How to solve predicate tableaux

Question

I can't figure out how to solve this particular tableaux. Especially I don't know how to treat the existence operator which comes first. I can't find a solution neither online or on books, so I don't know if my bet is right.

¬∃x ( ∃y P(x,y) → ∀z P(x,z) )

How I would procede:

( ∃y ¬P(a,y) → ∀z ¬P(a,z) )

now I split into two different fields:

∃y P(a,y) ------------------------------------------- ∀z ¬P(a,z)

P(a,b) --------------------------------------------- ¬P(a,c)

Is this right?

Thanks very much to anyone who will help.

Answer 1

No. On the left branch you should get $\neg \exists y P(a,y)$

Also, I wonder what happened to the negation from the very first statement in your post ... Or is that the statement you want to prove valid, i.e. have to negate at the start of the tableaux?

Oh, I see what is going on. Ok, no, that step you take in the beginning isn't right either. OK, then forget that first comment, and let's start from the very beginning:

$\neg \exists x (\exists y P(x,y) \rightarrow \forall z P(x,z))$

First step (assuming you have this as a separate rule... Some systems go straight to my second step)

$\forall x \neg (\exists y P(x,y) \rightarrow \forall z P(x,z))$

Now instantiate universal:

$\neg (\exists y P(a,y) \rightarrow \forall z P(a,z))$

Use negation of conditional rule:

$\exists y P(a,y)$

$\neg \forall z P(a,z)$

Instantiate existential:

$P(a,b)$

Bring negation in:

$\exists z \neg P(a,z)$

Instantiate existential

$\neg P(a,c)$

... And now you need to instantiate the universal with $b$ and $c$ ... which gives you $d$, and $e$, and $f$, ... and this will just continue without end ...