Transform $∀z(P(z) → Q(z)) ∧ ∃zS(z)$ into PNF. Then transform $∀z(P(z) → > Q(z)) ∧ ∃zS(z)$ into PNF without renaming the variables. Finally, state the result of evaluating your two transformed formulae given a domain consisting of $D=(a,b)$ and interpretation that makes $S(a)$, $Q(a)$ and $Q(b)$ true.
I've transformed to PNF, but have no clue what to do for the second part of the question.
PNF: $∀z∃y(¬P(z) ∨ Q(z)) ∧ S(y))$
PNF without renaming: $∀z∃z((¬P(z) ∨ Q(z)) ∧ S(z))$
Would appreciate some guidance.
Hint
Consider the outer quantifier : $∀z$ and consider in turn all elements of the domain $D$.
(i) When $z$ evalautes to $a$ we have $Q(a)$ true, and thus also $¬P(a)∨Q(a)$ is true. For this value of $z$ we can choose $a$ as value for $y$, and we have that $S(a)$ is true.
In conclusion, when $z$ evalautes to $a$, we have that $∃y \ [¬P(z)∨Q(z))∧S(y)]$ is true.
(ii) Now for $b$ as value for $z$. $Q(b)$ is true and again $¬P(b)∨Q(b)$ is. As above, we can choose $a$ as value for $y$, and thus, being $S(a)$ true, we have that also when $z$ evalautes to $b$, $∃y \ [¬P(z)∨Q(z))∧S(y)]$ is true.
Conclusion : for all possible value of $z$ in the domain $D$, $∃y \ [¬P(z)∨Q(z))∧S(y)]$ is true, i.e.
is true in $D$ with the said interpretation.