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When proving if... then... statements we can use the contradiction approach. We assume the if part and assume the negation of the then part after that we show that there is a contradiction which means that the negation of the then part is false, which means the then part is true.

I thought about it for a bit and something isn't clear to me. When we show that there is a contradiction why can't the if part (that we assumed) be false and the negation of the then part be true? Why do we conclude that the negation of the then part is false? Maybe it is the if part that is false.

mawaior
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  • First, to clear up your question's terminology. Take an implication $P\rightarrow Q$. If I understand correctly, what you're calling the "if part" is the antecedent $P$ and your "then part" is the consequent $Q$. You seem to be misunderstanding how contradiction works and what we show to be true/false. Contradiction assumes some antecedent $P$ and walks this through whatever it implies, $P\rightarrow Q$, to a certainly absurd, false consequent $Q$. Knowing $P\rightarrow Q$ and $\neg Q$, we can infer $\neg P$. (1/2) – Jam Dec 14 '23 at 20:51
  • We chiefly only take one assumption: the antecedent $P$. We don't need to assume anything about $Q$. Or equivalently, we could substitute $P$ for its negation $\neg P$ everywhere. Please see this question (2/2) – Jam Dec 14 '23 at 20:52
  • Does this answer your question? Wondering why proof by contradiction works – Jam Dec 14 '23 at 20:53
  • I meant that when we want to prove $P \rightarrow Q$ by contradiction. We assume $P$ and $\lnot Q$ and then we show that there is a contradiction so we conclude $Q$. My question is why not conclude $\lnot P$ @Jam – mawaior Dec 14 '23 at 20:58
  • Please reread my comment. That's not how contradiction works. We don't assume $\neg Q$, and we don't ultimately conclude $Q$, we only do so as a proof step. – Jam Dec 14 '23 at 20:59
  • This is what is done when proving contradiction what do you mean we don't do it? I assume $P$ and the negation of $Q$ ($\lnot Q$) and show that there is a contradiction. @Jam – mawaior Dec 14 '23 at 21:01
  • If you choose to prove an implication by contradiction, then it is entirely up to you how you gain the contradiction. However, it is generally much more informative when trying to prove a statement of the form $P \Rightarrow Q$, to assume $P$ and explore the consequences to see how they might lead to $Q$ rather than presuppose that a proof by contradiction is the right next step. – Rob Arthan Dec 14 '23 at 22:14

1 Answers1

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Use the "standard definition": $~~~P\implies Q ~\equiv ~ \neg (P \land \neg Q)$.

If you assume $P \land \neg Q~$ is true and subsequently obtain a contradiction without making any other assumptions, then you can infer that $\neg (P \land \neg Q)$ is true, or equivalently that $P \implies Q$.

Dan Christensen
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  • So just to be sure I understood. $P\implies Q \equiv \lnot (P \land \lnot Q)$ so when I want prove $P\implies Q$ is true, I assume $P\land \lnot Q$ and then get a contradiction which means that what I assumed ($P \land \lnot Q$) was false so $\lnot (P \land \lnot Q)$ is true. $\lnot (P \land \lnot Q) \iff (P\implies Q$) so $P\implies Q$ is also true. Am I right? – mawaior Dec 14 '23 at 22:09
  • That is correct. There are, of course, other ways to prove an implication. – Dan Christensen Dec 14 '23 at 22:11
  • Yes of course, I only asked about the contradiction approach. Thank you for you clear answer. – mawaior Dec 14 '23 at 22:29