# Boolean algebra simplification problem

I can't solve this equation:

$$(xy + x'yz)(xz + x'y') = xyz$$

After applying distribution I got this:

$$xyz + yz + x'z = xyz$$

I can't find the answer and have been thinking for hours now.

• Oh my god, facepalm I just scrapped both letters out of the equasion but leaved the rest in there, pretty retarded. Sorry for bothering you guys. – user3152069 Jan 11 '14 at 21:58

Except from $xyz$ every term contains a variable and its negation, which results in $0$.
$(xy+x'yz)(xz+x'y')=xxyz+xx'yy'+xx'yzz+x'x'yy'z=xyz+0 \times 0+0 \times yz + x' \times 0 z = xyz$
$\square$