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.
    – Dennis
    Jun 22, 2013 at 22:05

1 Answer 1


Answer to Question #1

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).

Answer to Question #2

Well, many interesting paraconsistent theories won't be consistent, so those theories certainly shouldn't be able to prove their own consistency--- that would be bad. I'm not quite sure what else you had in mind, but it is interesting to note that Tarski's corollary of Gödel's results--- Tarski's Undefinability Theorem ---is no longer much threat. If you look at the linked Shapiro article (in "Further Reading") you'll see that the theory he develops is a paraconsistent arithmetic (a dialetheist arithmetic to be more precise; I suspect that many of your questions about paraconsistent theories are really meant to be about dialetheist or otherwise inconsistent theories) which contains its own truth predicate. It can prove its own soundness and its own Gödel sentence.

Further Reading

SEP Article on Inconsistent Mathematics

IEP Article on Inconsistent Mathematics

Incompleteness and Inconsistency; by Stewart Shapiro

  • 1
    +1: nice answer, it clarified a few things for me. I'm not conceptually clear really about the differences between dialethism, paraconsistency and inconsistency. Its probably worth another question... Jun 23, 2013 at 1:50
  • @MoziburUllah roughly, paraconsistent just means that the theory rejects ex falso quodlibet and doesn't allow contradictions to "explode". Dialetheists accept true contradictions and typically go paraconsistent so as to not trivialize their theory. Inconsistency is simply the result of a theory containing a contadiction, but does not presume paraconsistency or dialetheism.
    – Dennis
    Jun 23, 2013 at 5:11

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