# How do inconsistent arithmetics get around Gödel incompleteness?

Gödel theorems imply that arithmetic can not be complete under recursive axiomatization and consistency assumptions. Unexpectedly, consistency can be dropped in a meaningful way by switching to a paraconsistent logic that allows contradictions but blocks the law of explosion ("contradiction implies anything"). Such arithmetics can be complete, have the truth predicate definable within them, and admit a procedure for deciding truth/provability of sentences. But it gets better, "a consequence of Meyer's construction was that within his arithmetic R# it was demonstrable by simple finitary means that whatever contradictions there might happen to be, they could not adversely affect any numerical calculations". What puzzles me here is "finitary means". That presumably should be the same regardless of the logic used, yet R# proves consistency of numerical computations, which Peano arithmetic can not do (I think). What gives?

If I understand correctly, paraconsistent arithmetics divide arithmetical sentences into three categories: pure truths (provable and not disprovable), pure falsehoods (disprovable and not provable), and "Liar sentences" (both provable and disprovable). Gödel sentences come out as Liars, but at least in some arithmetics so do simple statements about very large numbers (what is "large" depends on a particular version). For instance, there might be a number $n$ that satifies both $n=n+1$ and $n\neq n+1$, and there is the smallest among them! But for smaller numbers they agree with Peano on everything. Is the trick that even if we prove a sentence we may not be able to tell if it is a pure truth or a Liar?

Are there "Peano complete" paraconsistent arithmetics, where every theorem of Peano arithmetic is a pure truth? Is it possible to have an arithmetic where the entire trichotomy is decidable, i.e. there is a procedure that decides whether a sentence is a pure truth, a pure falsehood or a Liar. I guess this would be a formalization of "completeness modulo Gödel sentences" but it sounds too good to be true. Can Gödel's argument be modified to show that there can be no such thing?

• You need to read the abstract of Priest's paper carefully. It states that "The models are finite, but also verify all the truths of the standard model". There must be a lot of paraconsistent snake-oil to justify this claim: a logic that admits finite models for the Peano axioms is not coherent with everyday mathematical thought. – Rob Arthan Nov 11 '15 at 22:20
• @Rob Arthan It only means that all Peano theorems and even all Gödel sentences are theorems in (axiomatizations of) these models. But they are not necessarily pure truths, i.e. for some of them their negations are also theorems. So these models are not "Peano complete". The reason they are finite is that all properties of standard numbers above the smallest inconsistent number (which is its own successor) are collapsed to that number. If this number is larger than anything physically realizable the difference should not affect any practically relevant parts of arithmetic. It seems to, why? – Conifold Nov 11 '15 at 23:26
• The notion of an "inconsistent number" is incoherent: if you talk about a number $N$ that is its own successor, you have already conceded the existence of a list of more than $N$ symbols. Physically realisability is irrelevant and also incoherent: where do I find 1, or 42 or 1001 in the physical world? – Rob Arthan Nov 12 '15 at 0:50
• @Rob Arthan Talking about objects or symbols does not require their existence, we also talk about set of all sets, but I thought that the point of paraconsistent logics is that they allow one to talk about such incoherent notions while avoiding trivialization. 1, or 42 or 1001 can be realized by physical objects, which are ultimately what we use arithmetic for, but presumably not something that exceeds the number of atoms, etc. in the universe. – Conifold Nov 12 '15 at 1:10
• The abstractions denoted by $1$ and $42$ and $1001$ cannot (in my opinion) be realised by physical objects: they are abstractions like truth, beauty and humour. We can use physical signs or objects to represent these abstractions, but physical objects cannot realise them: to prove me wrong, please kick 42 over my garden fence. – Rob Arthan Nov 12 '15 at 2:10

## 1 Answer

We use definition from the question:

pure truths (provable and not disprovable),

to answer the question

Are there "Peano complete" paraconsistent arithmetics, where every theorem of Peano arithmetic is a pure truth?

For every formula $\phi$, the longer formula $E_\phi \equiv \lnot (\phi \land \lnot \phi)$ is a theorem of PA. In any inconsistent system, there is at least one $\phi$ such that $E_\phi$ is disprovable. Therefore, it is not possible for an inconsistent theorem to prove every theorem of PA while not disproving any theorem of PA.

Thus any paraconsistent system in which every theorem of PA is a pure truth must actually be consistent.

• This argument seems to show generally that paraconsistent extensions of consistent theories that preserve all of their theorems must be consistent. I also realized that my second question has a simple answer too: if it is decidable for every sentence if it is a theorem, it is also decidable if its negation is, so the trichotomy into pure truths, pure falsehooda and Liars is decidable. This is indeed the case in inconsistent arithmetics with finite models, but the trade off is that they make too many sentences into Liars, not just Godelian loops, but all statements about large enough numbers. – Conifold Nov 18 '15 at 2:06