Counterexample in propositional logic

Question

There is this lemma: Let $\Sigma\subset \textrm{Prop}(A)$ and $p, q \in \textrm{Prop}(A)$. Then $\Sigma\models p \implies \Sigma\models p\vee q$. I can't figure out a counterexample for the opposite implication ($\textrm{Prop(A)}$ denotes the set of propositions and $A$ is a set of propositional atoms.

Thanks for help.

-pizet

Answer 1

$${\bf \Sigma \models p \lor q \overset{?}\implies \Sigma \models p }\tag{${\bf converse}$}$$

What if $\;p\;$ is false and $\;q\;$ is true?: Suppose, e.g., $$\bf \text{ Suppose}\;\;\; q \;= \;\lnot p$$

$\quad$ Then your stated lemma: $\;\Sigma\models p \implies \Sigma\models p\vee q\;$ certainly holds.

But its converse (highlighted) certainly fails to hold.

Answer 2

Let $q=\lnot p$.${}{}{}{}{}{}{}{}$