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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 ?

Thanks in advance.

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    This is from the boolean algebra. x' and y' are the opposite of x & y. + = OR . = AND2010-08-13

3 Answers 3

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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 $\lnot x \lor x$.

I used

  • 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: $x\lor\lnot x$
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    With your notations I said x'+x(y+y')=x+x'=1. These notations make the law x+yz=(x+y)(x+z) look strange.2010-08-14