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I have to find X so that this $(p \Longleftrightarrow ¬q) ∧ (r ⇒ X) ∧ (¬r ⇒ ¬X)$ is a contradiction. Then I also have to find out whether or not I can find an X is a tautology.

What's the most efficient way of solving this? I'm clueless as to how to get this done. Should I draw a truth table? Is there a better way, because a truth table with 4 variables can take a lot of time + it's very easy to make a mistake. And even then, what do I do after I'm done with the truth table?

curious_mind
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peroxy
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  • $X$ cannot be a tautology. Assume that $X=TRUE$; then $(r⇒T) \land (¬r⇒¬T) \Leftrightarrow r$ (check it ...). Thus the complete formula is equivalent to : $(p \Leftrightarrow ¬q)∧ r$, which is not a contradiction. – Mauro ALLEGRANZA Jan 28 '15 at 08:54

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A trivial answer is to set $X = \neg r$. In fact, the proposition $(r \implies X) \land (\neg r \implies \neg X)$ has the same truth value of $(r \iff X)$. So, the whole proposition becomes $$ (p \iff \neg q) \land (r \iff X) $$ and, by setting $X=\neg r$, you obtain an easy contradiction $r \iff \neg r$ that makes the proposition a contradiction.