Let $a$ be a proposition (atomic or not), and let $F_a=\{c\mid a\vDash c\}$ is the set of all the propositions that are tautological consequence of $a$.
Prove that $a\equiv b \iff F_a=F_b$.
$\rightarrow$:
Let $c\in F_a$ such that $a\vDash c$, from assumption: $b\vDash c$ so $c\in F_b$ so $F_a\subseteq F_b$ and from symmetry: $F_a\supseteq F_b$ so $F_a=F_b$.
$\leftarrow$:
From $F_a=F_b$ we know that $\forall c\in F_a: a\vDash c \iff b\vDash c$.
Now I'm not sure what to do, we've seen that if the truth tables for two propositions are the same then the propositions are equivalent, will that be enough here? Since when $a,b$ are true, $c$ is true then they must have the same truth table, therefore $a\equiv b$ ?
Hint. Clearly $a\in F_a$, so if $F_a=F_b$ then $a\in F_b$, that is, $b\vDash a$. And by symmetry...