The proof of the Compactness Theorem of sentential logic (CTSL) was in two parts. First, a set $\Delta\supseteq\Gamma$ was constructed that for every wff would include either the wff or its negation. Second, a truth assignment was constructed that would satisfy all the wffs in $\Delta$ , and, hence, in $\Gamma$ . Without using the theorem itself, we could follow the logic of its proof here as well.

Step 1.$\Gamma\cup\Lambda$ is consistent. Indeed, by definition, $\Gamma\cup\Lambda\vdash\{\phi,\neg\phi\}$ iff $\Gamma\vdash\{\phi,\neg\phi\}$ . Note, that here we don’t use any facts about $\Lambda$ itself. Namely, if $\Lambda$ itself were inconsistent, it would simply mean that there does not exist any consistent set of wffs, but if we assume that a consistent set of wffs exists, this would automatically, by definition, imply that $\Lambda$ is consistent.

Step 2. We construct $\Delta$ . If we assume that the language is countable, we can effectively enumerate all formulas, $\phi_{1},\phi_{2},\ldots$ . Then, let $\Gamma_{0}=\Gamma\cup\Lambda$ , and for each $i\ge1$ , we construct $\Gamma_{i}=\Gamma_{i-1};\alpha$ , where $\alpha$ is either $\phi_{i}$ or $\neg\phi_{i}$ depending on which one keeps $\Gamma_{i}$ consistent (if both, we just add $\phi_{i}$ ). We then take $\Delta=\cup_{i\ge0}\Gamma_{i}$ .

Step 3. We need to verify that $\Delta$ is well-defined, i.e. that we can always construct $\Gamma_{i}$ . This would not be possible only if both $\Gamma_{i-1};\phi_{i}$ and $\Gamma_{i-1};\neg\phi_{i}$ were inconsistent, but then, by RAA, $\Gamma_{i-1}\vdash\neg\phi_{i}$ and $\Gamma_{i-1}\vdash\phi_{i}$ , implying that $\Gamma_{i-1}$ is inconsistent. So, by induction, we can show that, since $\Gamma_{0}=\Gamma\cup\Lambda$ is consistent, for every $i\ge1$ , $\Gamma_{i}$ can be constructed and $\Gamma_{i}$ is consistent (by its definition).

Step 4. We verify that $\Delta$ satisfies the conditions. It is clear that a) $\Gamma\subseteq\Gamma_{0}\subseteq\Delta$ , and b) for every wff $\alpha$ , exactly one of $\alpha\in\Delta$ and $\neg\alpha\in\Delta$ holds. Indeed, at least one of them should hold, as if $\alpha=\phi_{i}$ then $\Gamma_{i}$ , and, hence, $\Delta$ , contains at least one of $\alpha$ and $\neg\alpha$ , but they both cannot be in $\Delta$ , as this would imply that they are both in some finite $\Gamma_{i}$ contradicting its consistency. Similarly, we can argue that every finite subset of $\Delta$ is consistent.

We now only need to address consistency of $\Delta$ . We show that $\alpha\in\Delta$ iff $\Delta\vdash\alpha$ . From this, it follows that $\Delta$ is consistent (if $\Delta\vdash\{\gamma,\neg\gamma\}$ , then the finite inconsistent set $\{\gamma,\neg\gamma\}\subseteq\Delta$ ). If $\alpha\in\Delta$ then the deduction $<\alpha>$ from $\Delta$ proves $\alpha$ . Now, suppose that $\Delta\vdash\alpha$ . We show by induction that $\alpha\in\Delta$ . Indeed, if $\alpha\in\Delta$ or $\alpha\in\Lambda\subseteq\Delta$ then $\alpha\in\Delta$ . Now, by induction, if $\alpha$ is derived from $\beta$ and $\beta\rightarrow\alpha$ where $\beta,\beta\rightarrow\alpha\in\Delta$ , then, since $\{\neg\alpha,\beta,\beta\rightarrow\alpha\}$ is inconsistent ($<\beta,\beta\rightarrow\alpha,\alpha>$ and $<\neg\alpha>$ ), it cannot be a finite subset of $\Delta$ , but $\{\beta,\beta\rightarrow\alpha\}\subseteq\Delta$ , therefore, not $\neg\alpha$ but $\alpha$ is in $\Delta$ .