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 ?
Thanks in advance.
 A: 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.
A: 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.
A: 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: $x\lor\lnot x$

