[proofplan] Given an arbitrary proposition $P$, we construct two sets $A$ and $B$ whose equality depends on the truth value of $P$. The Axiom of Choice provides a choice function on $\{A, B\}$, and by analysing whether the chosen representatives agree, we decide $P$ or $\neg P$. [/proofplan] [step:Construct two sets whose equality encodes the proposition $P$] Let $P$ be an arbitrary proposition. Define \begin{align*} A &:= \{x \in \{0, 1\} : (x = 0) \lor P\}, \\ B &:= \{x \in \{0, 1\} : (x = 1) \lor P\}. \end{align*} We verify the key properties of $A$ and $B$: - $0 \in A$ (since $x = 0$ is satisfied) and $1 \in B$ (since $x = 1$ is satisfied). Therefore both $A$ and $B$ are nonempty. - If $P$ holds, then both defining conditions $(x = 0) \lor P$ and $(x = 1) \lor P$ are satisfied for every $x \in \{0, 1\}$, so $A = \{0, 1\} = B$. - We have $0 \in B$ if and only if $(0 = 1) \lor P$, which (since $0 \neq 1$) holds if and only if $P$. Similarly, $1 \in A$ if and only if $P$. So $A = B$ implies $0 \in B$ and $1 \in A$, which implies $P$. Therefore $A = B$ if and only if $P$. [guided] The construction is the heart of the argument. We need two sets that are equal precisely when $P$ holds. The idea: make each set "want" to contain both $0$ and $1$, but condition the extra element on $P$. The set $A$ always contains $0$ (the clause $x = 0$ guarantees this regardless of $P$). It contains $1$ only if $P$ holds (since $1 = 0$ is false, the only way $1 \in A$ is through the clause $P$). Symmetrically, $B$ always contains $1$ and contains $0$ only if $P$ holds. So without $P$: $A = \{0\}$ and $B = \{1\}$, distinct sets. With $P$: $A = \{0,1\} = B$. The logical equivalence $A = B \iff P$ is exact. This construction is valid in intuitionistic logic — we are using only the subset separation axiom (comprehension) applied to the decidable set $\{0,1\}$ with a formula involving $P$. [/guided] [/step] [step:Apply the Axiom of Choice to obtain a choice function on $\{A, B\}$] Consider the family $\mathcal{F} := \{A, B\}$. Each member of $\mathcal{F}$ is nonempty (since $0 \in A$ and $1 \in B$). By the Axiom of Choice, there exists a choice function \begin{align*} f: \mathcal{F} &\to A \cup B \\ S &\mapsto f(S) \in S. \end{align*} In particular, $f(A) \in A$ and $f(B) \in B$. Since $A, B \subset \{0, 1\}$, we have $f(A), f(B) \in \{0, 1\}$. [/step] [step:Decide $P \lor \neg P$ by comparing $f(A)$ and $f(B)$] We consider two cases based on whether $f(A) = f(B)$. **Case 1: $f(A) = f(B)$.** Call this common value $c \in \{0, 1\}$. Since $c = f(A) \in A$, the defining condition of $A$ gives $(c = 0) \lor P$. Since $c = f(B) \in B$, the defining condition of $B$ gives $(c = 1) \lor P$. Combining: \begin{align*} \bigl((c = 0) \lor P\bigr) \land \bigl((c = 1) \lor P\bigr). \end{align*} Distributing: $(c = 0 \land c = 1) \lor P$. Since $c \in \{0,1\}$ and $0 \neq 1$, the conjunction $c = 0 \land c = 1$ is false. Therefore $P$ holds. **Case 2: $f(A) \neq f(B)$.** Since $A = B$ implies $f(A) = f(B)$ (a function takes equal arguments to equal values), the contrapositive gives $A \neq B$. Since $A = B \iff P$ (established in the first step), we obtain $\neg P$. In both cases, we have established $P \lor \neg P$. Since $P$ was an arbitrary proposition, the law of excluded middle holds. [guided] The two cases are exhaustive because $f(A)$ and $f(B)$ are elements of $\{0,1\}$, a set with decidable equality (i.e., for any $a, b \in \{0,1\}$, either $a = b$ or $a \neq b$). This is a crucial subtlety: we are using the decidability of equality on $\{0,1\}$, which is valid even in intuitionistic logic. In Case 1, $f(A) = f(B)$ forces $P$ through a simple logical argument. The value $c$ cannot simultaneously equal $0$ and $1$, so at least one of the clauses in $(c = 0) \lor P$ and $(c = 1) \lor P$ must be satisfied through $P$ rather than through the equality. In fact, both disjunctions hold, and the only way both can hold when $c$ is a single value is if $P$ holds. In Case 2, $f(A) \neq f(B)$ tells us $A \neq B$ (since a function respects equality of its arguments). By contrapositive of $P \implies A = B$, we get $\neg P$. The role of the Axiom of Choice is essential: without it, we have no function $f$ to compare values of. The Axiom of Choice does not merely assert that $A$ and $B$ have elements (which we already know), but that there is a *uniform* selection — a single function choosing from both simultaneously. It is this uniformity that enables the comparison. [/guided] [/step]