 Solution

A propositional type is a set $\Sigma$ of propositional formulas; a Boolean valuation realizes it if every member is true, and omits it if at least one member is false. A consistent propositional theory $T$ locally omits $\Sigma$ if every formula $\theta$ which implies all members of $\Sigma$ modulo $T$ is refutable modulo $T$. Equivalently, whenever $T\cup \{\theta \}$ is consistent, there is a $\sigma \in \Sigma$ for which $T\cup \{\theta ,\neg \sigma \}$ is consistent. For a consistent propositional type, this says that it is a nonprincipal propositional type.

The extended omitting types theorem for propositional logic says that if a consistent $T$ locally omits each of a countable family ${\Sigma }_0,{\Sigma }_1,\dots$, there is one Boolean valuation satisfying $T$ and omitting them all. It also works inside any prescribed finite condition consistent with $T$.

To prove it, start with such a condition ${\theta }_0$, taking $\top$ when none is prescribed. At stage $i$, choose ${\sigma }_i\in {\Sigma }_i$ such thatSuch a choice exists by local omission. More explicitly, failure would imply $T⊢{\theta }_i\to \sigma$ for every $\sigma \in {\Sigma }_i$, hence $T⊢\neg {\theta }_i$, contradicting the stage invariant. Every finite subset of $T\cup \{{\theta }_0\}\cup \{\neg {\sigma }_i:i\in \mathbb{N}\}$ lies inside a consistent stage. By the propositional compactness theorem and the completeness theorem for propositional logic, it has a Boolean valuation. That Boolean valuation satisfies $T$ and makes the selected member ${\sigma }_i$ of every type false. All types are therefore omitted simultaneously. No effective test for consistency is assumed, and the argument does not require the ambient propositional language to be countable. Only the family of omission requirements is countable.