Diaconescu's theorem

In mathematical logic, Diaconescu's theorem, or the Goodman–Myhill theorem, states that the full axiom of choice is sufficient to derive the law of the excluded middle or restricted forms of it.

The theorem was discovered in 1975 by Radu Diaconescu and later by Goodman and Myhill. Already in 1967, Errett Bishop posed the theorem as an exercise (Problem 2 on page 58 in Foundations of constructive analysis).

Proof

The theorem is a forgone conclusion over classical logic, where law of the excluded middle is assumed, so the proof here is instead in the context of a constructive set theory. A core ingredient is the axiom of extensionality as well as an axiom of comprehension that allows for comprehension involving some proposition $P$.

Using comprehension, one considers two subsets of a set with two distinguishable members, such as $\{0,1\}$ with $0\neq 1$:

$U=\{x\in \{0,1\}\mid (x=0)\vee P\}$

and

$V=\{x\in \{0,1\}\mid (x=1)\lor P\}.$

If $P$ can be proven, then both of these sets equal $\{0,1\}$. In particular, $P\,\to \,U=V$ by extensionality. In turn, for any mathematical function $g$ that can take both of these sets as an argument, one finds $P\,\to \,g(U)=g(V)$, the contrapositive of which is $g(U)\neq g(V)\,\to \,\neg P$.

A choice function $f$ for the set $\{U,V\}$ into their union $\{0,1\}$ by definition fulfills

$f(U)\in U\,\land \,f(V)\in V.$

Using the definition of the two subsets and the function's established codomain, this reduces to

${\big (}f(U)=0\lor P{\big )}\land {\big (}f(V)=1\lor P{\big )}.$

Using the law of distributivity, this in turn implies $P\lor {\big (}f(U)\neq f(V){\big )}$. By the previous comment on functions, the existence of a choice function on this set thus implies (indeed is equivalent to) the disjunction $P\lor \neg P$.

Both $U$ and $V$ are inhabited, as witnessed by $0\in U$ and $1\in V$. So the axiom of choice, in particular granting a choice function on all sets of this form, implies excluded middle for all propositions.

Discussion

As noted, $P$ implies that both defined sets equal $\{0,1\}$. In that case, the pair $\{U,V\}$ equals the singleton set $\{\{0,1\}\}$ and there are two possible choice functions on that domain, picking either $0$ or $1$. If, instead, $P$ can be rejected, i.e. if $\neg P$ holds, then $U=\{0\}$ and $V=\{1\}$. So in that case $U\neq V$, and on the proper pair $\{\{0\},\{1\}\}$ there is only one possible choice function, picking the unique inhabitant of each singleton set. This last assignment "$U\mapsto 0$ and $V\mapsto 1$" is not viable if $P$ holds, as then the two inputs are actually the same. Similarly, the former two assignments are not viable if $\neg P$ holds, as then the two inputs share no common member. What can be said is that if a choice function exists at all, then there exists a choice function choosing $0$ from $U$, and one (possibly the same function) choosing $1$ from $V$.

For bivalent semantics, the above three explicit candidates are all the possible choice assignments.

One may define certain sets in terms of the proposition $P$ and, using excluded middle, in classical set theory prove that these sets constitute choice functions $f\colon \{U,V\}\to \{0,1\}$. Such a set represents an assignment conditioned on whether or not $P$ holds. If $P$ can be decided true or false, then such a set explicitly simplifies to one of the above three candidates.

But, in any case, neither $P$ nor $\neg P$ can necessarily be established. Indeed, they may even be independent of the theory at hand. Since the former two explicit candidates are each incompatible with the third, it is generally not possible to identify both of the choice function's return values, $f(U)$ and $f(V)$, among the terms $0$ and $1$. So it is not a function in the sense of the word that it could be explicitly evaluated into its codomain of distinguishable values.

Finitude

In a theory not assuming the disjunction $P\lor \neg P$ or any principle implying it, it cannot even be proven that a disjunction of the set equality statements above must be the case. In fact, constructively the two sets $U$ and $V$ are not even provably finite, in the usual sense of there existing a bijection with a natural number. (However, any finite ordinal injects into any Dedekind-infinite set and so a subset of a finite ordinal does validate the logically negative notion of Dedekind-finiteness. This is the case for both $U$ and $V$, into which $\{0,1,2\}$ cannot be injected. As an aside, it is consistent also with classical ${\mathsf {ZF}}$ that there exist sets which are neither Dedekind-infinite nor finite.)

In turn, the pairing $\{U,V\}$ is also elusive. It is in the surjective image of the domain $\{0,1\}$, but it is not known how explicit value-assignments for both $U$ and $V$ can be made, or even how many different assignments would have to be specified. So there is generally no (set) definition such that a constructive theory would prove that joint assignment (set) to be a choice function with domain $\{U,V\}$. Note that such a situation does not arise with the domain of choice functions granted by the weaker principles of countable and dependent choice, since in these cases the domain is always just $\mathbb {N}$, the trivially countable first infinite cardinal.

Adopting the full axiom of choice or classical logic formally implies that the cardinality of $\{U,V\}$ is either $1$ or $2$, which in turn implies that it is finite. But a postulate such as this mere function existence axiom still does not resolve the question what exact cardinality this domain has, nor does it determine the cardinality of the set of that functions possible output values.

Role of separation

In summary, functions are related to equality (by the definition of unique existence used in functionality), equality is related to membership (directly through the axiom of extensionality and also through the formalization of choice in sets) and membership is related to predicates (through an axiom of separation). Using the disjunctive syllogism, the statement $P$ ends up equivalent to the extensional equality of the two sets. And the excluded middle statement for it is equivalent to the existence of some choice function on $\{U,V\}$. Both goes through whenever $P$ can be used in a set separation principle.

In theories with only restricted forms of separation, the types of propositions $P$ for which excluded middle is implied by choice is also restricted. In particular, in the axiom schema of predicative separation only sentences with set bound quantifiers may be used. This restricted form of excluded middle is still not acceptable constructively. For example, when arithmetic has a model (when, relevantly, the infinite collection of natural numbers form a set one may quantify over), then set-bounded but undecidable propositions can be expressed.

Other frameworks

In constructive type theory, or in Heyting arithmetic extended with finite types, there is typically no separation principle at all - subsets of a type are given different treatments. A form of the axiom of choice is a theorem, yet excluded middle is not.

Topoi

In Diaconescu's original paper, the theorem is presented in terms of topos models of constructive set theory.