# Simplifying a boolean expression

I need to prove:

XY + ~XZ + YZ = XY + ~XZ

I cannot think how to do this.

I have tried factorising, but I just don't know of any rule that removes one of the terms like above.

All I can do is get to:

Y(X+Z) + ~XZ

• Since there are only three variables, writing out the two truth tables will do just fine. Commented Jan 6, 2015 at 23:19

Here is an ugly solution, found with the Karnaugh table and the equation : A = A.B + A./B

XY + /XZ + YZ

= X.Y.Z + X.Y./Z + /X.Y.Z + /X./Y.Z + X.Y.Z + /X.Y.Z

(removing two redundant terms)

= X.Y.Z + X.Y./Z + /X./Y.Z + /X.Y.Z

= XY + /X.Z

• I'm afraid this won't be a valid solution, I've been staring at this blankly for some time now. The question simply asks to use 'Boolean algebra' to prove ... Anyother ideas? Commented Jan 6, 2015 at 20:01

OK, answer is as follows, unsure if there's an easier method:

$XY+X'Z+YZ \\ XY+X'Z+YZ(1) \\ XY+X'Z+YZ(X+X') \\ XY+X'Z+XYZ+X'YZ \\ (XY+XYZ)+(X'Z+X'YZ) \\ XY(1+Z)+X'Z(1+Y) \\ XY(1)+X'Z(1) \\ XY+X'Z$

Here is a simple argument:

To prove $XY+X'Z+YZ = XY+X'Z$, we only need to show that $$(YZ = 1) \Rightarrow (XY + X'Z = 1)$$ which is obvious, because $$(YZ = 1) \Rightarrow (Y = 1 \land Z = 1) \Rightarrow (XY + X'Z = 1)$$ no matter whether $X = 0$ or $X = 1$.

Note that $$XY+X'Z = Y + Z$$.

Intuitively, because if X is true, then the second term must be false, which means whether the entire expression is true depends on $$Y$$. Likewise for $$Z$$. Then we get:

$$Y + Z + YZ$$

This statement is true when $$X$$ is true, or when $$Y$$ is true, or when both are true. But that is redundant, and we can reduce to:

$$Y + Z$$

Recall from earlier that we established that $$XY+X'Z = Y + Z$$. Thus we get:

$$XY + X'Z$$