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For arbitrary (first-order) theories T and S (where T needn't be a theory of arithmetic), is it meaningful to say "T proves the consistency of S"?

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2 Answers 2

No. The theory $T$ has to be able to express the consistency of $S$, otherwise we don't have a formula to talk about its provability.

However the theory doesn't need to be arithmetic, but the language should be strong enough to be able to express this. Also we usually require that some basic facts about the consistency statement can be proven inside the theory. The amount of machinery required for this is studied by people like Pavel Pudlak. You can Google for "sequential theory" if you are interested in learning more about this or check Petr Hajek and Pavel Pudlack, "Metamathematics of First-Order Arithmetic" (a free online version is available from Euclid here).

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If the question is "is there a commonly accepted meaning" then the answer is no as Kaveh says. However, one could also ask if it is possible to give it meaning in general. This is a fuzzy question, so it's hard to say, but I can think of one case where the answer is arguably still "no".

To give meaning to the statement "$T$ proves the consistency of $S$" we would have to formulate $\text{Con}(S)$ as a sentence in the language of $T$, in such a way that seeing a proof of $T \vdash \text{Con}(S)$ would convince us that $S$ really is a consistent theory. In the simple case where $T$ has only finitely many models (which must themselves be finite) then all that a proof from $T$ can do is to convince us of something that can be established by "inspection" of the models. Whatever $S$ happens to be, there is a priori no upper bound on the length of the shortest proof of contradiction from $S$, so there is no way that this "inspection" can convince us of the consistency of $S$.

I'm sure that there are some theories with infinitely many models that are still "too simple" to formulate consistency statements, but I don't know how to argue this.

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