Craig's Theorem

Proof of Craig’s Theorem: Assume that S is a deductively closed set of sentences, whose elements may be recursively enumerated thus F(0), F(1), ..., F(n), ..., where F is a recursive function from natural numbers to sentences (we assume that expressions, sentences, etc., have been Gödel-coded in some manner). The set of theorems of an axiomatic theory is automatically recursively enumerable. But, in general, a recursively enumerable set is not automatically recursive. An example of a recursively enumerable set which is non-recursive is the set of logical truths in a firstorder language with a single dyadic predicate (this result is known as Church’s Theorem). However, by a trick devised by Craig, we can define a recursive set Craig(S) whose deductive closure is S. Let A be a sentence and n a natural number. Let A be the (n+1)-fold conjunction A ∧ .... ∧ A. The sentence A is logically interdeducible with A. Next, consider sentences of the form F(n). Define Craig(S) to be {F(n): n ∈ N}. The deductive closure of Craig(S) must be S, since each element of Craig(S) is equivalent to an element of S. Next, we give an informal decision procedure for membership in Craig(S). Given a sentence A, to decide whether A ∈ Craig(S), first check if A has the form B, for some sentence B and number n. By unique readability, this is checkable, and if A is not of this form, then A ∉ Craig(S). So, suppose that A is of the form B. We calculate F(n), and if B is indeed F(n), then A ∈ Craig(S). And otherwise, A ∉ Craig(S). The existence of a decision procedure for membership in Craig(S) implies that Craig(S) is recursive. The set Craig(S) is therefore a recursive axiomatization of the theory S.