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Now I understand that strictly speaking the contrapositive of this statement would be $\neg r \implies \neg p\lor\neg q$, but what I would like to do instead is prove that $\neg r\land q \implies \neg p$.

What I'm asking is, can I still call this a contrapositive? Or does it have some other name I could use? Also since I'm doing this for an assignment would it be an acceptable proof of the statement? Or should I take the easy way out and simply assume all three statements and do a contradiction?

The exact wording of the question is "Given that $p$, and that $q$, show that $r$". I can give the actual statements if it helps but I imagine that would be superfluous information.

J.G.
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S. Teisseire
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    It's the contrapositive of the "curry'd" version of the statement: $p\to (q\to r).$ – spaceisdarkgreen Jan 12 '19 at 08:08
  • \lor produces $\lor$ and \land produces $\land$ (and \lnot produces $\lnot$), as far implications, I always felt that \to or \rightarrow, both of which produce $\to$, were better for propositional formulas compared to \implies. – Asaf Karagila Jan 12 '19 at 08:13
  • @drhab: Yes. Back when I was TA'ing basic logic in Beer-Sheva that was my approach: $p\to q$ is a proposition, $p\implies q$ is a statement about propositions (and the two are related, of course). – Asaf Karagila Jan 12 '19 at 09:07

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Hint

since $$\lnot r\Longrightarrow \lnot p\lor\lnot q$$then $$\lnot r\land q\Longrightarrow (\lnot p\lor\lnot q)\land q$$I leave to you to show that $$(\lnot p\lor\lnot q)\land q\equiv\lnot p$$

Mostafa Ayaz
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The contrapositive to $p \land q\implies r$ (which is what you are asking for in the title) is $\lnot r \implies \lnot(p \land q) = \lnot p \lor \lnot q$. In general, the contrapositive of a statement is that the negation of the consequent implies the negation of the antecedent, however copmlicated the expression (so not only in the case $p \implies q$).

This kind of proof is not nessecarily 'worse' in any sense; take this proof from math.stackexchange:

Proposition: $x^4 - x^3 + x^2 \neq 1$, then $x \neq 1.$

The easy and elegant way to prove this is by contrapositive:

If $x = 1$, then $x^4 - x^3 + x^2 = 1$.

If you only think about what happens when $x^4 - x^3 + x^2 \neq 1$, it's not as trivial that $x \neq 1.$ There are many cases where contrapositive is much easier than straightforwardly proving the implication; you will become more familiar with when this is so as you do and read more proofs. Also see https://math.stackexchange.com/a/650487/606584.

Steven
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