Undecidability and Recursive Inseparability

§1. Some Preliminary Theorems. we continue to let S be an arbitrary system, P be the set of Gödel numbers of the provable formulas of S and R be the set of Gödel numbers of the refutable formulas of S. Theorem 1. The set P̃* is not representable in S. Proof. This is the diagonal argument all over again. If H(v1) represents P̃* and h is the Gödel number of H(v1), the H[h̅] is provable in S iff h Ï p* iff d(h) ÏP iff H[h̅] is not provable in S, which is a contradiction. Theorem 1.1. If S is consistent, then P* is not definable in S. Proof. Suppose P* is definable in S. If S were consistent, then P* would be completely representable in S (cf. §3.1, Ch. 0). Hence P̃* would be representable in S, contrary to Theorem 1. Therefore, if S is consistent, then P* is not definable in S. Theorem 1.2. If the diagonal function d(x) is strongly definable in S and S is consistent, then P is not definable in S. Proof. Suppose d(x) is strongly definable in S. Since P* = d -1(P), then if P were definable in S, P* would be definable in S (by Th. 11.2, Ch. 0). Hence S would be inconsistent by Theorem 1.1. Exercise 1. Show that if S is consistent, then R* is not definable in S. Exercise 2. Show that if S is consistent, then no superset of R* disjoint from P* is definable in S, and no superset of P* disjoint from R* is definable in S. Exercise 3. Prove that if S is consistent and if the diagonal function is strongly definable in S, then no superset of P disjoint from R is definable in S. [This is stronger than Theorem 1.2.] §2. Undecidable Systems. A system S is said to be decidable (or to admit of a decision procedure) if the set P of Gödel numbers of the provable formulas of S is a recursive set. It is undecidable if P is not recursive. This meaning of ‘undecidable’ should not be confused with the meaning of ‘undecidable’ when applied to a particular formula (as being undecidable in a given system S).