Assume $L$ is countable and list all propositions as $L = \{t_1, t_2, t_3, \ldots\}$. **Step 1: Build a maximal consistent extension.** Set $S_0 = S$. Given a consistent $S_n$, apply the Consistent Extension Lemma: at least one of $S_n \cup \{t_{n+1}\}$ and $S_n \cup \{\neg t_{n+1}\}$ is consistent; choose one and call it $S_{n+1}$. Define \begin{align*} \bar{S} = \bigcup_{n=0}^{\infty} S_n. \end{align*} **Step 2: $\bar{S}$ is consistent.** If $\bar{S} \vdash \bot$, then since proofs are finite, some finite subset of $\bar{S}$ suffices. That finite subset is contained in some $S_N$ (as each step only adds finitely many propositions), so $S_N \vdash \bot$, contradicting the consistency of each $S_n$. **Step 3: $\bar{S}$ decides every proposition.** For each $t_k \in L$, by construction either $t_k \in S_k$ or $\neg t_k \in S_k$; in either case the chosen element belongs to $\bar{S}$. **Step 4: $\bar{S}$ is deductively closed.** If $\bar{S} \vdash p$ and $p \notin \bar{S}$, then $\neg p \in \bar{S}$. But then $\bar{S} \vdash p$ and $\bar{S} \vdash \neg p$, giving $\bar{S} \vdash \bot$, contradicting Step 2. **Step 5: Define a valuation.** Set \begin{align*} v(p) = \begin{cases} 1 & \text{if } p \in \bar{S}, \\ 0 & \text{if } p \notin \bar{S}. \end{cases} \end{align*} We check this is a valuation. First, $v(\bot) = 0$ since $\bar{S}$ is consistent, so $\bot \notin \bar{S}$. For $v(p \Rightarrow q)$: we check the three cases. - If $v(p) = 1$ and $v(q) = 0$: then $p \in \bar{S}$ and $q \notin \bar{S}$. If $p \Rightarrow q \in \bar{S}$, then by modus ponens $\bar{S} \vdash q$, and by deductive closure $q \in \bar{S}$, a contradiction. So $p \Rightarrow q \notin \bar{S}$ and $v(p \Rightarrow q) = 0$. Correct. - If $v(q) = 1$: then $q \in \bar{S}$. By (A1), $\vdash q \Rightarrow (p \Rightarrow q)$, so $\bar{S} \vdash p \Rightarrow q$, and by deductive closure $p \Rightarrow q \in \bar{S}$, giving $v(p \Rightarrow q) = 1$. Correct. - If $v(p) = 0$: then $p \notin \bar{S}$, so $\neg p \in \bar{S}$, i.e., $(p \Rightarrow \bot) \in \bar{S}$. We need $p \Rightarrow q \in \bar{S}$. By the Deduction Theorem, it suffices to derive $q$ from $\{p, \neg p\}$. Since $\{p, \neg p\} \vdash \bot$ (by modus ponens from $p$ and $(p \Rightarrow \bot)$), we apply (A1) to obtain $\vdash \bot \Rightarrow (\neg\neg q)$, so $\{p, \neg p\} \vdash \neg\neg q$. Then (A3) gives $\neg\neg q \Rightarrow q$, and modus ponens yields $\{p, \neg p\} \vdash q$. Hence $p \Rightarrow q \in \bar{S}$ and $v(p \Rightarrow q) = 1$. Correct. Thus $v$ is a valuation satisfying $v(s) = 1$ for all $s \in S_0 = S$.