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Providing concrete models is more-or-less impossible, since we are not sure about many things in real world. On the other side, abstract models are usually insufficient for proving consistency, since the consistency of the other axiomatic system (which we have constructed the model in) needs to be proved first and the problem arises again!

Another method is to count all theorems of the system and make sure that they do not contradict.

So I have two questions:

  1. Can an axiomatic system be proposed whose theorems can be counted completely? Do you know an example?
  2. Is there another method to prove absolute consistency of an axiomatic system without providing a model? If no, why not? And if yes, may you please provide an example?
Hamid Haghshenas
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  • Yes; see Gentzen's consistency proof and see Proof theory. – Mauro ALLEGRANZA Dec 06 '18 at 09:45
  • Thanks @MauroALLEGRANZA, but Gentzen's proof does not yet prove absolute consistency. Yes? And what about the first question? Can all theorems of a theory be counted to ensure its consistency? – Hamid Haghshenas Dec 08 '18 at 14:58
  • @HamidHaghshenas Nothing proves absolute consistency: even with a model "in hand," we need some ambient framework to verify that the structure is, in fact, a model (and to prove the soundness theorem, for that matter!). As to the first question, you're just asking about decidable theories, and there are lots of those - e.g. Presburger arithmetic. – Noah Schweber Dec 08 '18 at 18:56
  • @NoahSchweber Helpful, thanks. In a decidable theory, if the algorithm tells us that a proposition is provable but its negation is not, can it be said that we have a proof of consistency of the theory? – Hamid Haghshenas Dec 10 '18 at 06:14
  • @HamidHaghshenas Only if we also have a proof that the algorithm behaves correctly. But if so, yes. – Noah Schweber Dec 10 '18 at 14:53

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