Is there a tableau proof of $\exists x \forall y (P(x)\to P(y))$ I cannot figure out a counter model for this first order formula.
$\exists x \forall y (P(x)\to P(y))$
However I cannot come up with a tableau proof either.
Can somebody tell me if this formula is valid or not?
 A: The formula is not necessarily valid: if the domain of discourse is empty, then there does not exist any element; in particular none that satisfy the equation.
However, if the domain of discourse is non-empty, then the formula is valid. If there is some $x$ such that $P(x)$ is false, then $P(x) \rightarrow P(y)$ is true for every $y$. Conversely, if $P(x)$ is always true, then for any $x,y$, $P(x) \rightarrow P(y)$. So if there's one element in the domain, the statement is true.
A: 
Is there a tableau proof of $∃x∀y(P(x)→P(y))$ ?

The answer is YES :
1) $\lnot ∃x∀y(P(x)→P(y))$
2) $\lnot ∀y(P(a)→P(y))$ --- using constant $a$
3) $\lnot (P(a)→P(b))$ --- with $b$ new
4) $P(a)$
5) $\lnot P(b)$
6) $\lnot ∀y(P(b)→P(y))$ --- from 1)
7) $\lnot (P(b)→P(c))$ --- with $c$ new
8) $P(b)$
9) $\lnot P(c)$
But 5) and 8) close the tree: thus, formula 1) is unsatisfiable.
Being 1) the negation of the original formula : $∃x∀y(P(x)→P(y))$, we have that the original formula is valid.

As per Sambo's answer, the formula is invalid in an empty domain (as will be every existential formula).
In that case, what we can do is to prove : $\exists x (x= x) \vdash ∃x∀y(P(x)→P(y))$.
The corresponding tableau will start with step 0) $\exists x (x= x)$ and then we can instantiate it with $a$ to get : $a=a$. Having done this, we can use the constant $a$ in 2) above.
