jeff shelton 16 january 2015 - purdue engineering shelton β 16 january 2015 =πͺ1 +πͺ2...
TRANSCRIPT
Jeff Shelton β 16 January 2015
π₯ or π₯β² /π₯
β’
β’
π π
0 1
1 0
π
0 1
π 1 0
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Jeff Shelton β 16 January 2015
β’
β’
π π π β’ π
0 0 0
0 1 0
1 0 0
1 1 1
π AND π π
0 1
π 0 0 0
1 0 1
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Jeff Shelton β 16 January 2015
β’
β’
π π π + π
0 0 0
0 1 1
1 0 1
1 1 1
π OR π π
0 1
π 0 0 1
1 1 1
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Jeff Shelton β 16 January 2015
β
β’
β’
π π π β π
0 0 0
0 1 1
1 0 1
1 1 0
π XOR π π
0 1
π 0 0 1
1 1 0
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Jeff Shelton β 16 January 2015
π₯ + π¦ = π¦ + π₯ π₯ β π¦ = π¦ β π₯
π₯ + π¦ + π§ = π₯ + π¦ + π§ π₯ β π¦ β π§ = π₯ β π¦ β π§
π₯ β π¦ + π§ = π₯ β π¦ + π₯ β π§ π₯ + π¦ β π§ = π₯ + π¦ β π₯ + π§
π₯ + 0 = π₯ π₯ β 1 = π₯
π₯ β π₯ = 0 π₯ + π₯ = 1
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(Axioms define domain characteristics, from which other βtruthsβ can be derived.)
Jeff Shelton β 16 January 2015
π₯ + π₯ = π₯ π₯ β π₯ = π₯
π₯ + π₯ β π¦ = π₯ π₯ β (π₯ + π¦) = π₯
π₯ + π₯ β π¦ = π₯ + π¦
π₯ β π₯ + π¦ = π₯ β π¦
(π₯ + π¦) = π₯ β π¦
(π₯ β π¦) = π₯ + π¦
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(Theorems are proven through algebraic manipulation or exhaustive substitution.)
Jeff Shelton β 16 January 2015
π¦ = πͺ1 + πͺ2 +β―+ πͺπβ1 + πͺπ
πͺπ π
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0 0
0
1
Jeff Shelton β 16 January 2015
π¦ = π΅1 β π΅2 + π΅1 β π΅2 + π΅1 β π΅2 + π΅1 β π΅2
π¦ = π΅1 β π΅2 + π΅1 β π΅2 + π΅1 β π΅2 + π΅1 β π΅2
π¦ = π΅1 β π΅2
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0 0
0
1
Jeff Shelton β 16 January 2015
π¦ = π΅1 + π΅2 β π΅1 + π΅2 β π΅1 + π΅2 β π΅1 + π΅2
π¦ = π΅1 + π΅2 β π΅1 + π΅2 β π΅1 + π΅2 β π΅1 + π΅2
π¦ = π΅1 + π΅2 β π΅1 + π΅2 β π΅1 + π΅2
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0 0
0
1