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This is from the book by D.J. Velleman-"How to prove it?" Sec 3.5 Excercise 31:

Prove $\exists x(P(x)\rightarrow\forall y P(y))$

Suppose the universe of discourse is set of all men.

Let statement P(x) $:=$ $x$ is married. Then $\forall yP(y)$ means all men are married. Thus, there exists a man, if his marital status is married that means are all men are married. Am I interpreting the statement correctly? It is not abvious to me why this statement is true at all.

Ricky
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  • you might have to disprove it... – JMP Aug 07 '15 at 04:30
  • Hmm. How are you going to prove $\exists x(P(x)\to\forall y(P(y))$ considering a specific example? Are you simply trying to figure out how to create a useful linguistic analogue? – Daniel W. Farlow Aug 07 '15 at 04:31
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    This might be useful. – Winther Aug 07 '15 at 04:34
  • @Winther Nice link. Hadn't seen that before; apparently it was popularized by Smullyan. That guy is awesome. Thanks for the link. – Daniel W. Farlow Aug 07 '15 at 04:38
  • There is a man such that if he is awake, all men are awake. – mjqxxxx Aug 07 '15 at 04:38
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    Do it by cases. If all men are married, then who can you choose as $x$ to make the statement true? If there is an unmarried man, who can you choose as $x$ to make the statement true? – Akiva Weinberger Aug 07 '15 at 04:57
  • Note that, if one of the men marries or divorces, it is entirely possible that $x$ changes. At any given time, there is such an $x$, but it needn't always be the same $x$. (Usually $P(x)$ is a property that doesn't depend on time, though.) – Akiva Weinberger Aug 07 '15 at 04:58
  • This is an example of the so-called Drinker's Paradox. A set-theoretic version would be $\exists x:[x\in P \implies Q]$ where $P$ is an arbitrary set, and $Q$ an arbitrary proposition. For a detailed development, see conclusion of "The Drinker's Paradox: A Tale of Three Paradoxes" at my math blog at http://dcproof.wordpress.com/ Note that is not necessarily true that $\exists x: [x\in P] \implies Q$. – Dan Christensen Aug 07 '15 at 11:44

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