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an algebra that is applied only to binarynumbers
has a list of operations that it can use to simplifyits equations
Boolean Algebra * Property of STI
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a set of rules defined on how a set of numbers can bemanipulated
also called Huntington’s Postulates
used to derive certain theorems applicable assuming certainconditions
Huntington’s Postulates
I. There exists a set of K objects or elements, subject to anequivalence relation, denoted “= ”, which satisfies the principle ofsubstitution.
IIa. A rule of combination “+” is defined such that A + B is in K whenever both A and B are in K .
IIb. A rule of combination “·” is defined such that A · B (abbreviatedAB ) is in K whenever both A and B are in K .
IIIa. There exists an element 0 in K such that, for every A in K , A + 0 =A.
IIIb. There exists an element 1 in K such that, for every A in K, A · 1 =A.
IV. Commutative Property
Boolean Algebra * Property of STI
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a. on: + = +
b. Multiplication: A · B = B · A
V. Distributive Lawa. A + (B · C ) = (A + B ) · (A + C )
b. A · (B + C ) = (A · B ) + (A · C )
VI. For every element A in K , there exists an element A’ such that
A · A’ = 0
and
A + A’ = 1.
VII. There are at least two elements X and Y in K such that X = Y
Source: Hill, F. and Peterson, G., “Introduction to Switching Theory and Logical Design”, pp.42-43
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There exists a set of K ob ects or elements, sub ect
to an equivalence relation, denoted “=”, whichsatisfies the principle of substitution.
defines that variables can be defined using a(Boolean) algebraic expression by using thesymbol “=”
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A rule of combination “+” is defined such that A + B
is in K whenever both A and B are in K.
defines the addition of binary numbers
whenever two binary numbers are added, theirsum is also a binary number
Boolean Algebra * Property of STI
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A rule of combination ·” is defined such that A · B
(abbreviated AB) is in K whenever both A and Bare in K.
defines the multiplication of binary numbers
whenever two binary numbers are multiplied,their product is also a binary number
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There exists an element 0 in K such that, for ever
A in K, A + 0 = A.
defines the zero element in the binarynumbering system
whenever zero is added to any binary number,the value of the number to which zero was
added will not change
Boolean Algebra * Property of STI
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zero is defined such that when it is added to abinary number, the binary number will retain itsvalue
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There exists an element 1 in K such that, for ever
a in K, A · 1 = A.
defines the one element in the binarynumbering system
whenever one is multiplied to any binarynumber, the value of the number to which one
was multiplied will not change
Boolean Algebra * Property of STI
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one is defined such that when it is multiplied toa binary number, the binary number will retainits value
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Cummutative Pro ert of Addition: A + B = B + A.
reversing the order of two binary numbers inperforming addition is shown as not to effect theresultin sum
Boolean Algebra * Property of STI
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Cummutative Pro ert of Multi lication: A · B = B ·
A.
reversing the order of two binary numbers inerformin multi lication is shown as not to
Boolean Algebra * Property of STI
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effect the resulting product
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Distributive Pro ert of Addition: A + B · C = A +
B) · (A + C).
Boolean Algebra * Property of STI
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Distributive Property of Multiplication: A · (B + C) =(A · B) + (A · C).
Boolean Algebra * Property of STI
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For ever element A in K, there exists an element
A’ such that A · A’ = 0 and A + A’ = 1.
defines the complement (A’ ) of any variable Asuch that when A’ is multiplied to A, theresulting product is zero
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There are at least two elements X and Y in K such
that X ¹ Y.
defines at least two elements in K that are notequal
Assign the following expressions to variables X and Y :
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Boolean Algebra
1. valid only for a set having
(Decimal) Algebraordinary (decimal)
2. although this propertycould be derived orproven to hold true theHuntington’s postulatesdo not includeAssociative Property
3. the distributive property
of + over · is valid sinceits set have only twoelements (0 and 1)
4. Additive andMultiplicative Inversesare not defined
set having infinite numberof elements
not applicable
Boolean Algebra * Property of STI
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. operations are alsoundefined
6. the definition of acomplement is defined
7. especially defined tohave the property ofduality
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Theorem 1 (a) X + X = X (b) X · X = X
Theorem 2 (a) X + 1 = 1 (b) X · 0 = 0
eorem =
Theorem 4 (a) X+ (Y+Z ) = (X+Y) +Z (b) X(YZ) = (XY)Z
Theorem 5 (a) (X +Y)’ = X’Y’ (b) (XY)’ = X’+Y’
Theorem 6 (a) X + XY = X (b) X(X+Y) = X
Theorem 3 is sometimes called INVOLUTION and Theorems
6a&b are forms of ABSORPTION. Other forms of absorption
will be presented in a few examples later.
the number of combinations is directly related to
the number of variables
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2# of variables = # of combinations
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Solution:
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Boolean Algebra * Property of STI
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Theorem 1(a): X + X = X
Truth table:
Theorem 1(b): X · X = X
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Truth table:
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Theorem 2(a): X + 1 = 1
Truth table:
Theorem 2(b): X · 0 = 0
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Truth table:
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Theorem 3: (X’ )’ = X Truth table:
Y = X’
Y’ = (X’)’
Y’ = X = (X’)’
Theorem 4(a): X + (Y + Z ) = (X + Y ) + Z
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Theorem 4(b): X · (Y · Z ) = (X · Y ) · Z
X · (Y · Z) = X · (Y · Z + 0) Postulate 3a
= X · (Y + 0) · (Z + 0) Postulate 5a= ((X · Y) + (X · 0)) · Z Postulate 5b
Postulate 3a
= ((X · Y) + 0) · Z Theorem 2b
= (X · Y) · Z Postulate 3a
Truth table of X · (Y · Z ):
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Theorem 5(a): (X + Y )’ = X’Y’
(DeMorgan’s Theorem)
Postulate 6A · A’ = 0
and
A + A’ = 1
A = (X + Y )
A’ = (X + Y )’
X ’Y ’ = (X + Y )’
(X + Y) · (X + Y)’ = 0 (X + Y) · X’Y’ = 0
and
X + Y + X + Y ’ = 1 X + Y + X’Y’ = 1
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(X + Y) · X’Y’ = 1
X’Y’ · (X + Y) Postulate 4b
((X’Y’) · X)+((X’Y’) · Y) Postulate 5b
((X’X) · Y’)+((YY’) · X’) Postulate 4b
(0 · Y’) + (X’ · 0) Theorem 6
0 + 0 Theorem 2b
0 Postulate 3a
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(X + Y) + X’Y’ = 0
X + Y + X’ · X + Y +Y’ Postulate 5a
((X + X’)+ Y) · (X +(Y + Y’)) Postulate 4
(1 + Y) · (X+ 1) Postulate 6
1 · 1 Theorem 2a
1 Postulate 3b
Truth table:
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Theorem 5(b): (XY )’ = X’ + Y’
(DeMorgan’s Theorem)
Postulate 6(XY) · (XY)’ = 0 (XY) · X’ + Y’ = 0
and
(XY) + (XY)’ = 1 (XY) + X’ + Y’ = 1
(XY)·(X’ + Y’) = 0
((XY) · X’)+((XY) · Y’) Postulate 5b
((X’X) · Y)+((YY’) · X) Postulate 4b
(0 · Y) + (X · 0) Theorem 6
0 + 0 Theorem 2b
0 Postulate 3a
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(XY) + (X’ + Y’) = 1
((X’+ Y’)+ X) · ((X’+ Y’)+Y) Postulate 5a
((X+X’)+ Y’) · (X’+(Y + Y’)) Postulate 4a
(1+ Y’) · (X’ + 1) Postulate 6
1 · 1 Theorem 2a
1 Postulate 3b
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Truth table:
Theorem 6(a): X + XY = X X + XY = (X · 1) + (X · Y) Postulate 3b
= X · (1 + Y) Postulate 5b
= X · 1 Theorem 2a
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=
Truth table:
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Theorem 6(b): X (X + Y ) = X
X X+Y = X · X + X · Y Postulate 5b
= X + XY Theorem 1b
= X Theorem 6a
Truth table:
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Using the Postulates and Theorems presentedpreviously, prove the following Theorems and
write down their respective truth tables:
1. X +X ’Y =X +Y Theorem6c
2. X (X ’+Y )=XY Theorem6d
3. X ’+XY =X ’+Y Theorem6e
4. X ’(X +Y )=X ’Y ’ Theorem6f
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Solution:
1. X + X ’Y =X + Y
X + X ’Y = (X+XY)+X’Y Theorem 6a
= X+Y(X+X’) Distributive Property
(Postulate 5b)
= X+Y ·1 Postulate 6
= X+Y Postulate 3
Truth table:
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2. X (X ’+Y )=XY
X (X ’+Y ) = XX’+XY Distributive Property
(Postulate 5b)
= 0+XY Postulate 6
= XY Postulate 3
Truth table:
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3. X ’ + XY =X ’+Y
X ’ + XY = (X’+X’Y )+XY Theorem 6a
= X’+Y (X’+X) Postulate 5b
= X’+Y (1) Postulate 6
= X’+Y Postulate 3
Truth table:
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4. X + X ’Y =X ’Y ’
X + X ’Y = X’X+X’Y Distributive Property
(Postulate 5b)
= 0+X’Y Postulate 6
= X’Y Postulate 3
Truth table:
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Manipulate the following Boolean expressionsto minimize the number of terms or variables
used. Follow Operator Precedence and write-out a corresponding truth table that will showthe output values for every input combination.
5. F = (A’ + B’ + C) · (AB)’
6. F = X + (Y’ + XY’)
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Solution:
5. F = (A’ + B ’ + C ) · (AB )’
= ( A’ + B ’ + C ) · ( A’ + B ’) De Morgan’sTheorem
= ( A’ + B ’) · ( C + 1) Postulate 5b
= ( A’ + B ’) · 1 Theorem 2a
= A’ + B ’ Postulate 3b
Truth table:
6. F = X + (Y ’ + XY ’)
= X + Y ’ Theorem 6a
Truth table: