"Consider a language with a single object constant a, a single unary function constant s, and two unary relation constants p and q. We start with the premises shown below. We know that p is true of s(a) and only s(a). We know that q is also true of s(a), but we do not know whether it is true of anything else.
¬p(a) p(s(a)) ∀x:¬p(s(s(x))) q(s(a)) Prove ∀x.(p(x) ⇒ q(x)). Hint: Break the problem into two parts - first prove the result for s(x), and then use that intermediate conclusion to prove the overall result."
I know I have to reach AX:(p(s(X)) => q(s(X))) and p(a) => q(a) first, but I'm struggling getting started. I also don't know if my first steps are correct. Any help? Thank you.
The conclusion is not provable from your premises ... here is a counterexample:
Suppose you have $4$ objects in the Domain: $a$, $b$, $c$, and $d$.
Suppose $s(a)=b$, $s(b)=s(c)=s(d)=c$
Suppose $b$ and $d$ have property $p$, and $a$ and $c$ do not.
Suppose $b$ is the only object with property $q$.
With this, we clearly have $\neg p(a)$, $p(s(a))$ (because $p(b)$), $q(s(a))$ (because $q(b)$), and $\forall x \neg s(s(x))$ (because $s(s(x))=c$ for all objects $x$ in the domain, and we have $\neg q(c)$
However, we don't have $\forall x (p(x) \to q(x))$, because $d$ has property $p$, but not property $q$.
OK ... so what is going on? Well, as the counterexample shows, your first three premises do not rule out something other than $a$ to have property $p$. If you want $s(a)$ to be the one and only object to have property $p$, you 'll need something like $\forall x (p(x) \leftrightarrow x = s(a))$
Of course, with $\forall x (p(x) \leftrightarrow x = s(a))$ as a premise .. the conclusion becomes trivial to prove ... no 'two step' process needed .... so now I am confused ... did you come up with those premises, or were these premises given to you? But as demonstrated, with those premises this is impossible to prove ....