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LF1 Proving Theorem 5

# LF1 Proving Theorem 5 - x = xy ′ xy(Applied twice...

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Vignette: Proving Theorem 5 We want to prove that b a b a a + = + ' . We’ll show two different proofs. One will demonstrate the principle that sometimes Boolean algebra “simplification” requires steps that actually make an expression more complicated. The fact that we’ll see two proofs demonstrates that the same expression can be reduced to its most simple form using more than one procedure. The unfortunate corollary to the last notion is that multiple paths to the same result can, in fact, lead to multiple forms of the same expression – none of which end up being the most simple form of the expression. For our purposes, Boolean algebra does not represent a systematic means to perform logic simplification. Here’s the first proof: a + a b = ab + ab + a b T6(a): x = xy + xy = ab + ab + ab + a b T1(a): x = x + x = a + b T6(a):
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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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