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Prove $\exists x (P(x) \to \forall y P(y))$.

Let x = y. Suppose P(x) is true. Let y be arbitrary. Since P(x) is true, it must be that P(y) is true. Since y was arbitrary, we can conclude that $\forall y P(y)$. Therefore $P(x) \to \forall P(y)$. Since $P(x) \to \forall P(y)$, $\exists x (P(x) \to \forall y P(y))$.

Is my proof correct?

I'm not sure if I'm dealing with the $\forall y$ correctly. If the theorem were $\exists x (P(x) \to P(y))$, I think I would have a good handle on things.

Marty B.
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  • @copper.hat I thought something similar at first, but this doesn't in fact disprove the implication statement. The statement "If $2$ is odd then every integer is odd" is in fact true. – GPerez Jan 11 '15 at 01:42
  • Another case of the drinker's principle - P is either always true, in which case x is arbitrary, or there is some x where P(x) is false, in which case pick one such x. However y cannot be "arbitrary" - it must be all values. The trick here is that implies anything. – Joffan Jan 11 '15 at 01:44
  • @GPerez: Thanks for catching another of my egregious mistakes. I should know better... – copper.hat Jan 11 '15 at 01:47
  • If you can't convince yourself that a proof is correct, then it is not correct, even if you got lucky and picked the correct words. And in this case, phrases like "Let $x = y$ and "Suppose $P(x)$ is true" and "since $y$ was arbitrary" (even after you latched it to $x$ ... ?) seem pretty incoherent. I think it would be better if you started with something more basic than the drinker's paradox, and come back to this problem after you are comfortable with the basics of implication and quantification. – DanielV Jan 11 '15 at 01:51

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Like you, we give a semantic (model-theoretic) argument. There are also syntactic arguments, where the details depend very much on the setup (axioms, rules of inference).

Let $L$ be our formal language. We will show that the sentence $\exists x(P(x)\to \forall y P(y))$ is true in any $L$-structure $M$.

Suppose first that $\forall yP(y)$ is true in $M$. Then whatever $x$ we choose, the sentence $P(x)\to\forall y P(y)$ is true in $M$. So $\exists x(P(x)\to\forall yP(y))$

Suppose on the other hand that $\forall y P(y)$ is not true in $M$. Then there is an $x$ for which $P(x)$ is false. And then again, because of the truth table of implication, $\exists xP(x)\to\forall y P(y))$ is true in $M$.

Remark: The argument in the OP is no correct. You are letting $x=y$. What is $y$? We want to show that there exists an $x$ such that for $P(x)\to \forall y P(y)$. So the $x$ has to be universal, it has to work for all possible values of $y$.

André Nicolas
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