# Proof of the identity of a Boolean equation $Y+X'Z+XY' = X+Y+Z$

How to prove the following the identity of a Boolean equation?

$$Y+X'Z+XY'=X+Y+Z$$

I have tried :

$\space\space\space\space\space Y+X'Z+XY'\\ =X'Z+XY'+Y\\ =X'Z+XY'+Y(X+X')\\ =X'Z+XY'+XY+X'Y\\ =X'(Z+Y)+X(Y'+Y)\\ =X'(Z+Y)+X‧1\\ =X'(Z+Y)+X\\$

Then, how to continue?

• Use $A+BC=(A+B)(A+C)$ May 30 '14 at 6:08
• @EkaveeraKumarSharma How is (A+B)(A+C) = A+BC? LHS is A+AB+AC+BC. May 30 '14 at 6:22
• @tpb261: $A+AB+AC=A$ May 30 '14 at 6:31
• @user2357112 Ah.. yes.. stupid me :( May 30 '14 at 6:45

$$Y+X'Z+XY'$$ $$=(Y+Y')(Y+X)+X'Z$$ $$=1.(Y+X)+X'Z$$ $$=(X+X')(X+Z)+Y$$ $$=1.(X+Z)+Y$$ $$=X+Y+Z$$

Hence, proven.

• See, how better it looks now! Latex is simple. :-) Dec 4 '17 at 7:46

Logically,

• Y+X'Z+XY' is true whenever Y is true, irrespective of X and Z.
• Removing all inputs where Y is false, X'Z+XY' is true, when X is True (we know Y is false), irrespective of Z.
• And finally, when both X and Y are false, X'Z is true, hence the expression is true.
In all three cases, we were only concerned whether one parameter was true or
not, the others were either false or not affecting the result, so, it is same
as Y+X+Z.
Or you could expand Y=Y(X'Z'+X'Z+XZ+XZ') and then repeat some terms from this
expansion and group them with the other terms.