# Trying to prove Equivalency using Boolean Algebra

The question presented was to use boolean algebra to show that

XY’Z + X’Y’Z’ + XY’Z’ + X’YZ’ ≡ XYZ’ + XY’Z + XY’Z’ + XYZ’

I've tried using various laws of Boolean algebra, but the answer that I always end up with is

X' ≡ XY or X ≡ X' + Y'

Obviously neither are true. I tried plugging these functions into a truth table to verify the equivalence that way, but the outputs were different for both functions.

Is there an error in the question or am I missing something?

Here is the work I've done to reach my solution:

XY'Z + XY'Z' + X'YZ'+ X'Y'Z' == XYZ' + XYZ' + XY'Z + XY'Z'

X'YZ'+ X'Y'Z' == XYZ' by removing like terms

Z'(X'Y + X'Y') == Z'(XY) using distributive and associative law

X'Y + X'Y' = XY by removing like terms

X'Y + X'Y' = X' + Y' DeMorgan's

(X + Y')(X + Y) = X' + Y' DeMorgan's

X(Y + Y') = X' + Y' Distributive Law

X = X' + Y' by Inverse Law

X' = XY DeMorgan's

False, not equivalent.

• A string of OR's can be Absorbed. $X(X+Y) = X$ for instance. Likewise $X+(XY)=X$. Commented Jan 14, 2015 at 22:23
• Please fix your typo, the equation at the start of your question is not the same as the one in the first line of your answer. Commented Jan 14, 2015 at 22:26

$$X\overline {Y} Z + \overline {X}\ \overline {Y}\ \overline {Z} + X\ \overline {Y}\ \overline {Z} + \overline {X}Y\overline {Z} ≡ XY\overline {Z} + X\overline {Y}Z + X\overline {Y}\ \overline {Z} + XY\overline {Z}$$ They are equivalent if they reduce to the same minimum or expression on left can be transformed into expression on right or produce the same truth table.

Left side: $$X\overline {Y} Z + \overline {X}\ \overline {Y}\ \overline {Z} + X\ \overline {Y}\ \overline {Z} + \overline {X}Y\overline {Z}$$ Rearrange: $$(X\overline {Y} Z + X\ \overline {Y}\ \overline {Z}) + (\overline {X}\ \overline {Y}\ \overline {Z} + \overline {X}Y\overline {Z})$$ Complement Law: $A + \overline A = 1$ $$X\overline {Y} (Z + \overline {Z}) + \overline {X}\ \overline {Z} (\overline {Y} + Y)$$ $$X\overline {Y} + \overline {X}\ \overline {Z}$$ Right side: $$XY\overline {Z} + X\overline {Y}Z + X\overline {Y}\ \overline {Z} + XY\overline {Z}$$ Rearrange: $$XY\overline {Z} + XY \overline {Z} + X\overline {Y}Z + X\overline {Y}\ \overline {Z}$$ Idempotent Law $A + A = A$. Eliminate 2nd term + duplicate last term. $$XY\overline {Z} + X\overline {Y}Z + X\overline {Y}\ \overline {Z} + X\overline {Y}\ \overline {Z}$$ Rearrange: $$(XY\overline {Z} + X\overline {Y}\ \overline {Z}) + (X\overline {Y}Z + X\overline {Y}\ \overline {Z})$$ $$X\overline {Z} (Y+ \overline {Y}) + X\overline {Y} (Z + \overline {Z})$$ $$X\overline {Z} + X\overline {Y}$$ They are not equivalent. $$X\overline {Y} + \overline {X}\ \overline {Z}\ {\not\equiv}\ X\overline {Y} + X\overline {Z}$$ Which was your conclusion, but the math is incorrect. First term is common.