The question is to simplify the following expression:
$$(xz + y'x)' + y(x + y)(z' + y)$$
Rewritten using discrete mathematics symbols: $$ \neg((x \wedge z) \vee (\neg y \wedge x)) \vee y \wedge (x \vee y)(\neg z \vee y)$$
Whenever I attempt to solve it, I end up with y+x' or ($y\vee\neg x$). I'm not sure if this is the correct result as it omits the z variable.
I believe I'm making a mistake somewhere following this last expression:
$$\begin{align} (xz+y'x)'+y(x+y)(z'+y) &= (x'+z')(y+x')+(xy+yy)(z'+y) \nonumber \\ &= (x'y+x'x'+z'y+z'x')+(xy+y)(z'+y) \nonumber \\ &= x'y+x'+z'y+z'x'+xyz'+xyy+yz'+yy \nonumber \\ &= x'y+x'+z'y+z'x'+xyz'+xy+yz'+y \nonumber \\ \end{align}$$ The problem is that I'm unsure in which order to eliminate the variables (think x'y + x' = x'(y + 1) = x'(1) = x'). I'm only looking for a nudge in this instance on how best to proceed.
Reiterating Brian M. Scott's response (marking this post as answered), y + x' is indeed the correct simplified Boolean expression. The omission of z means that whatever truth value z takes on is irrelevant when considering the truth value of the entire expression.