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By Godel's incompleteness theorems, a formula expressing consistency of a theory that can contain Peano arithmetic cannot be derived or contained from/in the theory.

Godel's completeness theorem states that if a formula $\phi$ is true in every model, then finite deduction for truth of $\phi$ can be made.

With that said, for some reasons, especially by the method of infinitely many deduction steps, we realize that the theory $T$ has to be consistent, which means that formula $\phi_1$ that expresses consistency of the theory $T$ is true. This would mean that every model will satisfy $\phi_1$. However, by Godel's completeness theorem, this would mean that there exists the method of finite deduction that shows $\phi_1$ is true, which would mean that Godel's incompleteness theorem cannot be maintained.

Of course this is obvious, as containing $\phi_1$ would lead to inconsistency as predicted by Godel's incompleteness theorem. My question is, then when theory $T$ can never be shown to be consistent regardless of whether deduction is finite or not, how can we be so sure to believe that $T$ is consistent? Is it because asking whether $\neg \phi_1$ is true is meaningless because if so, theory $T$ itself is completely messed up? (only some deduction/model example would prove that $T$ is inconsistent) Isn't there possibility that the whole argument above shows that $T$ is indeed inconsistent?

By the way, of course infinite steps of deduction is not currently possible. I am rather assuming "as if that is possible, what would happen?"

Finite deduction steps would refer to ordinary finite algorithm.

user56292
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    What's with the downvotes? This looks like a completely legitimate question... – 5xum Jul 30 '14 at 07:16
  • What does "With that said, for some reasons, especially by the method of infinite deductions, we realize that the theory T has to be consistent, which means that formula ϕ1 that expresses consistency of the theory T is true." suppose to mean? – William Jul 30 '14 at 07:19
  • See this post for explanation of the relationship between G's theorems. – Mauro ALLEGRANZA Jul 30 '14 at 07:20
  • @William what I am saying is let's say that we are now able to do more than what we did by deduction methods without adding more deduction rules. Hypothetical example would be infinite number of deductions. Then by those methods we somehow realize that $T$ is consistent and then.. – user56292 Jul 30 '14 at 07:21
  • I am not asking for explanation of relationship between different Godel theorems. I am rather trying to combine two, and by assuming current impossible deductions without adding new deduction rules - here, infinite deduction steps - what would happen to the theory $T$ if it turns out that $T$ can be shown to be consistent. – user56292 Jul 30 '14 at 07:25
  • Godel's Theorems talk about first-order logic. If you change your deduction rules, you're no longer doing first-order logic. You would have to reformulate the theorems and the notions of semantic vs. syntactical truth in your new logic (as has been done, for example, in intuitionistic logic) – Dorebell Jul 30 '14 at 07:27
  • I do get that first-order logic is about finite deduction and so on. However, that does not seem to mean truth/falsity of a statement is affected by whether finite deduction or infinite deduction is used. And there is no new deduction rule. So without adding a new deduction rule to first-order logic, suppose that for whatever reason, we known $\phi_1$ that expresses consistency of $T$ is known to be true. Then the rest that is in the question seems to follow. – user56292 Jul 30 '14 at 07:33
  • Or Godel's completeness theorem should be understood differently - but then what is the point of saying that "if $\phi_1$ is true in every model, then $\phi_1$ can be deduced by finite algorithm deduction"? Then the whole point of finite algorithm is meaningless... – user56292 Jul 30 '14 at 07:36
  • You can see Infinitary Logic. The system $L(ω_1,ω)$ admits deductions of countable length. It has a Completeness Theorem but the compactness theorem fails. Due to the fact that the weak second-order language is translatable into $L(ω_1,ω)$, it seems to me - but I need an expert - that we can have in it the "enough arithmetic" needed for the proof of G's Incompleteness Th. – Mauro ALLEGRANZA Jul 30 '14 at 07:45

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Suppose that :

we realize that the theory $T$ has to be consistent, which means that formula $\varphi_1$ that expresses consistency of the theory $T$ is true. This would mean that every model will satisfy $\varphi_1$.

It is not clear from the above discussion the "intended meaning" of : "the method of infinitely many deduction steps", unless you are referring to Infinitary Logic.

If we stay in the usual context of first-order logic, we know that Peano arithmetic is consistent. It has a model $\mathfrak N$ : the standard one, with domain $\mathbb N$.

Thus, a formula $\varphi_1$ expressing in the language of the theory the consistency of the theory itself is true in $\mathfrak N$.

But - and this is the result of the "interplay" between G's Incompleteness Th and G's Completeness Th - it is not true in all models.

I.e. there are some non-standard model in which the formula $\varphi_1$ is false.

Mauro ALLEGRANZA
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