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Consider the propositional language $L$ with denumerably many sentence letters $S_1,S_2,S_3,\ldots$ and the two connectives $\lnot,\lor$. Suppose that the set of sentences $\Gamma$ is a formal theory in $L$ and that for each sentence letter $S_i$, either $S_i\in\Gamma$ or $\lnot S_i\in\Gamma$. Prove by induction on structural complexity that $\Gamma$ is complete.

I'm familiar with how proof by induction on structural complexity works, but I'm unsure how to apply it in this context so I'm having trouble getting started.

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Let $\varphi$ be some sentence in the language you just described. We prove by induction on the complexity of $\varphi$ that either $\Gamma \vdash \varphi$ or $\Gamma \vdash \neg \varphi$.

The base case is just atomic sentences. That is, $\varphi$ is of the form $S_i$ for some $i$ (or maybe $\top$ or $\bot$, depending on your definitions). So this case is covered by the assumption that either $S_i \in \Gamma$ or $\neg S_i \in \Gamma$, which translates into $\Gamma \vdash S_i$ or $\Gamma \vdash \neg S_i$.

If you want to try the rest of the induction for yourself, you may want to consider to stop reading here.

There are two induction steps: one where $\varphi$ is of the form $\neg \psi$ and one where $\varphi$ is of the form $\psi_1 \lor \psi_2$. Let us consider each case.

If $\varphi$ is of the form $\neg \psi$, then by induction hypothesis we have $\Gamma \vdash \psi$ or $\Gamma \vdash \neg \psi$. In the first case we thus have $\Gamma \vdash \neg \varphi$, and in the second case we find $\Gamma \vdash \varphi$. So again, either $\Gamma \vdash \varphi$ or $\Gamma \vdash \neg \varphi$.

If $\varphi$ is of the form $\psi_1 \lor \psi_2$, we can again use the induction hypothesis to find a two different cases. The first case is that $\Gamma \vdash \psi_1$ or $\Gamma \vdash \psi_2$ (or both), which gives us $\Gamma \vdash \varphi$. The other case is that $\Gamma \vdash \neg \psi_1$ and $\Gamma \vdash \neg \psi_2$, which means that $\Gamma \vdash \neg \varphi$. Once more concluding that indeed $\Gamma \vdash \varphi$ or $\Gamma \vdash \neg \varphi$.

This completes the induction, and so we can conclude that $\Gamma$ is complete.

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