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Unformatted text preview: x = xy + xy (Applied twice.) Heres 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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This note was uploaded on 01/17/2011 for the course ECE 3504 at Virginia Tech.
 '06
 JSThweatt

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