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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?

Thank you for your help.

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    $\begingroup$ Use $A+BC=(A+B)(A+C)$ $\endgroup$ – Ekaveera Kumar Sharma May 30 '14 at 6:08
  • $\begingroup$ @EkaveeraKumarSharma How is (A+B)(A+C) = A+BC? LHS is A+AB+AC+BC. $\endgroup$ – tpb261 May 30 '14 at 6:22
  • $\begingroup$ @tpb261: $A+AB+AC=A$ $\endgroup$ – user2357112 supports Monica May 30 '14 at 6:31
  • $\begingroup$ @user2357112 Ah.. yes.. stupid me :( $\endgroup$ – tpb261 May 30 '14 at 6:45
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$$Y+X'Z+XY'$$ $$=(Y+Y')(Y+X)+X'Z$$ $$=1.(Y+X)+X'Z$$ $$=(X+X')(X+Z)+Y$$ $$=1.(X+Z)+Y$$ $$=X+Y+Z$$

Hence, proven.

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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.
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