# Prove that (not(p rarr q) ^^ (p harr notq)) harr ((p ^^ notq)) ?

Feb 20, 2017

We can build a truth table for each of the statements.

Since the truth values for:

$\neg \left(p \rightarrow q\right) \wedge \left(p \leftrightarrow \neg q\right)$ and $\left(p \wedge \neg q\right)$

are exactly the same for all possible combinations of truth values of $p$ and $q$, then

$\left(\neg \left(p \rightarrow q\right) \wedge \left(p \leftrightarrow \neg q\right)\right) \leftrightarrow \left(\left(p \wedge \neg q\right)\right)$

is a tautology and therefore the two propositions are equivalent. QED