I am trying to prove the following claim:
let $sk(\alpha)$ be the sentence received from the skolemization of a given sentence $\alpha$.
Prove : $\vDash sk(\alpha) \implies \vDash \alpha$
I tried to prove an equivalent claim : $\nvDash a \implies \nvDash sk(\alpha)$
but I had one problem in the proof of this claim : can I say the following :
Lets assume $\alpha$ is not valid but it is satisfied by some structure $\\M$. then from the definition of skolemization we know that $sk(\alpha)$ is satisfied by some other structure M'.
In addition , there exists another structure $\hat{M}$ s.t $\hat{M}\nvDash \alpha$ which implies that $\hat{M} \vDash \neg{\alpha}$ . Therefore , $sk(\neg\alpha)$ is satisfied by some structure $M_1$. Does it say that $M_1$ satisfy $\neg sk(\alpha)$?
If it does, how can i prove it?
Thanks.
NO; saying that $\nvDash \alpha$ means that there is at least one structure $\mathcal M$ such that $\mathcal M \nvDash \alpha$.
Hint
Assume that $\alpha$ is $\forall x \exists y \phi(x,y)$; thus its "skolemization" must be : $sk(\alpha) := \forall x \phi(x,f(x))$.
From $\nvDash \alpha$ we have that, for some $\mathcal M$, with domain $M=|\mathcal M|$, we have some $a \in M$ such that $\mathcal M \nvDash \exists y \phi(x,y)[a]$ [intuitively, if $\forall x\psi(x)$ is not true in $\mathcal M$, we have that, for some "value" $a$ of $x$, $\psi$ does not hold of $a$].
But $\mathcal M \nvDash \exists y \phi(x,y)[a]$ means that for every $b \in M$ : $\mathcal M \nvDash \phi(x,y)[a,b]$.
This means that we have "no way" to define a function $f^M : M \to M$ such that : $f^M(a)$ satisfy in $\mathcal M$ : $\phi(x,y)[a,f^M(a)]$, and this implies that $\mathcal M \nvDash \forall x \phi(x,f(x))$.
Thus, having found a structure $\mathcal M$ such that $\mathcal M \nvDash \forall x \phi(x,f(x))$, we conclude with :