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Task: Prove this theorem: $ \exists x (P(x) \Rightarrow \forall yP(y) )$.

I got this far: I figured out this is equivalent to $\exists x (\neg P(x) \lor \forall yP(y) )$.

I don't understand however how I can go about completing the proof without any other extra information. I can't really assume anything about the statement $P$ can I?

pseudomarvin
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3 Answers3

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You can't assume anything about $P$, nor do you have to. Note that because $x$ is not free in $\forall y P(y)$, $$ \exists x(\neg P(x)\lor \forall y P(y)) \equiv (\exists x \neg P(x)\lor \forall y P(y)). \tag{1} $$ Now recall that $\exists\neg \equiv \neg\forall$, so the righthand side of (1) is equivalent to $$ \text{(1)}_{RHS} \equiv (\neg \forall x P(x)\lor \forall y P(y)).\tag{2} $$ By changing variables, $$ \text{(2)}_{RHS} \equiv (\neg \forall y P(y)\lor \forall y P(y)),\tag{3} $$ which is a tautology, hence provable and valid. These are all equivalent to the lefthand side of (1), which, as you noticed, is equivalent to the formula you wish to prove.

BrianO
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  • I should have been clearer in the question I apologize for that. The goal is not proving that $ \exists x (P(x) \Rightarrow \forall yP(y) )$ is equivalent to $\exists x (\neg P(x) \lor \forall yP(y) )$ (what I assume you thought was the goal). I have edited the question to more clearly reflect that. – pseudomarvin Mar 07 '16 at 13:00
  • You're right, I did assume you were asking that different question. I emended my answer to address your actual question. – BrianO Mar 07 '16 at 18:13
  • Great answer, thanks. So at point $(2)$ it is perfectly all right to change the variable in the statement $\neg \forall xP(x)$ because it is "independent"? (There is probably a formal term for what is going on that I don't know.) – pseudomarvin Mar 08 '16 at 23:41
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    That's right. "Independent" isn't quite a standard term but you'll be understood. There's no other variable in the scope of the $\forall x$ quantifier, either bound or free, so you can change it to any variable, even $y$. ¶ You're welcome:) & thank you. – BrianO Mar 08 '16 at 23:55
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$A\implies B$ which is logically equivalent to $\neg A \vee B$. Therefore $$ \exists x(A\implies B) $$ is logically equivalent to $$ \exists x(\neg A \vee B) $$ no matter what the statements $A$ and $B$ are.

MMI
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  • I'm sorry but I probably was not clear enough about the goal of the proof. I have edited the question to remedy this. – pseudomarvin Mar 07 '16 at 13:02
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Using a tableau expansion:

  1. Start with: $ \neg (∃x(P(x)⇒∀yP(y)))$
  2. The leading connective is $\neg$ then $∃$: $∀x \neg (P(x)⇒∀yP(y))$
  3. Instantiate: $ \neg (P(a)⇒∀yP(y))$
  4. The leading connective is $\neg$ then $⇒$: $ (P(a)\wedge \neg ∀yP(y))$
  5. We now have three formulas: $∀x \neg (P(x)⇒∀yP(y))$, $ P(a)$ and $\neg ∀yP(y)$
  6. Or: $∀x \neg (P(x)⇒∀yP(y))$, $ P(a)$ and $∃y \neg P(y)$
  7. Introduce a witness term: $\neg P(b)$
  8. Instantiate:$ \neg (P(b)⇒∀yP(y))$
  9. Or: $ (P(b)\wedge \neg ∀yP(y))$
  10. P(b) and $\neg P(b)$
  11. Close the branch, hence the tableau.

This theorem is by the way the drinker's theorem. See Drinker paradox -Wikipedia

user155673
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