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Russells 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 fully formal one. It has the positive advantage of proving Choice and disproving the Continuum Hypothesis.

Now Godels 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.

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

b. Since the theory is paraconsistent, then Godels 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?)

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    $\begingroup$ A theory such as PA, which is consistent under normal inference rules, is also consistent under any smaller, paraconsistent set of inference rules, and so again it cannot be complete. $\endgroup$ Jun 21, 2013 at 11:40
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    $\begingroup$ @Mozibur: Which paraconsistent set theory disproves CH? Could you provide a sketch of the proof? $\endgroup$ Jul 7, 2013 at 6:07
  • $\begingroup$ @ThomasBenjamin The reference seems to be Transfinite numbers in paraconsistent set theory by Zach Weber, in the Review of Symbolic Logic. $\endgroup$ Feb 8, 2021 at 15:03
  • $\begingroup$ @TimothyChow; Just looked at the abstract--I believe it is. Thanks. $\endgroup$ Mar 9, 2021 at 21:41

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Gödel’s first incompleteness theorem has not really much to do with the background logic, but with computability. One way to formulate it is that there is no formal system $T$ such that

  1. The provable statements of $T$ are recursively enumerable.

  2. $T$ can express basic arithmetic to the extent that one can formulate in the system the statement that a particular Diophantine equation has no solution.

  3. $T$ is complete in the sense that for any Diophantine equation $f(x_1,\dots,x_n)=0$, this equation has no solution if and only if $T$ proves this fact.

While there are other ways to state the incompleteness theorem that may or may not apply to systems using unconventional logic, the formulation above applies quite universally, and it is faithful to the intended interpretation that no axiomatic system can completely capture all mathematical truths.

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  • $\begingroup$ Do you have references to literature where these matters are explained more fully? $\endgroup$ Nov 3, 2015 at 21:46
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The answer to your question „a. Is it correct to say then that a paraconsistent theory will always be complete?” is in general negative.

As an example, we could take a previous version of Hewitt's Direct Logic. Hewitt himself proved that theories of Direct Logic are incomplete.

The above proof applied a self-refutation rule (a concept proposed by Hewitt). However, as shown by Kao, self-refutation leads to explosiveness. Now, Hewitt is pursuing a different version of his Direct Logic without self-refutation and claims that it is not “fundamentally ‘incomplete’”.

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