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Let us assume that we have a statement $P \rightarrow Q$. In this case, what would $P \rightarrow \lnot Q$ be called?

The reason why I want to know is that I want to show that $P$ is true by contradiction, proving that both $\lnot P \rightarrow \lnot Q$ and $\lnot P \rightarrow Q$ are true to conclude that $\lnot P$ is false. I initially framed this by stating that $\lnot P \rightarrow \lnot Q$ is a contradiction to $\lnot P \rightarrow Q$, but my supervisor said that this was incorrect, since both theorems are technically true.

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  • $\begingroup$ $P\to\lnot Q$ is the same thing as $\lnot(P\wedge Q)$. $\endgroup$ – Don Thousand Dec 9 '19 at 6:11
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    $\begingroup$ In computer science it is Nand(P,Q). That is, "not (P and Q)". $\endgroup$ – DanielWainfleet Dec 9 '19 at 6:23
  • $\begingroup$ There are lots of names for statements related to implications, e.g. inverse, converse, contrapositive, but this one does not seem to have a name. $\endgroup$ – user856 Dec 9 '19 at 6:27
  • $\begingroup$ "proving that both ¬P→Q and ¬P→Q are true to conclude that P is false" um.... huh? $\endgroup$ – fleablood Dec 9 '19 at 6:27
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    $\begingroup$ "Sorry, I meant "both ¬P→Q and ¬P→¬Q"" Did you also mean "P is true". If $\lnot P \implies Q$ and $\lnot P\implies Q$ that (which is not a contradiction) would mean $\lnot P$ is false. $\endgroup$ – fleablood Dec 9 '19 at 6:32
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There seems to be no name for $P \rightarrow \lnot Q$.

The statement "$\neg P \to \neg Q$ is a contradiction to $\neg P \to Q$" is indeed wrong, since both these implications are true, they don't contradict each other.

The correct way to state this was to use modus ponens. You have already proven $\vdash \neg P \to \neg Q$ and $\vdash \neg P \to Q$. Since you assume $\neg P$, you then get $\vdash Q$ and $\vdash \neg Q$, and this is the contradiction you've been looking for.

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