# Boolean Algebra Proof

I'm new at learning boolean algebra and understand the basic laws. But now I'm trying to proof:

$x'z+ xyz + xy'z = z$

Can someone help me with this please ?

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What is this from? What are x' and y' ? Factor out a z from each term on the left and you have reduced your problem to proving x' + xy + xy' =1. –  Tom Stephens Aug 13 '10 at 13:15
This is from the boolean algebra. x' and y' are the opposite of x & y. + = OR . = AND –  Anonymous Aug 13 '10 at 13:18

Are you still stuck? I'll use your terminology, which seems to be used in electrical engineering.

1) $x'z + xyz + xy'z$

2) $x'z + xz(y+y')$

Since $y+y' = 1$, and $xz1 = xz$ then

3) $x'z + xz$

4) $(x'+x)z$

5) $z$

Hope that helps.

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If you think of the Boolean operations as complement, intersection and union of sets, which is equivalent since every Boolean algebra is isomorphic to a field of sets, then the equation is obvious.

Namely, on the right side we have the set $z$. On the left side, we have the elements of $z$ separated into three categories: those that are not in $x$, those that are in $x$ and also in $y$, and those that are in $x$ but not in $y$. Since this accounts for all possible elements of $z$, we may deduce the equality.

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Assuming you mean what is more commonly written as $(\lnot x\land z)\lor(x\land y\land z)\lor(x\land\lnot y\land z)=z$, then it is equivalent, as Tom Stephens points out, to proving that $\lnot x\lor(x\land(y\lor\lnot y))$ is valid (that is, evaluates to true for all x and y). But that's clearly the same as $\lnot x \lor x$.
• distributivity: $x\lor(y\land z)=(x\lor y)\land(x\lor z)$ and also $x\land(y\lor z)=(x\land y)\lor(x\land z)$
• excluded middle: $x\lor\lnot x$