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Lets say I want to prove or disprove the statement $p \to q$. If I show that $p\to \lnot q$ for one particular case, isn't that a counterexample that shows that the statement is false?

amWhy
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    Yes, certainly, although I would express it as $P$ and not Q. – saulspatz Mar 20 '20 at 16:38
  • @saulspatz $p \to \lnot q \not\equiv (p \land \lnot q) \equiv \lnot(p \to q)$. So the OP's "counterexample" does not necessarily contradict $p\to q$. Please see my answer for further clarification. – amWhy Mar 20 '20 at 17:15
  • @amWhy Yes, I know that, but what the OP has shown in "one particular case" is surely $p\wedge\lnot q$. – saulspatz Mar 20 '20 at 18:16
  • No, @saulspatz, the OP says if they find a case where $p \to \lnot q \equiv \lnot p \lor \lnot q \equiv \lnot (p \land q),$ is that a counterexample. (Not necessarily, because if p is false it doesn't matter what the truth value of q is. They did NOT say that they found a case where $p \to q$ is false. – amWhy Mar 20 '20 at 18:19
  • To yield a contradiction across the board, we need $\lnot (p \to q) \equiv (p \land \lnot q) \not\equiv (p \to \lnot q),$ as the OP proposed. – amWhy Mar 20 '20 at 18:24
  • So your comment/answer "yes certainly" is wrong and misleading. See both answers below. – amWhy Mar 20 '20 at 18:26

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No, that is not correct in general. For example, if $P$ is false, then you can prove that $P\implies \neg Q$, but you can also prove $P\implies Q$.

Maybe your question is abstracted from a problem in which this is not the case. You would need to give more details, in that case. As a general logical deduction, proving $P\implies \neg Q$, doesn't imply $\neg(P\implies Q)$.