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How can I show using formal logic that $\lnot p \to (p \to q)$? I'm looking for hints towards a propositional calculus proof, and would not accept an answer consisting solely of truth tables.

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closed as off-topic by 3SAT, Jack's wasted life, Najib Idrissi, user228113, Daniel Fischer Feb 17 '16 at 14:05

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    $\begingroup$ If you desire a "propositional calculus proof", you have to specify the calculus you are using: Tableau, Natural Deduction, Hilbert-style (and in this case, what axioms...). $\endgroup$ – Mauro ALLEGRANZA Feb 17 '16 at 7:51
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Without knowing the details of the deduction system your class uses, I'll assume you can do proofs by assumption. In that case, assume $\neg p$. Then assume $p$. Then for proof by contradiction assume $\neg q$. Then you can show $p\land \neg p$ so that you have a contradiction, and then you can infer $q$. Closing this proof by assumption out you get $p\rightarrow q$ at the end of a proof by assumption starting with $\neg p$. So again close the proof by assumption and derive $\neg p \rightarrow (p \rightarrow q)$.

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We know that the inverse of $a \rightarrow b$ is $a \wedge \neg b$. Therefore if $\neg p \rightarrow (p \rightarrow q)$ is not correct then we must have $ \neg p \wedge \neg (p \rightarrow q)$ which is $ \neg p \wedge p \wedge \neg q $, but this would be a contradiction as we can not have $p$ and $-p$ at the same time.

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First, prove the following equivalence: $$ (A\to(B\to C)) \equiv ((A\land B)\to C). $$ From this, you can see that your formula is equivalent to $(\neg p\land p)\to q$. The antecedent $(\neg p\land p)$ is a contradiction, and somehow in your deductive system you can show that a contradiction implies anything ($q$).

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