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hi im having a lot of trouble proving this boolean expression. Im getting many differing answers so I assume I must be going about it in the wrong way.

To explain, I'm trying to negate the whole LHS to split up the brackets and then trying to use the rules to reduce the expression from there. I must be using de morgans wrong or something since my answers vary. Can anyone give a hand.

$$((A\land B)\lor C\lor(\lnot D\land E))\land((A\land D)\lor\lnot C\lor (D\land E))\land((A\land B)\lor(A\land D)\lor E)\equiv((A\land B)\lor C\lor(\lnot D\land E))\land((A\land D)\lor\lnot C\lor(D\land E))$$

Senex Ægypti Parvi
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dmnte
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2 Answers2

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You do not need deMorgan's.

This is an example of consensus. $(X + Y) • (\overline X + Z) • (Y + Z) = (X + Y) • (\overline X + Z)$

Take the right side of the equation.

$$\color {red}{(}(A\land B)\lor C\lor(\lnot D\land E)\color {red}{)}\land\color {red}{(}(A\land D)\lor\lnot C\lor (D\land E)\color {red}{)}$$ $${(}AB+ C+\overline D E)\ (AD+ \overline C + DE)$$

Notice $C$ and $\overline C$. Apply consensus. Remove $C$ and $\overline C$ and simplify. $$(AB+ C+\overline D E)\ (AD+ \overline C + DE )\ (AB + \overline D E + AD + DE )$$ $$(AB+ C+\overline D E)\ (AD+ \overline C + DE )\ (AB + AD + (\overline D + D)E )$$ $$(AB+ C+\overline D E)\ (AD+ \overline C + DE )\ (AB + AD + E)$$

$$(AB+ C+\overline D E)\ (AD+ \overline C + DE )\ (AB + AD + E) = (AB+ C+\overline D E)\ (AD+ \overline C + DE )$$

Laws and Theorems of Boolean Algebra

StainlessSteelRat
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  • thanks for your help i didn't even know about that law =/, without knowing about that law is it possible to get the answer any other way ? – dmnte May 16 '16 at 16:04
  • You'd have to reduce left and right sides to get the same minimized answer. The last term is redundant. – StainlessSteelRat May 16 '16 at 16:28
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$((A\land B)\lor C\lor(\lnot D\land E))\quad =\quad P$
$((A\land D)\lor\lnot C\lor (D\land E))\quad =\quad Q$
$((A\land B)\lor(A\land D)\lor E)\,\;\;\quad =\quad R$

$(P\land Q\land R)\equiv(P\land Q)$

$(P\land Q\land R)\equiv(P\land Q\land 1[true])$

The only way the whole shebang is true is if $R\quad$ END OF SENTENCE

Senex Ægypti Parvi
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