriptive, nor schematic. Sothey are another set of reasoning structures that we mustautomatize.

8. We present them all at once to provide all possible tools.Facility with the tools will come with practice.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?5. So to understand proofs, we need to collect tautologies that

can serve as templates for proofs.6. Thankfully there is a fairly small number of such

“templates”.7. But we must be careful, because the use of these templates

in actual proofs is neither prescriptive, nor schematic. Sothey are another set of reasoning structures that we mustautomatize.

8. We present them all at once to provide all possible tools.

Facility with the tools will come with practice.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

What is a Proof?5. So to understand proofs, we need to collect tautologies that

can serve as templates for proofs.6. Thankfully there is a fairly small number of such

“templates”.7. But we must be careful, because the use of these templates

in actual proofs is neither prescriptive, nor schematic. Sothey are another set of reasoning structures that we mustautomatize.

8. We present them all at once to provide all possible tools.Facility with the tools will come with practice.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof

1. The statement[p∧ (p⇒ q)

]⇒ q is a tautology.

2. So if we know that p is true and that p⇒ q is true, then wecan infer that q is true.

3. This method is called a direct proof, or the law ofdetachment (because we detach q from the implication),or modus ponens (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof1. The statement

[p∧ (p⇒ q)

]⇒ q is a tautology.

2. So if we know that p is true and that p⇒ q is true, then wecan infer that q is true.

3. This method is called a direct proof, or the law ofdetachment (because we detach q from the implication),or modus ponens (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof1. The statement

[p∧ (p⇒ q)

]⇒ q is a tautology.

2. So if we know that p is true and that p⇒ q is true, then wecan infer that q is true.

3. This method is called a direct proof, or the law ofdetachment (because we detach q from the implication),or modus ponens (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof1. The statement

[p∧ (p⇒ q)

]⇒ q is a tautology.

2. So if we know that p is true and that p⇒ q is true, then wecan infer that q is true.

3. This method is called a direct proof, or the law ofdetachment (because we detach q from the implication),or modus ponens (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

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We will use that the sum of the angles in a triangle is 180◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

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We will use that the sum of the angles in a triangle is 180◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

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We will use that the sum of the angles in a triangle is 180◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

We will use that the sum of the angles in a triangle is 180◦.

Letthe vertices be A

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

We will use that the sum of the angles in a triangle is 180◦. Letthe vertices be A

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

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B

We will use that the sum of the angles in a triangle is 180◦. Letthe vertices be A, B

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

We will use that the sum of the angles in a triangle is 180◦. Letthe vertices be A, B, C

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

We will use that the sum of the angles in a triangle is 180◦. Letthe vertices be A, B, C and D

, with A opposite C.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

We will use that the sum of the angles in a triangle is 180◦. Letthe vertices be A, B, C and D, with A opposite C.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

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��

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���A

B

C

D

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

α

The angle at each vertex is labeled by the corresponding Greekletter: α

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

α

The angle at each vertex is labeled by the corresponding Greekletter: α

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

α

The angle at each vertex is labeled by the corresponding Greekletter: α , β

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

γ

α

The angle at each vertex is labeled by the corresponding Greekletter: α , β , γ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

The angle at each vertex is labeled by the corresponding Greekletter: α , β , γ and δ .

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

The line through A and C splits the quadrilateral into twotriangles.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

The line through A and C splits the quadrilateral into twotriangles.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

Label the new angles at A as α1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

Label the new angles at A as α1 and α2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

γ1

Label the new angles at C as γ1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

γ1

Label the new angles at C as γ1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

γ1

γ2

Label the new angles at C as γ1 and γ2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

γ1

γ2

Now α1 +α2 = α and γ1 + γ2 = γ .ABC is a triangle, so α2 +β + γ2 = 180◦. ACD is a triangle, soα1 +δ + γ1 = 180◦. Hence α +β + γ +δ = 360◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

γ1

γ2

Now α1 +α2 = α

and γ1 + γ2 = γ .ABC is a triangle, so α2 +β + γ2 = 180◦. ACD is a triangle, soα1 +δ + γ1 = 180◦. Hence α +β + γ +δ = 360◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

γ1

γ2

Now α1 +α2 = α and γ1 + γ2 = γ .

ABC is a triangle, so α2 +β + γ2 = 180◦. ACD is a triangle, soα1 +δ + γ1 = 180◦. Hence α +β + γ +δ = 360◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

γ1

γ2

Now α1 +α2 = α and γ1 + γ2 = γ .ABC is a triangle, so α2 +β + γ2 = 180◦.

ACD is a triangle, soα1 +δ + γ1 = 180◦. Hence α +β + γ +δ = 360◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

γ1

γ2

Now α1 +α2 = α and γ1 + γ2 = γ .ABC is a triangle, so α2 +β + γ2 = 180◦. ACD is a triangle, soα1 +δ + γ1 = 180◦.

Hence α +β + γ +δ = 360◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

γ1

γ2

Now α1 +α2 = α and γ1 + γ2 = γ .ABC is a triangle, so α2 +β + γ2 = 180◦. ACD is a triangle, soα1 +δ + γ1 = 180◦. Hence α +β + γ +δ = 360◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Direct Proof That the Sum of the Angles in aConvex Quadrilateral is 360◦

ZZ

ZZ

ZZ�

�����

@@

@@

@@�

��

��

���A

B

C

D

β

δ

γ

α

α1

α2

γ1

γ2

Now α1 +α2 = α and γ1 + γ2 = γ .ABC is a triangle, so α2 +β + γ2 = 180◦. ACD is a triangle, soα1 +δ + γ1 = 180◦. Hence α +β + γ +δ = 360◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction

1. The statement[p∧

((p∧¬q)⇒ FALSE

)]⇒ q is a

tautology.2. So if we know that p is true and if assuming that q is false

leads to a contradiction (a statement that is always false),then we can infer that q is true.

3. Basically the idea is that because a true statement cannotimply a false statement, p∧¬q must be false. Hence,because p is true, ¬q must be false. Therefore q must betrue.

4. This method is called a proof by contradiction, orreductio ad absurdum (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction1. The statement

[p∧

((p∧¬q)⇒ FALSE

)]⇒ q is a

tautology.

2. So if we know that p is true and if assuming that q is falseleads to a contradiction (a statement that is always false),then we can infer that q is true.

3. Basically the idea is that because a true statement cannotimply a false statement, p∧¬q must be false. Hence,because p is true, ¬q must be false. Therefore q must betrue.

4. This method is called a proof by contradiction, orreductio ad absurdum (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction1. The statement

[p∧

((p∧¬q)⇒ FALSE

)]⇒ q is a

tautology.2. So if we know that p is true and if assuming that q is false

leads to a contradiction (a statement that is always false),then we can infer that q is true.

3. Basically the idea is that because a true statement cannotimply a false statement, p∧¬q must be false. Hence,because p is true, ¬q must be false. Therefore q must betrue.

4. This method is called a proof by contradiction, orreductio ad absurdum (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction1. The statement

[p∧

((p∧¬q)⇒ FALSE

)]⇒ q is a

tautology.2. So if we know that p is true and if assuming that q is false

leads to a contradiction (a statement that is always false),then we can infer that q is true.

3. Basically the idea is that because a true statement cannotimply a false statement, p∧¬q must be false.

Hence,because p is true, ¬q must be false. Therefore q must betrue.

4. This method is called a proof by contradiction, orreductio ad absurdum (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction1. The statement

[p∧

((p∧¬q)⇒ FALSE

)]⇒ q is a

tautology.2. So if we know that p is true and if assuming that q is false

leads to a contradiction (a statement that is always false),then we can infer that q is true.

3. Basically the idea is that because a true statement cannotimply a false statement, p∧¬q must be false. Hence,because p is true, ¬q must be false.

Therefore q must betrue.

4. This method is called a proof by contradiction, orreductio ad absurdum (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction1. The statement

[p∧

((p∧¬q)⇒ FALSE

)]⇒ q is a

tautology.2. So if we know that p is true and if assuming that q is false

leads to a contradiction (a statement that is always false),then we can infer that q is true.

3. Basically the idea is that because a true statement cannotimply a false statement, p∧¬q must be false. Hence,because p is true, ¬q must be false. Therefore q must betrue.

4. This method is called a proof by contradiction, orreductio ad absurdum (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction1. The statement

[p∧

((p∧¬q)⇒ FALSE

)]⇒ q is a

tautology.2. So if we know that p is true and if assuming that q is false

leads to a contradiction (a statement that is always false),then we can infer that q is true.

3. Basically the idea is that because a true statement cannotimply a false statement, p∧¬q must be false. Hence,because p is true, ¬q must be false. Therefore q must betrue.

4. This method is called a proof by contradiction, orreductio ad absurdum (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific Method

1. In science, experiments and observations are used to testhypotheses/theories.

2. Hypotheses/theories that do not accurately predict theexperiment or which contradict the observation areconsidered false and must be discarded.

3. This is exactly the structure of a proof by contradiction:We assume the hypothesis/theory is correct. Then wederive a prediction p. If the prediction p does not match theresults of the experiment/observation, then we have theclassical contradiction p∧¬p.This establishes that the hypothesis/theory is not correct.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific Method1. In science, experiments and observations are used to test

hypotheses/theories.

2. Hypotheses/theories that do not accurately predict theexperiment or which contradict the observation areconsidered false and must be discarded.

3. This is exactly the structure of a proof by contradiction:We assume the hypothesis/theory is correct. Then wederive a prediction p. If the prediction p does not match theresults of the experiment/observation, then we have theclassical contradiction p∧¬p.This establishes that the hypothesis/theory is not correct.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific Method1. In science, experiments and observations are used to test

hypotheses/theories.2. Hypotheses/theories that do not accurately predict the

experiment or which contradict the observation areconsidered false and must be discarded.

3. This is exactly the structure of a proof by contradiction:We assume the hypothesis/theory is correct. Then wederive a prediction p. If the prediction p does not match theresults of the experiment/observation, then we have theclassical contradiction p∧¬p.This establishes that the hypothesis/theory is not correct.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific Method1. In science, experiments and observations are used to test

hypotheses/theories.2. Hypotheses/theories that do not accurately predict the

experiment or which contradict the observation areconsidered false and must be discarded.

3. This is exactly the structure of a proof by contradiction

:We assume the hypothesis/theory is correct. Then wederive a prediction p. If the prediction p does not match theresults of the experiment/observation, then we have theclassical contradiction p∧¬p.This establishes that the hypothesis/theory is not correct.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific Method1. In science, experiments and observations are used to test

hypotheses/theories.2. Hypotheses/theories that do not accurately predict the

experiment or which contradict the observation areconsidered false and must be discarded.

3. This is exactly the structure of a proof by contradiction:We assume the hypothesis/theory is correct.

Then wederive a prediction p. If the prediction p does not match theresults of the experiment/observation, then we have theclassical contradiction p∧¬p.This establishes that the hypothesis/theory is not correct.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific Method1. In science, experiments and observations are used to test

hypotheses/theories.2. Hypotheses/theories that do not accurately predict the

experiment or which contradict the observation areconsidered false and must be discarded.

3. This is exactly the structure of a proof by contradiction:We assume the hypothesis/theory is correct. Then wederive a prediction p.

If the prediction p does not match theresults of the experiment/observation, then we have theclassical contradiction p∧¬p.This establishes that the hypothesis/theory is not correct.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific Method1. In science, experiments and observations are used to test

hypotheses/theories.2. Hypotheses/theories that do not accurately predict the

experiment or which contradict the observation areconsidered false and must be discarded.

3. This is exactly the structure of a proof by contradiction:We assume the hypothesis/theory is correct. Then wederive a prediction p. If the prediction p does not match theresults of the experiment/observation

, then we have theclassical contradiction p∧¬p.This establishes that the hypothesis/theory is not correct.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific Method1. In science, experiments and observations are used to test

hypotheses/theories.2. Hypotheses/theories that do not accurately predict the

experiment or which contradict the observation areconsidered false and must be discarded.

3. This is exactly the structure of a proof by contradiction:We assume the hypothesis/theory is correct. Then wederive a prediction p. If the prediction p does not match theresults of the experiment/observation, then we have theclassical contradiction p∧¬p.

This establishes that the hypothesis/theory is not correct.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific Method1. In science, experiments and observations are used to test

hypotheses/theories.2. Hypotheses/theories that do not accurately predict the

experiment or which contradict the observation areconsidered false and must be discarded.

3. This is exactly the structure of a proof by contradiction:We assume the hypothesis/theory is correct. Then wederive a prediction p. If the prediction p does not match theresults of the experiment/observation, then we have theclassical contradiction p∧¬p.This establishes that the hypothesis/theory is not correct.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific Method

Incorrect hypotheses and theories that have been dismissed inthis fashion include the assumption that the Earth is flat (it canbe circumnavigated) or that the Earth is the center of the solarsystem, with all planets and the sun revolving around it(contradicts observations).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific MethodIncorrect hypotheses and theories that have been dismissed inthis fashion include the assumption that the Earth is flat

(it canbe circumnavigated) or that the Earth is the center of the solarsystem, with all planets and the sun revolving around it(contradicts observations).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific MethodIncorrect hypotheses and theories that have been dismissed inthis fashion include the assumption that the Earth is flat (it canbe circumnavigated)

or that the Earth is the center of the solarsystem, with all planets and the sun revolving around it(contradicts observations).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific MethodIncorrect hypotheses and theories that have been dismissed inthis fashion include the assumption that the Earth is flat (it canbe circumnavigated) or that the Earth is the center of the solarsystem, with all planets and the sun revolving around it

(contradicts observations).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

The Scientific MethodIncorrect hypotheses and theories that have been dismissed inthis fashion include the assumption that the Earth is flat (it canbe circumnavigated) or that the Earth is the center of the solarsystem, with all planets and the sun revolving around it(contradicts observations).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction (Justification)

[p∧

((p∧¬q)⇒ FALSE

)]⇒ q

= ¬[p∧

((¬p∨q)∨FALSE

)]∨q

= ¬[p∧ (¬p∨q)

]∨q

= ¬[(p∧¬p)︸ ︷︷ ︸=FALSE

∨(p∧q)]∨q

= ¬(p∧q)∨q= (¬p∨¬q)∨q= ¬p∨ (¬q∨q)︸ ︷︷ ︸

=TRUE= TRUE

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction (Justification)[p∧

((p∧¬q)⇒ FALSE

)]⇒ q

= ¬[p∧

((¬p∨q)∨FALSE

)]∨q

= ¬[p∧ (¬p∨q)

]∨q

= ¬[(p∧¬p)︸ ︷︷ ︸=FALSE

∨(p∧q)]∨q

= ¬(p∧q)∨q= (¬p∨¬q)∨q= ¬p∨ (¬q∨q)︸ ︷︷ ︸

=TRUE= TRUE

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction (Justification)[p∧

((p∧¬q)⇒ FALSE

)]⇒ q

= ¬[p∧

((¬p∨q)∨FALSE

)]∨q

= ¬[p∧ (¬p∨q)

]∨q

= ¬[(p∧¬p)︸ ︷︷ ︸=FALSE

∨(p∧q)]∨q

= ¬(p∧q)∨q= (¬p∨¬q)∨q= ¬p∨ (¬q∨q)︸ ︷︷ ︸

=TRUE= TRUE

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction (Justification)[p∧

((p∧¬q)⇒ FALSE

)]⇒ q

= ¬[p∧

((¬p∨q)∨FALSE

)]∨q

= ¬[p∧ (¬p∨q)

]∨q

= ¬[(p∧¬p)︸ ︷︷ ︸=FALSE

∨(p∧q)]∨q

= ¬(p∧q)∨q= (¬p∨¬q)∨q= ¬p∨ (¬q∨q)︸ ︷︷ ︸

=TRUE= TRUE

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction (Justification)[p∧

((p∧¬q)⇒ FALSE

)]⇒ q

= ¬[p∧

((¬p∨q)∨FALSE

)]∨q

= ¬[p∧ (¬p∨q)

]∨q

= ¬[(p∧¬p)︸ ︷︷ ︸=FALSE

∨(p∧q)]∨q

= ¬(p∧q)∨q= (¬p∨¬q)∨q= ¬p∨ (¬q∨q)︸ ︷︷ ︸

=TRUE= TRUE

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction (Justification)[p∧

((p∧¬q)⇒ FALSE

)]⇒ q

= ¬[p∧

((¬p∨q)∨FALSE

)]∨q

= ¬[p∧ (¬p∨q)

]∨q

= ¬[(p∧¬p)︸ ︷︷ ︸=FALSE

∨(p∧q)]∨q

= ¬(p∧q)∨q

= (¬p∨¬q)∨q= ¬p∨ (¬q∨q)︸ ︷︷ ︸

=TRUE= TRUE

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction (Justification)[p∧

((p∧¬q)⇒ FALSE

)]⇒ q

= ¬[p∧

((¬p∨q)∨FALSE

)]∨q

= ¬[p∧ (¬p∨q)

]∨q

= ¬[(p∧¬p)︸ ︷︷ ︸=FALSE

∨(p∧q)]∨q

= ¬(p∧q)∨q= (¬p∨¬q)∨q

= ¬p∨ (¬q∨q)︸ ︷︷ ︸=TRUE

= TRUE

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction (Justification)[p∧

((p∧¬q)⇒ FALSE

)]⇒ q

= ¬[p∧

((¬p∨q)∨FALSE

)]∨q

= ¬[p∧ (¬p∨q)

]∨q

= ¬[(p∧¬p)︸ ︷︷ ︸=FALSE

∨(p∧q)]∨q

= ¬(p∧q)∨q= (¬p∨¬q)∨q= ¬p∨ (¬q∨q)︸ ︷︷ ︸

=TRUE

= TRUE

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction (Justification)[p∧

((p∧¬q)⇒ FALSE

)]⇒ q

= ¬[p∧

((¬p∨q)∨FALSE

)]∨q

= ¬[p∧ (¬p∨q)

]∨q

= ¬[(p∧¬p)︸ ︷︷ ︸=FALSE

∨(p∧q)]∨q

= ¬(p∧q)∨q= (¬p∨¬q)∨q= ¬p∨ (¬q∨q)︸ ︷︷ ︸

=TRUE= TRUE

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction That√

2 is Not an Integer

Suppose for a contradiction that r =√

2 is an integer. Becausesquare roots are nonnegative and 02 = 0, r must be positive. Butthen r ≥ 1, which means r2 ≥ r. Now 12 = 1 6= 2, so r ≥ 2.Moreover, 22 = 4 6= 2, so r ≥ 3. But then r2 ≥ r ≥ 3 impliesthat r2 6= 2, contradicting the assumption that r2 = 2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction That√

2 is Not an IntegerSuppose for a contradiction that r =

√2 is an integer.

Becausesquare roots are nonnegative and 02 = 0, r must be positive. Butthen r ≥ 1, which means r2 ≥ r. Now 12 = 1 6= 2, so r ≥ 2.Moreover, 22 = 4 6= 2, so r ≥ 3. But then r2 ≥ r ≥ 3 impliesthat r2 6= 2, contradicting the assumption that r2 = 2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction That√

2 is Not an IntegerSuppose for a contradiction that r =

√2 is an integer. Because

square roots are nonnegative and 02 = 0, r must be positive.

Butthen r ≥ 1, which means r2 ≥ r. Now 12 = 1 6= 2, so r ≥ 2.Moreover, 22 = 4 6= 2, so r ≥ 3. But then r2 ≥ r ≥ 3 impliesthat r2 6= 2, contradicting the assumption that r2 = 2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction That√

2 is Not an IntegerSuppose for a contradiction that r =

√2 is an integer. Because

square roots are nonnegative and 02 = 0, r must be positive. Butthen r ≥ 1, which means r2 ≥ r.

Now 12 = 1 6= 2, so r ≥ 2.Moreover, 22 = 4 6= 2, so r ≥ 3. But then r2 ≥ r ≥ 3 impliesthat r2 6= 2, contradicting the assumption that r2 = 2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction That√

2 is Not an IntegerSuppose for a contradiction that r =

√2 is an integer. Because

square roots are nonnegative and 02 = 0, r must be positive. Butthen r ≥ 1, which means r2 ≥ r. Now 12 = 1 6= 2, so r ≥ 2.

Moreover, 22 = 4 6= 2, so r ≥ 3. But then r2 ≥ r ≥ 3 impliesthat r2 6= 2, contradicting the assumption that r2 = 2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction That√

2 is Not an IntegerSuppose for a contradiction that r =

√2 is an integer. Because

square roots are nonnegative and 02 = 0, r must be positive. Butthen r ≥ 1, which means r2 ≥ r. Now 12 = 1 6= 2, so r ≥ 2.Moreover, 22 = 4 6= 2, so r ≥ 3.

But then r2 ≥ r ≥ 3 impliesthat r2 6= 2, contradicting the assumption that r2 = 2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction That√

2 is Not an IntegerSuppose for a contradiction that r =

√2 is an integer. Because

square roots are nonnegative and 02 = 0, r must be positive. Butthen r ≥ 1, which means r2 ≥ r. Now 12 = 1 6= 2, so r ≥ 2.Moreover, 22 = 4 6= 2, so r ≥ 3. But then r2 ≥ r ≥ 3 impliesthat r2 6= 2

, contradicting the assumption that r2 = 2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction That√

2 is Not an IntegerSuppose for a contradiction that r =

√2 is an integer. Because

square roots are nonnegative and 02 = 0, r must be positive. Butthen r ≥ 1, which means r2 ≥ r. Now 12 = 1 6= 2, so r ≥ 2.Moreover, 22 = 4 6= 2, so r ≥ 3. But then r2 ≥ r ≥ 3 impliesthat r2 6= 2, contradicting the assumption that r2 = 2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Contradiction That√

2 is Not an IntegerSuppose for a contradiction that r =

√2 is an integer. Because

square roots are nonnegative and 02 = 0, r must be positive. Butthen r ≥ 1, which means r2 ≥ r. Now 12 = 1 6= 2, so r ≥ 2.Moreover, 22 = 4 6= 2, so r ≥ 3. But then r2 ≥ r ≥ 3 impliesthat r2 6= 2, contradicting the assumption that r2 = 2.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction

1. The statement[p∧ (A∨B)∧

((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.2. Basically, if we can prove a result under an additional

hypothesis A, if we can also prove the result under theadditional hypothesis B and if A or B is true, then q is true.

3. In a typical proof by case distinction A∨B is a tautology,that is, one of A and B is always true. But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases. We could distinguish any number of cases, using asimilar structure. We will not state the correspondingtautology. (Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction1. The statement[

p∧ (A∨B)∧((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.

2. Basically, if we can prove a result under an additionalhypothesis A, if we can also prove the result under theadditional hypothesis B and if A or B is true, then q is true.

3. In a typical proof by case distinction A∨B is a tautology,that is, one of A and B is always true. But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases. We could distinguish any number of cases, using asimilar structure. We will not state the correspondingtautology. (Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction1. The statement[

p∧ (A∨B)∧((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.2. Basically, if we can prove a result under an additional

hypothesis A

, if we can also prove the result under theadditional hypothesis B and if A or B is true, then q is true.

3. In a typical proof by case distinction A∨B is a tautology,that is, one of A and B is always true. But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases. We could distinguish any number of cases, using asimilar structure. We will not state the correspondingtautology. (Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction1. The statement[

p∧ (A∨B)∧((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.2. Basically, if we can prove a result under an additional

hypothesis A, if we can also prove the result under theadditional hypothesis B

and if A or B is true, then q is true.3. In a typical proof by case distinction A∨B is a tautology,

that is, one of A and B is always true. But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases. We could distinguish any number of cases, using asimilar structure. We will not state the correspondingtautology. (Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction1. The statement[

p∧ (A∨B)∧((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.2. Basically, if we can prove a result under an additional

hypothesis A, if we can also prove the result under theadditional hypothesis B and if A or B is true, then q is true.

3. In a typical proof by case distinction A∨B is a tautology,that is, one of A and B is always true. But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases. We could distinguish any number of cases, using asimilar structure. We will not state the correspondingtautology. (Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction1. The statement[

p∧ (A∨B)∧((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.2. Basically, if we can prove a result under an additional

hypothesis A, if we can also prove the result under theadditional hypothesis B and if A or B is true, then q is true.

3. In a typical proof by case distinction A∨B is a tautology

,that is, one of A and B is always true. But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases. We could distinguish any number of cases, using asimilar structure. We will not state the correspondingtautology. (Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction1. The statement[

p∧ (A∨B)∧((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.2. Basically, if we can prove a result under an additional

hypothesis A, if we can also prove the result under theadditional hypothesis B and if A or B is true, then q is true.

3. In a typical proof by case distinction A∨B is a tautology,that is, one of A and B is always true.

But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases. We could distinguish any number of cases, using asimilar structure. We will not state the correspondingtautology. (Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction1. The statement[

p∧ (A∨B)∧((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.2. Basically, if we can prove a result under an additional

hypothesis A, if we can also prove the result under theadditional hypothesis B and if A or B is true, then q is true.

3. In a typical proof by case distinction A∨B is a tautology,that is, one of A and B is always true. But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases. We could distinguish any number of cases, using asimilar structure. We will not state the correspondingtautology. (Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction1. The statement[

p∧ (A∨B)∧((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.2. Basically, if we can prove a result under an additional

hypothesis A, if we can also prove the result under theadditional hypothesis B and if A or B is true, then q is true.

3. In a typical proof by case distinction A∨B is a tautology,that is, one of A and B is always true. But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases.

We could distinguish any number of cases, using asimilar structure. We will not state the correspondingtautology. (Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction1. The statement[

p∧ (A∨B)∧((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.2. Basically, if we can prove a result under an additional

hypothesis A, if we can also prove the result under theadditional hypothesis B and if A or B is true, then q is true.

3. In a typical proof by case distinction A∨B is a tautology,that is, one of A and B is always true. But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases. We could distinguish any number of cases, using asimilar structure.

We will not state the correspondingtautology. (Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction1. The statement[

p∧ (A∨B)∧((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.2. Basically, if we can prove a result under an additional

hypothesis A, if we can also prove the result under theadditional hypothesis B and if A or B is true, then q is true.

3. In a typical proof by case distinction A∨B is a tautology,that is, one of A and B is always true. But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases. We could distinguish any number of cases, using asimilar structure. We will not state the correspondingtautology.

(Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction1. The statement[

p∧ (A∨B)∧((p∧A)⇒ q

)∧

((p∧B)⇒ q

)]⇒ q

is a tautology.2. Basically, if we can prove a result under an additional

hypothesis A, if we can also prove the result under theadditional hypothesis B and if A or B is true, then q is true.

3. In a typical proof by case distinction A∨B is a tautology,that is, one of A and B is always true. But note that A and Bcan overlap.

4. Also note that we are not limited to a distinction into twocases. We could distinguish any number of cases, using asimilar structure. We will not state the correspondingtautology. (Thank goodness.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

We have two cases: The quadrilateral can be convex or not.Convex quadrilaterals have already been handled.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

We have two cases: The quadrilateral can be convex or not.Convex quadrilaterals have already been handled.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

We have two cases: The quadrilateral can be convex or not.

Convex quadrilaterals have already been handled.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

We have two cases: The quadrilateral can be convex or not.Convex quadrilaterals have already been handled.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

For non-convex quadrilaterals, we argue as follows.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

For non-convex quadrilaterals, we argue as follows.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

Label the vertex at which the angle is greater than 180◦ as A

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

Label the vertex at which the angle is greater than 180◦ as A

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

Label the vertex at which the angle is greater than 180◦ as A,label the remaining vertices B

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

Label the vertex at which the angle is greater than 180◦ as A,label the remaining vertices B, C

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

Label the vertex at which the angle is greater than 180◦ as A,label the remaining vertices B, C and D.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

Label the angles α

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

Label the angles α

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

β

Label the angles α , β

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

β

γ

Label the angles α , β , γ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

β

γ

δ

Label the angles α , β , γ and δ .

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

β

γ

δ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

β

γ

δ

The line segment from A to C stays inside the quadrilateral. Itsplits the quadrilateral into the two triangles ABC and ACD.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

β

γ

δ

The line segment from A to C stays inside the quadrilateral.

Itsplits the quadrilateral into the two triangles ABC and ACD.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

β

γ

δ

The line segment from A to C stays inside the quadrilateral. Itsplits the quadrilateral into the two triangles ABC and ACD.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

β

γ

δ

The same argument as in the convex case now shows thatα +β + γ +δ = 360◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

β

γ

δ

The same argument as in the convex case now shows thatα +β + γ +δ = 360◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof by Case Distinction That the Sum of Anglesin a Quadrilateral is 360◦

SS

SS

SS�

��

����

��

��

���

A

B

C

D

α

β

γ

δ

The same argument as in the convex case now shows thatα +β + γ +δ = 360◦.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Breaking Up a Biconditional

1. The statement (p⇔ q)⇔((p⇒ q)∧ (q⇒ p)

)is a

tautology.2. So to prove that p and q are equivalent, we can prove that p

implies q and q implies p. In this fashion, one proof breaksup into two (hopefully simpler) proofs.

3. The tautology is also called the biconditional law.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Breaking Up a Biconditional1. The statement (p⇔ q)⇔

((p⇒ q)∧ (q⇒ p)

)is a

tautology.

2. So to prove that p and q are equivalent, we can prove that pimplies q and q implies p. In this fashion, one proof breaksup into two (hopefully simpler) proofs.

3. The tautology is also called the biconditional law.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Breaking Up a Biconditional1. The statement (p⇔ q)⇔

((p⇒ q)∧ (q⇒ p)

)is a

tautology.2. So to prove that p and q are equivalent, we can prove that p

implies q and q implies p.

In this fashion, one proof breaksup into two (hopefully simpler) proofs.

3. The tautology is also called the biconditional law.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Breaking Up a Biconditional1. The statement (p⇔ q)⇔

((p⇒ q)∧ (q⇒ p)

)is a

tautology.2. So to prove that p and q are equivalent, we can prove that p

implies q and q implies p. In this fashion, one proof breaksup into two (hopefully simpler) proofs.

3. The tautology is also called the biconditional law.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Breaking Up a Biconditional1. The statement (p⇔ q)⇔

((p⇒ q)∧ (q⇒ p)

)is a

tautology.2. So to prove that p and q are equivalent, we can prove that p

implies q and q implies p. In this fashion, one proof breaksup into two (hopefully simpler) proofs.

3. The tautology is also called the biconditional law.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple.

It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.

“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”:

Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number.

Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1.

Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.

“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐:

Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.

Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even.

Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k.

But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1

, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2.

This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving The Biconditional “A prime number p isodd iff p≥ 3” by Breaking It Up

The statement looks deceptively simple. It really is simple, butattempts to preserve the “iff” throughout the argument makesthings harder than they need to be.“⇒”: Let p be an odd prime number. Then p is odd and notequal to 1. Thus p≥ 3.“⇐: Let p be a prime number that is greater than or equal to 3.Suppose for a contradiction that p is even. Then p = 2k forsome k. But because p is prime, k would need to be 1, implyingk = 1 and p = 2. This is a contradiction to p≥ 3.

(Note that we used a proof by contradiction inside the proof of abiconditional.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition

1. The statement (p⇒ q)⇔ (¬q⇒¬p) is a tautology.2. Basically, if the conclusion looks better to you than the

hypothesis, check if the negation of the conclusion lookspromising, too. If so, try to use it to prove the negation ofthe hypothesis.

3. The tautology is also called the law of contraposition.4. The proof method is also called modus tollens (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition1. The statement (p⇒ q)⇔ (¬q⇒¬p) is a tautology.

2. Basically, if the conclusion looks better to you than thehypothesis, check if the negation of the conclusion lookspromising, too. If so, try to use it to prove the negation ofthe hypothesis.

3. The tautology is also called the law of contraposition.4. The proof method is also called modus tollens (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition1. The statement (p⇒ q)⇔ (¬q⇒¬p) is a tautology.2. Basically, if the conclusion looks better to you than the

hypothesis

, check if the negation of the conclusion lookspromising, too. If so, try to use it to prove the negation ofthe hypothesis.

3. The tautology is also called the law of contraposition.4. The proof method is also called modus tollens (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition1. The statement (p⇒ q)⇔ (¬q⇒¬p) is a tautology.2. Basically, if the conclusion looks better to you than the

hypothesis, check if the negation of the conclusion lookspromising, too.

If so, try to use it to prove the negation ofthe hypothesis.

3. The tautology is also called the law of contraposition.4. The proof method is also called modus tollens (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition1. The statement (p⇒ q)⇔ (¬q⇒¬p) is a tautology.2. Basically, if the conclusion looks better to you than the

hypothesis, check if the negation of the conclusion lookspromising, too. If so, try to use it to prove the negation ofthe hypothesis.

3. The tautology is also called the law of contraposition.4. The proof method is also called modus tollens (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition1. The statement (p⇒ q)⇔ (¬q⇒¬p) is a tautology.2. Basically, if the conclusion looks better to you than the

hypothesis, check if the negation of the conclusion lookspromising, too. If so, try to use it to prove the negation ofthe hypothesis.

3. The tautology is also called the law of contraposition.

4. The proof method is also called modus tollens (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition1. The statement (p⇒ q)⇔ (¬q⇒¬p) is a tautology.2. Basically, if the conclusion looks better to you than the

hypothesis, check if the negation of the conclusion lookspromising, too. If so, try to use it to prove the negation ofthe hypothesis.

3. The tautology is also called the law of contraposition.4. The proof method is also called modus tollens (Latin).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition That If 3 is Not a Factorof n Or 5 is Not a Factor of n, Then 15 Is Not aFactor of n

The contrapositive is “If 15 is a factor of n, then 3 is a factor ofn and 5 is a factor of n.” (Good exercise in using DeMorgan’sLaws.) Let n ∈ N be so that 15 is a factor of n. Thenn = 15 · k = 3 ·5 · k for some k ∈ N. Hence 3 is a factor of n and5 is a factor of n.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition That If 3 is Not a Factorof n Or 5 is Not a Factor of n, Then 15 Is Not aFactor of n

The contrapositive is “If 15 is a factor of n, then 3 is a factor ofn and 5 is a factor of n.”

(Good exercise in using DeMorgan’sLaws.) Let n ∈ N be so that 15 is a factor of n. Thenn = 15 · k = 3 ·5 · k for some k ∈ N. Hence 3 is a factor of n and5 is a factor of n.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition That If 3 is Not a Factorof n Or 5 is Not a Factor of n, Then 15 Is Not aFactor of n

The contrapositive is “If 15 is a factor of n, then 3 is a factor ofn and 5 is a factor of n.” (Good exercise in using DeMorgan’sLaws.)

Let n ∈ N be so that 15 is a factor of n. Thenn = 15 · k = 3 ·5 · k for some k ∈ N. Hence 3 is a factor of n and5 is a factor of n.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition That If 3 is Not a Factorof n Or 5 is Not a Factor of n, Then 15 Is Not aFactor of n

The contrapositive is “If 15 is a factor of n, then 3 is a factor ofn and 5 is a factor of n.” (Good exercise in using DeMorgan’sLaws.) Let n ∈ N be so that 15 is a factor of n.

Thenn = 15 · k = 3 ·5 · k for some k ∈ N. Hence 3 is a factor of n and5 is a factor of n.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition That If 3 is Not a Factorof n Or 5 is Not a Factor of n, Then 15 Is Not aFactor of n

The contrapositive is “If 15 is a factor of n, then 3 is a factor ofn and 5 is a factor of n.” (Good exercise in using DeMorgan’sLaws.) Let n ∈ N be so that 15 is a factor of n. Thenn = 15 · k = 3 ·5 · k for some k ∈ N.

Hence 3 is a factor of n and5 is a factor of n.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition That If 3 is Not a Factorof n Or 5 is Not a Factor of n, Then 15 Is Not aFactor of n

The contrapositive is “If 15 is a factor of n, then 3 is a factor ofn and 5 is a factor of n.” (Good exercise in using DeMorgan’sLaws.) Let n ∈ N be so that 15 is a factor of n. Thenn = 15 · k = 3 ·5 · k for some k ∈ N. Hence 3 is a factor of n and5 is a factor of n.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proof By Contraposition That If 3 is Not a Factorof n Or 5 is Not a Factor of n, Then 15 Is Not aFactor of n

The contrapositive is “If 15 is a factor of n, then 3 is a factor ofn and 5 is a factor of n.” (Good exercise in using DeMorgan’sLaws.) Let n ∈ N be so that 15 is a factor of n. Thenn = 15 · k = 3 ·5 · k for some k ∈ N. Hence 3 is a factor of n and5 is a factor of n.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Disproof By Counterexample

1. The statement[¬(p⇒ q)

]⇔ (p∧¬q) is a tautology.

2. If we need to show that an implication does not hold, wemust prove that p and ¬q can be true simultaneously. Thisis usually done by finding some structure that satisfies both(hence the name).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Disproof By Counterexample1. The statement

[¬(p⇒ q)

]⇔ (p∧¬q) is a tautology.

2. If we need to show that an implication does not hold, wemust prove that p and ¬q can be true simultaneously. Thisis usually done by finding some structure that satisfies both(hence the name).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Disproof By Counterexample1. The statement

[¬(p⇒ q)

]⇔ (p∧¬q) is a tautology.

2. If we need to show that an implication does not hold, wemust prove that p and ¬q can be true simultaneously.

Thisis usually done by finding some structure that satisfies both(hence the name).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Disproof By Counterexample1. The statement

[¬(p⇒ q)

]⇔ (p∧¬q) is a tautology.

2. If we need to show that an implication does not hold, wemust prove that p and ¬q can be true simultaneously. Thisis usually done by finding some structure that satisfies both(hence the name).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Disproving The Statement “If the Function f IsContinuous, Then It Is Differentiable”

The function f (x) = |x| is continuous and not differentiable. Soit is a counterexample.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Disproving The Statement “If the Function f IsContinuous, Then It Is Differentiable”

The function f (x) = |x| is continuous and not differentiable.

Soit is a counterexample.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Disproving The Statement “If the Function f IsContinuous, Then It Is Differentiable”

The function f (x) = |x| is continuous and not differentiable. Soit is a counterexample.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Disproving The Statement “If the Function f IsContinuous, Then It Is Differentiable”

The function f (x) = |x| is continuous and not differentiable. Soit is a counterexample.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving Universally Quantified Statements

1. To prove that ∀x ∈ S : p(x) is true, we first pick anarbitrary, but fixed, x ∈ S.

2. Because x was fixed, we can prove the statement p(x).3. After that, because x was arbitrary in S, we conclude that

p(x) holds for all x ∈ S.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving Universally Quantified Statements1. To prove that ∀x ∈ S : p(x) is true, we first pick an

arbitrary, but fixed, x ∈ S.

2. Because x was fixed, we can prove the statement p(x).3. After that, because x was arbitrary in S, we conclude that

p(x) holds for all x ∈ S.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving Universally Quantified Statements1. To prove that ∀x ∈ S : p(x) is true, we first pick an

arbitrary, but fixed, x ∈ S.2. Because x was fixed, we can prove the statement p(x).

3. After that, because x was arbitrary in S, we conclude thatp(x) holds for all x ∈ S.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving Universally Quantified Statements1. To prove that ∀x ∈ S : p(x) is true, we first pick an

arbitrary, but fixed, x ∈ S.2. Because x was fixed, we can prove the statement p(x).3. After that, because x was arbitrary in S, we conclude that

p(x) holds for all x ∈ S.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3. Hence, overall, k must be divisible by6.Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary

but fixed natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3. Hence, overall, k must be divisible by6.Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed

natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3. Hence, overall, k must be divisible by6.Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed natural number so that k−1 andk +1 are prime.

Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3. Hence, overall, k must be divisible by6.Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2.

Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3. Hence, overall, k must be divisible by6.Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3.

Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3. Hence, overall, k must be divisible by6.Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3.

Thusk must be divisible by 3. Hence, overall, k must be divisible by6.Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3.

Hence, overall, k must be divisible by6.Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3. Hence, overall, k must be divisible by6.

Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3. Hence, overall, k must be divisible by6.Now, because k was arbitrary

(that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3. Hence, overall, k must be divisible by6.Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3)

the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3. Hence, overall, k must be divisible by6.Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That Every Natural Number between aPair of Twin Prime Numbers Greater Than 3 IsDivisible By 6

Let k be an arbitrary but fixed natural number so that k−1 andk +1 are prime. Because all twin prime numbers are odd, kmust be even, that is, k is divisible by 2. Moreover, one of k−1,k and k +1 must be divisible by 3. Because k−1 and k +1 areprime numbers greater than 3, they are not divisible by 3. Thusk must be divisible by 3. Hence, overall, k must be divisible by6.Now, because k was arbitrary (that is, we did not use anyspecial properties of k, except that it was between two primesgreater than 3) the claimed result must hold.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving Existentially Quantified Statements

1. To prove that ∃x ∈ S : p(x) is true, we usually must find anx ∈ S for which p(x) is true.

2. This may look like trying to prove by example, which is atypical mistake. (Just because your faithful narrator is agoofball does not mean that all German-born Americansare.)

3. The technique works because we are looking for just oneexample.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving Existentially Quantified Statements1. To prove that ∃x ∈ S : p(x) is true, we usually must find an

x ∈ S for which p(x) is true.

2. This may look like trying to prove by example, which is atypical mistake. (Just because your faithful narrator is agoofball does not mean that all German-born Americansare.)

3. The technique works because we are looking for just oneexample.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving Existentially Quantified Statements1. To prove that ∃x ∈ S : p(x) is true, we usually must find an

x ∈ S for which p(x) is true.2. This may look like trying to prove by example, which is a

typical mistake.

(Just because your faithful narrator is agoofball does not mean that all German-born Americansare.)

3. The technique works because we are looking for just oneexample.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving Existentially Quantified Statements1. To prove that ∃x ∈ S : p(x) is true, we usually must find an

x ∈ S for which p(x) is true.2. This may look like trying to prove by example, which is a

typical mistake. (Just because your faithful narrator is agoofball does not mean that all German-born Americansare.)

3. The technique works because we are looking for just oneexample.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving Existentially Quantified Statements1. To prove that ∃x ∈ S : p(x) is true, we usually must find an

x ∈ S for which p(x) is true.2. This may look like trying to prove by example, which is a

typical mistake. (Just because your faithful narrator is agoofball does not mean that all German-born Americansare.)

3. The technique works because we are looking for just oneexample.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That There Is a Twin Prime NumberBetween 100 and 110

101 is prime. 103 is prime. That’s it.

(Typically these examples are harder to construct.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That There Is a Twin Prime NumberBetween 100 and 110

101 is prime.

103 is prime. That’s it.

(Typically these examples are harder to construct.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That There Is a Twin Prime NumberBetween 100 and 110

101 is prime. 103 is prime.

That’s it.

(Typically these examples are harder to construct.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That There Is a Twin Prime NumberBetween 100 and 110

101 is prime. 103 is prime. That’s it.

(Typically these examples are harder to construct.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That There Is a Twin Prime NumberBetween 100 and 110

101 is prime. 103 is prime. That’s it.

(Typically these examples are harder to construct.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Proving That There Is a Twin Prime NumberBetween 100 and 110

101 is prime. 103 is prime. That’s it.

(Typically these examples are harder to construct.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies

1. (p∧q)⇒ q (law of simplification)If we know two statements are true, we know that each oneof them is true. Typically used when a strong theoremprovides more than we need.

2. p⇒ (p∨q) (law of addition)If we know a statement is true, than it or anything else istrue. Typically used when a theorem has an OR in thehypothesis and we have verified one of the two statementsin the OR.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies1. (p∧q)⇒ q (law of simplification)

If we know two statements are true, we know that each oneof them is true. Typically used when a strong theoremprovides more than we need.

2. p⇒ (p∨q) (law of addition)If we know a statement is true, than it or anything else istrue. Typically used when a theorem has an OR in thehypothesis and we have verified one of the two statementsin the OR.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies1. (p∧q)⇒ q (law of simplification)

If we know two statements are true, we know that each oneof them is true.

Typically used when a strong theoremprovides more than we need.

2. p⇒ (p∨q) (law of addition)If we know a statement is true, than it or anything else istrue. Typically used when a theorem has an OR in thehypothesis and we have verified one of the two statementsin the OR.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies1. (p∧q)⇒ q (law of simplification)

If we know two statements are true, we know that each oneof them is true. Typically used when a strong theoremprovides more than we need.

2. p⇒ (p∨q) (law of addition)If we know a statement is true, than it or anything else istrue. Typically used when a theorem has an OR in thehypothesis and we have verified one of the two statementsin the OR.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies1. (p∧q)⇒ q (law of simplification)

If we know two statements are true, we know that each oneof them is true. Typically used when a strong theoremprovides more than we need.

2. p⇒ (p∨q) (law of addition)

If we know a statement is true, than it or anything else istrue. Typically used when a theorem has an OR in thehypothesis and we have verified one of the two statementsin the OR.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies1. (p∧q)⇒ q (law of simplification)

If we know two statements are true, we know that each oneof them is true. Typically used when a strong theoremprovides more than we need.

2. p⇒ (p∨q) (law of addition)If we know a statement is true, than it or anything else istrue.

Typically used when a theorem has an OR in thehypothesis and we have verified one of the two statementsin the OR.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies1. (p∧q)⇒ q (law of simplification)

If we know two statements are true, we know that each oneof them is true. Typically used when a strong theoremprovides more than we need.

2. p⇒ (p∨q) (law of addition)If we know a statement is true, than it or anything else istrue. Typically used when a theorem has an OR in thehypothesis and we have verified one of the two statementsin the OR.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies

3.((p⇒ q)∧ (q⇒ r)

)⇒ (p⇒ r) (law of syllogism)

If we can reach a halfway point between two statementsand then get from the halfway point to the finish, we haveproved the implication. Very typical for proofs. We usuallyestablish a chain of intermediate results. Also calledtransitivity of implications.

4.(¬p∧ (p∨q)

)⇒ q (law of disjunction, also, modus

tollendo ponens)If one of two statements is true and one of them is false,then the one that is not false must be true. Typically usedwhen we can arrive at one of two conclusions and we onlywant one of them: We exclude the other.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies3.

((p⇒ q)∧ (q⇒ r)

)⇒ (p⇒ r) (law of syllogism)

If we can reach a halfway point between two statementsand then get from the halfway point to the finish, we haveproved the implication. Very typical for proofs. We usuallyestablish a chain of intermediate results. Also calledtransitivity of implications.

4.(¬p∧ (p∨q)

)⇒ q (law of disjunction, also, modus

tollendo ponens)If one of two statements is true and one of them is false,then the one that is not false must be true. Typically usedwhen we can arrive at one of two conclusions and we onlywant one of them: We exclude the other.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies3.

((p⇒ q)∧ (q⇒ r)

)⇒ (p⇒ r) (law of syllogism)

If we can reach a halfway point between two statementsand then get from the halfway point to the finish, we haveproved the implication.

Very typical for proofs. We usuallyestablish a chain of intermediate results. Also calledtransitivity of implications.

4.(¬p∧ (p∨q)

)⇒ q (law of disjunction, also, modus

tollendo ponens)If one of two statements is true and one of them is false,then the one that is not false must be true. Typically usedwhen we can arrive at one of two conclusions and we onlywant one of them: We exclude the other.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies3.

((p⇒ q)∧ (q⇒ r)

)⇒ (p⇒ r) (law of syllogism)

If we can reach a halfway point between two statementsand then get from the halfway point to the finish, we haveproved the implication. Very typical for proofs. We usuallyestablish a chain of intermediate results.

Also calledtransitivity of implications.

4.(¬p∧ (p∨q)

)⇒ q (law of disjunction, also, modus

tollendo ponens)If one of two statements is true and one of them is false,then the one that is not false must be true. Typically usedwhen we can arrive at one of two conclusions and we onlywant one of them: We exclude the other.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies3.

((p⇒ q)∧ (q⇒ r)

)⇒ (p⇒ r) (law of syllogism)

If we can reach a halfway point between two statementsand then get from the halfway point to the finish, we haveproved the implication. Very typical for proofs. We usuallyestablish a chain of intermediate results. Also calledtransitivity of implications.

4.(¬p∧ (p∨q)

)⇒ q (law of disjunction, also, modus

tollendo ponens)If one of two statements is true and one of them is false,then the one that is not false must be true. Typically usedwhen we can arrive at one of two conclusions and we onlywant one of them: We exclude the other.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies3.

((p⇒ q)∧ (q⇒ r)

)⇒ (p⇒ r) (law of syllogism)

If we can reach a halfway point between two statementsand then get from the halfway point to the finish, we haveproved the implication. Very typical for proofs. We usuallyestablish a chain of intermediate results. Also calledtransitivity of implications.

4.(¬p∧ (p∨q)

)⇒ q (law of disjunction, also, modus

tollendo ponens)

If one of two statements is true and one of them is false,then the one that is not false must be true. Typically usedwhen we can arrive at one of two conclusions and we onlywant one of them: We exclude the other.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies3.

((p⇒ q)∧ (q⇒ r)

)⇒ (p⇒ r) (law of syllogism)

If we can reach a halfway point between two statementsand then get from the halfway point to the finish, we haveproved the implication. Very typical for proofs. We usuallyestablish a chain of intermediate results. Also calledtransitivity of implications.

4.(¬p∧ (p∨q)

)⇒ q (law of disjunction, also, modus

tollendo ponens)If one of two statements is true and one of them is false,then the one that is not false must be true.

Typically usedwhen we can arrive at one of two conclusions and we onlywant one of them: We exclude the other.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

More Tautologies3.

((p⇒ q)∧ (q⇒ r)

)⇒ (p⇒ r) (law of syllogism)

If we can reach a halfway point between two statementsand then get from the halfway point to the finish, we haveproved the implication. Very typical for proofs. We usuallyestablish a chain of intermediate results. Also calledtransitivity of implications.

4.(¬p∧ (p∨q)

)⇒ q (law of disjunction, also, modus

tollendo ponens)If one of two statements is true and one of them is false,then the one that is not false must be true. Typically usedwhen we can arrive at one of two conclusions and we onlywant one of them: We exclude the other.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Good Advice

1. Only use what is given in the hypothesis, in axioms andwhat you have already proved.

2. Restating the hypothesis at the start and the conclusion atthe end can help.

3. Write complete sentences without abbreviations orunnecessary symbols.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

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Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Good Advice1. Only use what is given in the hypothesis, in axioms and

what you have already proved.

2. Restating the hypothesis at the start and the conclusion atthe end can help.

3. Write complete sentences without abbreviations orunnecessary symbols.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Good Advice1. Only use what is given in the hypothesis, in axioms and

what you have already proved.2. Restating the hypothesis at the start and the conclusion at

the end can help.

3. Write complete sentences without abbreviations orunnecessary symbols.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs

logo1

Introduction Direct Contradiction Cases Biconditionals Contraposition Counterexamples ∀ ∃ Useful Steps

Good Advice1. Only use what is given in the hypothesis, in axioms and

what you have already proved.2. Restating the hypothesis at the start and the conclusion at

the end can help.3. Write complete sentences without abbreviations or

unnecessary symbols.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proofs