Simple proof involving quantifiers 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?
 A: 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.
A: $A\implies B$ 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.
A: Using a tableau expansion:


*

*Start with: $ \neg (∃x(P(x)⇒∀yP(y)))$

*The leading connective is $\neg$ then $∃$: $∀x \neg (P(x)⇒∀yP(y))$

*Instantiate: $ \neg (P(a)⇒∀yP(y))$

*The leading connective is $\neg$ then $⇒$: $  (P(a)\wedge \neg ∀yP(y))$

*We now have three formulas:  $∀x \neg (P(x)⇒∀yP(y))$, $  P(a)$ and  $\neg ∀yP(y)$

*Or: $∀x \neg (P(x)⇒∀yP(y))$, $  P(a)$ and  $∃y \neg P(y)$

*Introduce a witness term: $\neg P(b)$

*Instantiate:$ \neg (P(b)⇒∀yP(y))$ 

*Or: $  (P(b)\wedge \neg ∀yP(y))$

*P(b) and $\neg P(b)$ 

*Close the branch, hence the tableau. 


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