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I want to understand where am I wrong in the following reasoning.

It is known that $A(x)\vdash\forall x A(x)$ and that $ \forall x A (x) \vdash A (t) $ if $ t $ is free for $ x $ in $ A (x ) $, which implies that $ \forall x A (x) \vdash A (x) $. Moreover, there is an axiom $ \forall x A (x) \supset A (t) $ with the same condition on $ t $, which also implies $ \forall x A (x) \supset A (x) $, but at the same time as far as I understood, the converse, that $ A (x) \supset \forall x A (x) $ is not always true in predicate calculus, namely, it may not be true if x varies in the output of $ A (x) \vdash \forall x A (x) $. I do not understand why this is so, if, in fact, there is no difference between the statements $ A (x) $ and $ \forall x A (x) $ except that in the first statement $ x $ can be contained freely, but in the second it cannot. That is, these statements should be equivalent. What's the catch?

Thank you

Paul Ivanov
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  • See this post: the issue is that $Ax \to \forall x Ax$ is not valid. Thus, we have to "mix" axioms and rules (and the way they interact in the Deduction Theorem) in order to avoid that the formula is provable, because we want that the predicate calculus is sound. – Mauro ALLEGRANZA Jul 11 '21 at 13:08
  • I'm not sure if I've understood correctly. Could you confirm my conclusion? In Kleene's system A(x)⊢∀xA(x) is valid and A(x)→∀xA(x) is not always valid. (If we talk about reasons and how it sound I've got it) – Paul Ivanov Jul 11 '21 at 14:29
  • $Ax \vdash \forall x Ax$ is a "correct move" in system with Generalization rule of inference. If the rule is not available, what we have is: if $\vdash Ax$, then $\vdash \forall x Ax$. – Mauro ALLEGRANZA Jul 11 '21 at 15:33
  • You can see also this post – Mauro ALLEGRANZA Jul 11 '21 at 15:36
  • Ok, then let's consider system as having this rule in every next discussion. But, if system has this rule, then A(x)→∀xA(x) is also sound, because of the same reason (even if we can't deduce it) and in this case I don't see, why A(x) ="x=0" is counter example as it was described at the link you've sent. – Paul Ivanov Jul 11 '21 at 20:32
  • And I still don't understand what do we need quantifier ∀ for at all if semantically there is no difference between A(x) and ∀x A(x) – Paul Ivanov Jul 11 '21 at 20:41
  • It is confusing. Here is how I think about it. Basically, in a proof, if you introduce a new free variable $x$ with the assumption $P(x)$ and, making no other assumptions and do not introduce other free variables, you can derive $Q(x)$, then you can generalize $\forall x: [P(x) \to Q(x)]$. It simplifies matters if you don't introduce new free variables by universal specification. Yes, it is possible to deal with various complications, but this is the basic idea. – Dan Christensen Jul 11 '21 at 21:04
  • There is a difference, the same difference that we have when we assert "it is red" (when pointing with my finger to a red pen on the table) and asserting "everything on my table is red". – Mauro ALLEGRANZA Jul 12 '21 at 05:58
  • Well, I'm still little bit confused. According to the fact that it sounds illogical that x=0→∀x (x=0) we can say the same thing about x=0 ⊢∀x (x=0), but according to Kleene the last statement works. So it looks like totally logical formal system, that semantically sounds dissonantly. My concluding question is "where can I find the proof, that In common case A(x)→∀xA(x) is not provable according to Kleene's system"? Answer like contra example that sounds is now not an answer for me. For example how can we prove that x=0→∀x(x=0) is not provable and can we and how prove that not x=0→∀x(x=0) – Paul Ivanov Jul 12 '21 at 10:26

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Universal generalization states that-" In order to show a universal formula, which is to say a formula of the form for all v F[v], it is sufficient to show a substitution instance F[n] of F[v]. where n is any new name, which is to say that n does not appear anywhere earlier in the derivation.

Let v be any variable, let F[v] be any formula containing v, and let n be any name. Then a substitution instance of the formula F[v] is any formula F[n] obtained from F[v] by substituting occurrences of the name n for each and every occurrence of the variable v that is free in F[v]. Universal eliminatiomn states that- Let v be any variable, let F[v] be any formula containing v, and let n be any name. Then a substitution instance of the formula F[v] is any formula F[n] obtained from F[v] by substituting occurrences of the name n for each and every occurrence of the variable v that is free in F[v]