Unformatted text preview: x = xy ′ + xy (Applied twice.) Here’s the second proof: a + a ′ b = ( a + a ′ )( a + b ) P5(a): x + yz = ( x + y )( x + z ) = 1( a + b ) P6(a): x + x ′ = 1 = a + b P2(b): x • 1 = x Take note of the various substitutions used in the second proof, especially in the first step. As a final bonus note, remember that in the application of any of these axioms, substitution can be used to transform an expression into a form that is identical to one of the axioms. Here is Theorem 5 applied in an instance of substitution. abc + a ′ d + b ′ d + c ′ d = abc + ( a ′ + b ′ + c ′ ) d P5(b): x ( y + z ) = xy + xz (Applied to three terms.) = abc + ( abc ) ′ d T8(b): x ′ + y ′ = ( xy ) ′ = abc + d T5(a): x + x ′ y = x + y Note the substitution in step 3....
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 '06
 JSThweatt
 Logic, Mathematical logic, simple form

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