 Solution

Suppose both extensions were inconsistent. By the explosion argument in (i), the extension by $\varphi$ proves $\neg \varphi$, and the extension by $\neg \varphi$ proves $\varphi$. The deduction theorem for propositional logic therefore yieldsThe identity theorem gives $⊢\neg \varphi \to \neg \varphi$ and $⊢\varphi \to \varphi$. Apply E to the first displayed logical implication and the identity logical implication with antecedent $\neg \varphi$ to obtain $\Gamma ⊢\neg \varphi$. Apply E to the identity logical implication with antecedent $\varphi$ and the second displayed logical implication to obtain $\Gamma ⊢\varphi$. This contradicts propositional consistency of $\Gamma$.

Consequently at least one of the two one-formula extensions is propositionally consistent. The formal proof uses the source's own calculus, including its rule E, rather than assuming an unproved completeness theorem for this presentation.