Boolean Algebra: How does (x * z) + (!x * y) + (z * y) simplify to (x * z) + (!x * y)

Question

How does the equation:

(x * z) + (!x * y) + (z * y)

simplify to:

(x * z) + (!x * y)

using boolean algebra?

I've been looking at the various theorems for quite awhile, but I can't figure out how this study guide I found online got to that answer.

Answer 1

If a possible transformation of one expression into the other one is not obvious, you can still blow up both expressions to their full disjunctive form:

$$xz + \bar x y + zy = x(y+\bar y)z + \bar xy(z+\bar z) + (x+\bar x)yz\\=xyz + x\bar yz + \bar xyz + \bar xy\bar z + xyz + \bar xyz\\ = xyz + \bar xyz + x\bar yz + \bar xy\bar z$$ and $$xz + \bar xy = x(y+\bar y)z + \bar xy(z+\bar z)\\ = xyz + x\bar yz + \bar xyz + \bar xy\bar z$$

As both expressions are equal, you have now a transformation of one expression into the other one.

Answer 2

You can draw a truth table and compare the output of the two expressions.

Basically for $z = y$, both expressions will output $z$ ($=y$).

For $z = 1, y = 0$, both will output $x$.

For $z = 0, y = 1$, both will output $!x$.