Jeff Shelton – 16 January 2015

Jeff Shelton – 16 January 2015 2

Jeff Shelton – 16 January 2015

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Jeff Shelton – 16 January 2015

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Jeff Shelton – 16 January 2015

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Jeff Shelton – 16 January 2015

•

•

•

•

•

8

Jeff Shelton – 16 January 2015

•

•

•

•

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Jeff Shelton – 16 January 2015

510 = 1012

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Jeff Shelton – 16 January 2015

Jeff Shelton – 16 January 2015

𝑥 or 𝑥′ /𝑥

•

•

𝒙 𝒙

0 1

1 0

𝒙

0 1

𝒙 1 0

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Jeff Shelton – 16 January 2015

•

•

𝒙 𝒚 𝒙 • 𝒚

0 0 0

0 1 0

1 0 0

1 1 1

𝒙 AND 𝒚 𝒙

0 1

𝒚 0 0 0

1 0 1

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Jeff Shelton – 16 January 2015

•

•

𝒙 𝒚 𝒙 + 𝒚

0 0 0

0 1 1

1 0 1

1 1 1

𝒙 OR 𝒚 𝒙

0 1

𝒚 0 0 1

1 1 1

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Jeff Shelton – 16 January 2015

⊕

•

•

𝒙 𝒚 𝒙 ⊕ 𝒚

0 0 0

0 1 1

1 0 1

1 1 0

𝒙 XOR 𝒚 𝒙

0 1

𝒚 0 0 1

1 1 0

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Jeff Shelton – 16 January 2015

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Jeff Shelton – 16 January 2015

𝑥 + 𝑦 = 𝑦 + 𝑥 𝑥 ⋅ 𝑦 = 𝑦 ⋅ 𝑥

𝑥 + 𝑦 + 𝑧 = 𝑥 + 𝑦 + 𝑧 𝑥 ⋅ 𝑦 ⋅ 𝑧 = 𝑥 ⋅ 𝑦 ⋅ 𝑧

𝑥 ⋅ 𝑦 + 𝑧 = 𝑥 ⋅ 𝑦 + 𝑥 ⋅ 𝑧 𝑥 + 𝑦 ⋅ 𝑧 = 𝑥 + 𝑦 ⋅ 𝑥 + 𝑧

𝑥 + 0 = 𝑥 𝑥 ⋅ 1 = 𝑥

𝑥 ⋅ 𝑥 = 0 𝑥 + 𝑥 = 1

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(Axioms define domain characteristics, from which other ‘truths’ can be derived.)

Jeff Shelton – 16 January 2015

𝑥 + 𝑥 = 𝑥 𝑥 ⋅ 𝑥 = 𝑥

𝑥 + 𝑥 ⋅ 𝑦 = 𝑥 𝑥 ⋅ (𝑥 + 𝑦) = 𝑥

𝑥 + 𝑥 ⋅ 𝑦 = 𝑥 + 𝑦

𝑥 ⋅ 𝑥 + 𝑦 = 𝑥 ⋅ 𝑦

(𝑥 + 𝑦) = 𝑥 ⋅ 𝑦

(𝑥 ⋅ 𝑦) = 𝑥 + 𝑦

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(Theorems are proven through algebraic manipulation or exhaustive substitution.)

Jeff Shelton – 16 January 2015

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Jeff Shelton – 16 January 2015

0 0

0

1

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Jeff Shelton – 16 January 2015

𝑦 = 𝑪1 + 𝑪2 +⋯+ 𝑪𝑛−1 + 𝑪𝑛

𝑪𝑖 𝑛

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0 0

0

1

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Jeff Shelton – 16 January 2015

𝑦 = 𝐵1 ⋅ 𝐵2 + 𝐵1 ⋅ 𝐵2 + 𝐵1 ⋅ 𝐵2 + 𝐵1 ⋅ 𝐵2

𝑦 = 𝐵1 ⋅ 𝐵2 + 𝐵1 ⋅ 𝐵2 + 𝐵1 ⋅ 𝐵2 + 𝐵1 ⋅ 𝐵2

𝑦 = 𝐵1 ⋅ 𝐵2

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0 0

0

1

Jeff Shelton – 16 January 2015

𝑦 = 𝐵1 + 𝐵2 ⋅ 𝐵1 + 𝐵2 ⋅ 𝐵1 + 𝐵2 ⋅ 𝐵1 + 𝐵2

𝑦 = 𝐵1 + 𝐵2 ⋅ 𝐵1 + 𝐵2 ⋅ 𝐵1 + 𝐵2 ⋅ 𝐵1 + 𝐵2

𝑦 = 𝐵1 + 𝐵2 ⋅ 𝐵1 + 𝐵2 ⋅ 𝐵1 + 𝐵2

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0 0

0

1

Jeff Shelton – 16 January 2015

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