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I have always thought that according to Gödel's incompleteness problems, every inconsistent theory would be decidable. This is indicated, here for example: https://en.wikipedia.org/wiki/Decidability_(logic)

There are several basic results about decidability of theories. Every inconsistent theory is decidable, as every formula in the signature of the theory will be a logical consequence of, and thus a member of, the theory.

But I was having a conversation with a mathematician and he told me quite the contrary.

We were having a discussion and I mentioned that inconsistent theories would be decidable. He said:

Undecidability requires first-order logic, and that's it. No Paraconsistent logic, Trivialism, or even Deviant logic. Just plain old first-order logic.
I think you are sort of kind of in the right neighborhood, but there are a few technical issues we need as a prerequisite to continue talking. For instance, you say that "decidability implies inconsistency". Actually, decidability implies the exact opposite, it implies consistency

I thought that a truly inconsistent theory would be completely decidable, as everything would be provable. But now I am doubting...

But is this right? Am I completely wrong? Can't there be inconsistent and decidable theories? And can there be inconsistent and undecidable theories?

jerard
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  • Pure first-order logic with no function or relation symbols is both decidable and consistent. But once you allow either a binary function or a trinary relation, then it's undecidable (for example because then it's powerful enough to express any instance of the group word problem). – Daniel Schepler Jun 19 '19 at 22:58
  • @DanielSchepler: A binary relation symbol (which we could choose to call $\in$...) ought to be enough. – hmakholm left over Monica Jun 19 '19 at 22:59

2 Answers2

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It looks like your mathematician got carried away a bit at the end, because of course decidability does not imply consistency. As you point out, an inconsistent theory is decidable because it proves everything. (If you want to refer to Gödel for that, it's the completeness theorem you want, though). And inconsistent theories certainly exist.

On the other hand, we can have a consistent theory that is decidable such as in first-order logic with equality but no other non-logical symbol the theory with the single axiom $$ \forall x\forall y(x=y) $$

It is consistent because it has a model (with a single object); it is decidable because this is the only model and it is easy to evaluate any wff to a truth value in that model.

We can also have a consistent theory that is undecidable -- at least we hope PA and ZFC are such theories.

So the only nontrivial implication among these properties and their negation is yours: inconsistency implies decidability (or contraposed: undecidability implies consistency).

hmakholm left over Monica
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    +1: nice answer: In addition to PA and ZFC (where consistency may be a source of concern to some people), we also have examples like the theory of groups which must be consistent (because we can exhibit finite models) and are provably undecidable. – Rob Arthan Jun 19 '19 at 23:20
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    @HenningMakholm thank you for your answer. What about inconsistent and undecidable models? Are there any? – jerard Jun 19 '19 at 23:51
  • @jerard HenningMakholm explicitly stated that inconsistency implies decidability. And it's the logic that is inconsistent/decidable, not the models. The proof of this implication is straightforward. If a logic is inconsistent then an effective method to decide theoremhood is to return true for everything since everything is a theorem in that logic. – Derek Elkins left SE Jun 20 '19 at 00:06
  • @DerekElkins I was asking whether there can be models that are both logically inconsistent and undecidable because of this: (link.springer.com/chapter/10.1007%2F978-3-540-92687-0_30). This says "We study computational aspects of a probabilistic logic based on a well-known model of induction by Valiant. We prove that for this paraconsistent logic the set of valid formulas is undecidable." Because of this I thought that there could be models that are both inconsistent and undecidable – jerard Jun 20 '19 at 01:02
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    @jerard If you consider a paraconsistent logic, then you need to be clearer about what you mean by "inconsistent". Generally, if you don't specify, people will assume you intend classical logic. Gödel's theorems are talking about classical first-order logic. If we define "inconsistency" as meaning $\vdash A$ and $\vdash\neg A$ for some (closed) formula $A$, then it is possible to have paraconsistent logics where there is still some $B$ such that $\nvdash B$ so determining theoremhood is at least not trivial. You can probably formulate an example where it is undecidable. – Derek Elkins left SE Jun 20 '19 at 01:14
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You're right that every inconsistent theory is decidable. In fact, here is the decision procedure for any inconsistent theory:

Input: [any statement]

Output: "Yes, that's a theorem!"

So, inconsistency implies decidablity

But this is the converse of what your teacher claims you are saying:

For instance, you say that "decidability implies inconsistency".

??Huh? You are not saying that at all

and when your teacher says:

Actually, decidability implies the exact opposite, it implies consistency

Well, that's just plain wrong, for that would mean that inconsistency implies undecidability and, as we just saw, inconsistency instead implies decidability

Bram28
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