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Let's say we're given two conclusions A and B and we want to show that only one can be true at once. Can I prove this by assuming A and showing that B cannot be true and vice versa?

mathnerd
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  • Yes, that method of proof would work. – Robert Shore Jan 26 '23 at 02:24
  • Awesome, thank you! – mathnerd Jan 26 '23 at 02:25
  • @mathnerd Note that the title says "exactly one", but the questions says "only one". These are not the same, and what you said is only good for the "only one" case. – Yanko Jan 26 '23 at 02:51
  • I didn't even realize I mistyped that. What is the difference, and how would one go about proving the "exactly one" case? – mathnerd Jan 26 '23 at 03:44
  • To prove the exactly one case, you need to prove that $$A \implies \neg B ~\text{and}~ \neg A \implies B.$$ To prove the no more than one case, it is sufficient to prove (for example) merely that $$A \implies \neg B.$$ Alternatively, in this second interpretation you could prove that it is not the case that $$A ~\text{and}~ B.$$ – user2661923 Jan 26 '23 at 04:07

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