Boolean algebra simplification with DeMorgan Laws

Question

I can't simplify the following expression:

$$x·y'+z+(x'+y)·z'$$

I've tried to multiply the last term with the guys in the parentheses but I can't go any longer.

Thanks in advance.

Answer 1

Since $xy'+z+(x'+y)z'$ is $1$ when $z=1$, the factor $z'$ in the last summand can be omitted, so it becomes $$ xy'+z+x'+y. $$ By the same reasoning, the factor $x$ in the first summand can now be omitted, so it becomes $$ y'+z+x'+y. $$ Now $y'+y=1$ so the expression is $1$ always, we conclude that $$ xy'+z+(x'+y)z' = 1. $$