I wanted to ask you if the following claim is correct, assuming the principle of bivalence:
"If $B$ is not a tautology and $A \models B$, then $ \lnot A \not\models B$"
My reasoning is the following, indicating Models of $A, B$ with $Ma,Mb$ and with $Ca$ the countermodels of $A$.
if $A \models B$, then $Ma \subseteq Mb$; B is not a tautology, so there is a valutation in which it is false.
Supposing that $ \lnot A \models B$, then $Ca \subseteq Mb$.
Since $Ca \cup Ma$ are all the possibile valutations of the language, and are all models of $B$, $B$ must be a tautology. But this contradicts the assumption.
Is the proof correct?
