Show the following statement in PA
$\forall v_1\dotso\forall v_k\,(\exists v_0\,\varphi\to\exists v_0(\varphi\wedge\forall v_{k+1}<v_0\,\neg\varphi^{v_0}_{v_{k+1}}))$
With $v_0, v_1, \ldots, v_k$ are free variables in $\varphi$
Hello, I want to proof this in PA, but I do not know how to start. I already showed in PA, that the addition is commutative and associative.
Can you help me? Thanks in advance.
This has nothing to do with addition being commutative and associative. Except, perhaps, that both proofs use induction.
If you are familiar with the usual proof that induction is equivalent to the well-ordering principle of the natural numbers, this is just the internalization of this proof. If $\varphi$ defines a set, then if it is non-empty, it must have a least element.
If $0$ satisfies $\varphi$, then we are done, as $0$ is the minimum. Now show that if no $y\leq x$ satisfies $\varphi$, then either this means that $S(x)$ satisfies $\varphi$, or by the appropriate induction axiom no $x$ satisfies $\varphi$.