# 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 $X+(XY)=X$. – Felix Castor Jan 14 '15 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. – Dale M Jan 14 '15 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.