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?
Does it mean that all such interpretations $\mathcal{I}$ of the LHS also have to model $\phi$ and $\{\neg \psi\}$?
So $\phi \cup \{\neg \psi\} \Leftrightarrow \phi \wedge \neg \psi$ ?
I read a few other related posts.
First one which I changed w.r.t. my proof:
Let $\phi$ be a set of propositional formulas and $\psi$ a propositional formula.
$\Rightarrow:$
Assume $\phi\models \psi$ and $\phi \cup \{\neg \psi\}$ satisfiable
Then there is a variable assignment s.t. $\forall \varphi:(\beta \models \varphi) \wedge (\beta \models \neg\psi)$
It follows that $\beta \nvDash \psi$. Therefore $\psi$ not a consequence of $\phi$.
$\Leftarrow$:
Assume $\psi \cup \{\neg \psi\}$ not satisfiable and $\beta$ any variable assignment s.t. $\beta \models \phi$.
To be shown $\beta \models \psi$.
For contradiction assume $\beta \nvDash \psi$.
Then $\beta \models \neg \psi$, whch implies $\beta \models \phi \cup \{\neg \psi\}$.
So $\phi \cup \{ \neg \psi\}$ satisfiable, which is a contradiction.
Is this correct?
Another proof with natural deduction.
$\frac{\psi \cup \{\phi\} \models \bot}{\phi \models \neg \psi}$
And
$\frac{\phi \models \psi}{\phi \cup \{ \neg \psi\}\models \bot}$
$\Rightarrow:$
If $\phi \models \psi$ then $\phi \cup \{\neg \psi\} \models \psi$.
But $\phi \cup \{\neg \psi\} \models \{ \neg \psi\}$.
Can't be satisfied, because it proves contradiction.
$\Leftarrow:$
If $\phi \cup \{ \neg \psi\} \models \phi$ can't be satisfied, then it has no models and so $\phi \cup \{ \neg \psi\} \models \bot$
By usage of semantic completeness theorem we get $\phi \cup \{ \neg \psi \} \models \bot$.
Which by usage of natural deduction results in $\phi \models \psi$.
In the deductions I changed $\vdash$ to $\models$, is this still correct?
Best Answer
We have to be careful with symbols. $a\models b$ could mean one of two things:
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.
$\phi\wedge\neg\psi$ makese no sense, unless $\phi$ is a single formula instead of a (potentially infinite) set of formulas.
"$\forall, \wedge$" 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 $\varphi\in\phi$?
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.