How can I tell if this boolean expression is a tautology without proving it?

Question

((X'+ Y)·(Z'+ Y'))' + (Z'+ X')

So I proved this to be a tautology. And a follow up question I received was how can I know it's a tautology simply by looking the original boolean expression. I can't seem to see what makes it always evaluate to 1.

Answer 1

I don't know exactly what they mean, but here is an attempt: If the expression is to be $0$, then by the second term, $X'$ and $Z'$ must both be $0$. Now you can simplify the first term to $(Y\cdot Y')'$, which is clearly $1$.

Answer 2

Your proof may look like a slightly better written version of the below:

''' The phrase on the right (Z'+ X') is a catch all that lets you know that: If Z' is 1, the whole expression is 1. And, If X' is 1, the whole expression is 1.

So we only need to see that ((X'+Y).(Z'+Y'))' is 1 when X.Z is 1: If X.Z is 1; X' is 0; so (X'+Y) is Y If X.Z is 1; Z' is 0; so (Z'+Y') is Y' Therefore, if X.Z is 1; ((X'+Y).(Z'+Y')) is (Y.Y')' which is 1.

As the expression is true for X', Z' and X.Z it is a Tautology. '''

But you can see where the proof comes from relatively immediately, if you look for the hidden Tautology. That is (Y.Y')'.

((X'+Y).(Z'+Y'))' is a thinly veiled version of (Y.Y')', which creates exceptions for the cases of X'.Y' and Y.Z'. But these are then covered by the expression (Z'+ X').

To conclude, the left expression can be read as nearly a Tautology but with exceptions created for a limited number of cases. And the right can be read as a list of the cases in which the left may be False. Leaving the whole thing to be a Tautology.