I want to prove $\phi \models \psi$ iff $\phi \cup \{\neg \psi\}$ unsatisfiable.
What's the meaning of the union operator w.r.t. satisfiability in FOL?
We have to be careful with symbols. $a\models b$ could mean one of two things:
- $a$ is a model, and the formula $b$ is true for $a$.
- $a$ is a set of formulas, and for every model $\mathcal M$ that satisfies $a$, it must satisfy $b$.
In your case, it seems you mean $\phi$ is either a formula or a set of formulas. If $\phi$ is a set of (well-formed) formulas that are understood as axioms, then $\cup$ is literally the union of the two sets, that is to add $\neg\psi$ as a new axiom.
Now the statement is almost trivial: if $\phi\models \psi$, i.e. for any model $\mathcal M$ that satisfies $\phi$, it must satisfy $\psi$, hence it cannot satisfy $\neg\psi$, so $\phi\cup\{\neg\psi\}$ cannot be satisfied. If $\mathcal M$ satisfies $\phi\cup\{\neg\psi\}$, it satifies $\phi$ but not $\psi$, which contradicts the meaning of $\phi\models\psi$.
This is trivial precisely because $\vDash$ is about semantics. Only "$\phi\vdash\psi$ iff $\phi\cup\{\neg\psi\}$ cannot be satisfied" needs the Godel completeness theorem. See the difference between $\vdash$ and $\vDash$.
There seem to be many notational problems. Just to name a few, but you probably should read entire sections of your study materials more carefully.
So $\phi \cup \{\neg \psi\} \Leftrightarrow \phi \wedge \neg \psi$ ?
$\phi\wedge\neg\psi$ makese no sense, unless $\phi$ is a single formula instead of a (potentially infinite) set of formulas.
$\forall \varphi:(\beta\models \varphi) \wedge (\beta \models \neg\psi)$.
"$\forall, \wedge$" are reserved symbols in the FOL, so better to avoid using them on the meta-level. If $\beta$ is actually a model/interpretation (together with assignment for free variables if necessary), it's correct to use $\vDash$ here, but this $\vDash$ has a different meaning from the one in $\phi\vDash\psi$. Also, are you suggesting $\varphi\in\phi$?
If $\phi \cup \{ \neg \psi\} \models \phi$ can't be satisfied, then it has no models and so $\phi \cup \{ \neg \psi\} \models \bot$
It makes no sense to say whether "$a\vDash b$" can be satisfied, which is either true or false objectively. Only a set of sentences can be satisfied or not.
Best Answer
I've found a counter example:
Let $\sigma=\{c,p\}$
Let $\psi=p(y)\vee\neg p(x)$
So:
$\vdash_{FOL}\exists x\forall y\ (p(y)\vee\neg p(x))$)
Because $\exists x\forall y\ (p(y)\vee\neg p(x)) \equiv (\forall y \ p(y))\vee (\neg \forall x \ p(x))$
But:
$\nvdash_{FOL}p(y)\vee\neg p(c)$
Because if we define:
$M=<\{0,1\},I>$ such that $I[p]=\{1\}\ ,I[c]=1$
and an assignment $v$ such that $v[y]=0$,
then $v[p(y)\vee\neg p(c)]=$f
and $c$ is the only closed term in the language