LARA

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Complete Recursive Axiomatizations

Theorem: Let a set of formulas $Ax$ be a recursive axiomatization for a complete and consistent theory, that is:

  • $Ax$ is recursive: there exists a an algorithm for checking, given $F$, whether $F \in Ax$
  • $Conseq(Ax)$ is complete: for each FOL sentence $F$, either $F \in Conseq(Ax)$ or $(\lnot F) \in Conseq(Ax)$

Then there exists an algorithm for checking, given $F$, whether $F \in Conseq(Ax)$.

In other words, if a complete theory has a recursive axiomatization, then this theory is decidable.

(Note: a finite axiomatization is recursive. Typical axiomatizations that use “axiom schemas” are also recursive.)

Conversely: if a theory is undecidable (there is no algorithm for deciding whether a sentence is true or false), then the theory does not have a recursive axiomatization.

Proof. Suppose $Ax$ is a complete recursive axiomatization. There are two cases, depending on whether $Ax$ is consistent.

Case 1): The set $Ax$ is inconsistent, that is, there are not models for $Ax$. Then $Conseq(Ax)$ is the set of all first-order sentences and there is a trivial algorithm for checking whether $F \in Conseq(Ax)$: always return true.

Case 2): The set $Ax$ is consistent. Given $F$, to check whether $F \in Conseq(Ax)$ we run in parallel two complete theorem provers (which exist by Herbrand theorem), the first one trying to prove the formula $F$, the second one trying to prove the formula $\lnot F$; the procedure terminates if any of these theorem provers succeed. We show that this is an algorithm that decides $F \in Conseq(Ax)$.

  • Because either $F \in Conseq(Ax)$ or $(\lnot F) \in Conseq(Ax)$ and theorem provers are complete one of these theorem provers will eventually halt. The procedure is therefore an algorithm.
  • If $F$ is proved, then by soundness of theorem prover, $F \in Conseq(Ax)$.
  • If $(\lnot F)$ is proved, then by soundness of theorem prover $(\lnot F) \in Conseq(Ax)$. Because $Ax$ is consistent, there is a model $I$ for $Ax$. Then $(\lnot F)$ is true in $I$, so $F$ is false in $I$. Because there is a model where $F$ does not hold, we have $F \notin Conseq(Ax)$.

End of Proof.

Example: the theory of integers with multiplication and quantifiers is undecidable

  • consequently, there are no complete axiomatizations for it, no decidable set of axioms from which the truth value of facts about natural numbers follows
  • this result is one part of Goedel's incompleteness theorem

References