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

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Answer 1 · 2017-02-20T22:31:52

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

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Since the truth values for:

$not(p rarr q) ^^ (p harr notq)$ and $(p ^^ notq)$

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

$(not(p rarr q) ^^ (p harr notq)) harr ((p ^^ notq))$

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

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