Boolean simplification

Question

I'm trying to simplify the following expression:

(x+ ~x)~(~(~z xor y) + (yx))
   (1)~(~(~z xor y) + (yx))
   ~(~(~z xor y) + (yx))
   (~z xor y) + ~(yx)
   (~z xor y) + (~y + ~x)
   (~y~z + yz) + (~y + ~x) <---- im stuck here

How can I simplify this more? and once it is simplified how do i calculate the number of gate delays

Answer 1

Observe that $+$ is both associative and commutative, which allows you to drop parentheses and rearrange terms.