Diaconescu-Goodman-Myhill Theorem

Theorem
The axiom of choice implies the law of excluded middle.

Proof
Let $\mathbb{B} = \{ 0, 1 \}$.

Now, for any proposition $p$, define the following two sets:
 * $A = \{ x \in \mathbb{B} \, \vert \, x = 0 \vee p \} $
 * $B = \{ x \in \mathbb{B} \, \vert \, x = 1 \vee p \} $

Then the set $X = \{A, B\}$ is a (finite) family of inhabited sets $\left( \varnothing \notin X \text{ because } 0 \in A \text{ and } 1 \in B \right)$.

Via the axiom of choice, we now get a choice function $f : X \to \mathbb{B}$ since $\left( \bigcup X = \mathbb{B} \right)$.

Now, there are only four possible cases to consider:


 * 1) $f(A) = f(B) = 0$. This means that $0 \in B$ must hold. But for that to happen, $(0 = 1) \vee p$ must be true. So $p$ holds (since $0 = 1$ is false).
 * 2) $f(A) = f(B) = 1$. Symmetrically, this means $p$ holds as well (since $1 \in A$ must hold).
 * 3) $f(A) = 1 \ne f(B) = 0$. This means that $A \neq B$ (or otherwise f would pick the same element). But if $p$ is true, that implies that $A = B = \mathbb{B}$ (contradiction). Therefore, $\neg p$
 * 4) $f(A) = 0 \ne f(B) = 1$. Symmetrically, this means $\neg p$ holds as well.

So in all cases either $(p \vee \neg p)$ holds. That is the Law of Excluded Middle.