I need to convert the following formula into Prenex normal form:
$ (∃z)S(z) ∧ (∃x)(∀y)\{ (∀z)[S(z) → P(y,z)] → R(x,y) \} $
Now I started by doing this:
$(∃z)S(z) ∧ (∃x)(∀y)\{ \neg(∀z)[S(z) → P(y,z)] \vee R(x,y) \} $
By using the formula $ P\rightarrow Q = \neg P \vee Q $
and then:
$(∃z)S(z) ∧ (∃x)(∀y)\{ (\exists z)[S(z) \land \neg P(y,z)] \vee R(x,y) \} $
By using the formula $\neg \forall z P = \exists z \neg P $
From here I extracted the $\exists z$
and got:
$(∃z)S(z) ∧ (∃x)(∀y)(\exists z)\{[S(z) \land \neg P(y,z)] \vee R(x,y) \} $
and now took $\exists z$ outside:
$(∃z)(∃x)(∀y)\{S(z) \land [S(z) \land \neg P(y,z)] \vee R(x,y) \} $
and now I received a prenex normal form, although this is not the answer and I do not know where have I mistaken.
Thanks for the help.