Question: Prove or disprove (in first order logic) given that a is a formula, $T$ is a set of formulae: $T \nvDash ^t a \iff$ there exists a structure s.t. that $T \cup \{\neg a\}$ is valid in that structure.
Tries: I've successfully proved the $\Leftarrow$ side of the proof, but I just feel that I don't have enough information in order to prove the other side, which leads me to search for a counterexample. I was wondering if the counterexample should s.t. $T$ is the empty set (so every structure $M$ will satisfy it) and a should be some contradiction?
Thanks in advance for your help.
For $\Rightarrow$ : it holds for sentences, i.e. closed formuale.
For a counter-example, consider that $T⊭a$, means that, for some structure $M$ and variable assignment $v$ :
We assume that a formula $a$ is valid in a structure $M$ if $M,v \vDash a$, for any $v$.
Thus consider as $T$ the Peano axioms for arithmetic and consider as $a$ the formula : $x=0$.
Then $T \nvDash (x=0)$ becuase with a $v$ such that $v(x)=1$ we have that $M,v \nvDash (x=0)$ for any structure $M$ that satisfies $T$.
But for such $M$, $T \cup \{ \lnot (x = 0) \}$ is not valid, because for $v'$ such that $v'(x)=0$ we have $M, v' \nvDash \lnot (x=0)$.