natrual deduction is this model wrong answer?

Question

This model entails premise but not the conclusion

But in my opinion if we use the b also in conclusion we have a proof that this premise entails conclusion or?

Answer 1

No, the answer given is correct.

If I understand the picture correctly, the extracted countermodel is

$\mathcal{M} = \langle \{a, b\}, P \mapsto \{a\}, S \mapsto \text{False} \rangle$.

Then $\mathcal{M} \models \exists x (P(x) \to S)$ because with $x \mapsto b$, $P(x)$ is false so $P(x) \to S$ is true so $\exists x (P(x) \to S)$ is true, but $\mathcal{M} \not \models \exists x P(x) \to S$ because with $x \mapsto a$, $P(x)$ is true so $\exists x P(x)$ is true but $S$ is false so $\exists x P(x) \to S$ is false.

For $x \mapsto b$, sure, $P(x)$ is false. But there exists an $x$ such that $P(x)$ is true, so $\exists x P(x)$ is true, whereas $S$ is false, so the conclusion $\exists x P(x) \to S$ is still false in that model.

Even if we find a model that makes both the premise and the conclusion true, e.g. by interpreting $S$ as $\text{True}$, that doesn't make the premise entail the conclusion. Entailment means that in all models in which the premises are true, the conclusion must be true as well. The existence of a single model in which the conclusion is false despite the premises being true already makes the sequent invalid; that there may be other models which do satisfy the conclusion is irrelevant. The one counter example above is enough to show the invalidity and hence unprovability of the sequent.