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So I'm trying to simplify an equation using boolean algebra. I'm still very confused as to how all the rules work and such, but I feel like I'm very close to the answer.

The original equation is (in canonical SOP form):

Z'X'Y' + Z'X'Y + Z'XY' + Z'XY + ZX'Y' + ZXY

So far I've gotten this far:

Distributive Rule 12a: Z'X'(Y + Y') + Z'X(Y' + Y) + ZX'Y' + ZXY

Identity Rule 6a: Z'X' + Z'X + ZX'Y' + ZXY

Rule 12a again: Z'(X' + X) + ZX'Y' + ZXY

Rule 6a again: Z' + ZX'Y' + ZXY

My issue is that the answer is Z' + X'Y' + XY, but how do I get there? Because doesn't ZX'Y' + ZXY simplify to Z(X'Y' + XY) = Z?

What am I doing wrong? Are the steps I have so far correct?

Thanks.

ESM
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2 Answers2

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Because doesn't $ZX'Y'+ZXY$ simplify to $Z(X'Y'+XY)=Z$?

No, because $(XY)'\ne X'Y'$, so in parentheses you do NOT have something of the form "$A+A'$". (In fact, by De Morgan's laws, $(XY)'=X'+Y'$.)

My issue is that the answer is $Z'+X'Y'+XY$, but how do I get there?

There are some additional useful properties that often help simplify boolean expressions. Here are two of them.

1) $\color{red}{A+AB=A}$. Proof: $$A+AB=A1+AB=A(1+B)=A1=A.$$

2) $\color{red}{A+A'B=A+B}$. Proof: $$A+A'B=A+AB+A'B=A+(A+A')B=A+1B=A+B.$$

Now, $Z'+ZX'Y'+ZXY=Z'+Z(X'Y'+XY)=Z'+X'Y'+XY$, where the last step is an application of the second property above (with $A=Z'$ and $B=X'Y'+XY$).

zipirovich
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  • Ohhh I gotcha. Yea that must be using DeMorgan's Theorem then. I never thought to have the Z' and Z = A and the XY = B. Thank you so much man! That definitely helps to clear things up for me :) – ESM Jan 31 '19 at 05:55
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You need to immediately add the following equivalence to your toolbox:

Reduction

$P + P'Q=P+Q$

That is, the $P$ term 'reduces' the $P'Q$ term to just $Q$

With that, you can see how in your expression, the $Z'$ term reduces the $ZX'Y'$ to $X'Y'$, and also reduces the $ZXY$ to $XY$. So there:

$Z'+ZX'Y'+ZXY \overset{Reduction \ x \ 2}{=} Z'+X'Y'+XY$

Bram28
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