# What is the impact of paraconsistency on Gödel's theorem?

Russell's paradox forced a restriction of the natural abstraction principle (that every predicate determines a set) so that set theory could be consistent; the standard one being ZF.

However paraconsistency allows one to retain the natural abstraction principle by allowing a degree of inconsistency in the logic which allows a revival of naive set theory as a fully formal one. It has the positive advantage of proving the axiom of choice and disproving the continuum hypothesis.

Now Gödel's incompleteness theorem says that one cannot have a theory that is both complete and consistent. One must be given up. Usually this is completeness. But in view of paraconsistency, consistency can be given up.

1. Is it correct to say then that a paraconsistent theory will always be complete?

2. Since the theory is paraconsistent, Gödel's second theorem about not being able to prove the consistency of a theory loses its traction. (Or does it? Should one be able to prove paraconsistency?)

• I changed the link to an article on paraconsistent logic rather than inconsistent mathematics. My answer will contain a link to the previously linked article on inconsistent mathematics. Commented Jun 22, 2013 at 22:05

Paraconsistent mathematical theories will not always be complete. Depending on what the theory takes to be true, and the strength of its deductive system, there might well be unprovable truths. As I'm sure you are by now aware, all paraconsistent theories give up ex falso quodlibet (the rule that allows you to derive anything from a contradiction) as well as principles that entail it (like disjunctive syllogism: from "A or B" and "not-B" deduce "A"). This means that inconsistencies within these theories will not "explode" allowing the proof of any statement of the language. Thus, paraconsistency is no guarantee of completeness. Embracing inconsistency does, however, open the door to the possibility of a complete theory whose classical counterpart would be essentially incomplete. For a toy example, a paraconsistent theory which keeps ex falso quodlibet (though, such a theory wouldn't really be paraconsistent anymore) as an admissible inference will be trivially complete (I imagine this is something like what you had in mind).