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I'm reading Roger Penrose's Shadow of the mind, in page-92, the following is said:

It is one of the essential properties of a formal system that there must indeed be an algorithmic (i.e, computational procedure $F$ for checking whether or not the rules of $\mathbb{F}$ have been correctly applied).

Long ago , I had asked if one could have an infinite axiom system, the answer I got was affirmative (ref). Intuitively speaking, it doesn't make sense that if we have a formal system with infinite axiom, we could possibly write a proof of infinite steps. If the steps are truly infinite, then how could a computer ever check all of them?

tryst with freedom
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  • That is the whole point. Every proof is finite. Thus, even if there are infinitely many axioms, only finitely many of them will be used in the proof, and the proof can be formally checked. –  May 14 '22 at 08:25

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I think you've got a bit confused. You can certainly have infinitely big axiom schemas, and that's what Noah's reply in your linked question affirmed. But e.g. in ZFC (which has infinitely many axioms) we still only consider finite-length proofs. This is no more problematic than the notion of a "finite list of natural numbers", of which there are also infinitely many.

Patrick Stevens
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  • ZFC has infinite axioms??? When I learned , the list was finite – tryst with freedom May 14 '22 at 08:26
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    No, you learned a finite list of axiom schemas. Some of those axiom schemas, such as Comprehension, are templates into which you can slot infinitely many formulae, thereby creating infinitely many axioms. – Patrick Stevens May 14 '22 at 08:27
  • Even Peano arithmetic is not finitely axiomatisable; the induction schema cannot be represented by finitely many axioms. – Patrick Stevens May 14 '22 at 08:29
  • Hmmmm I am a bit confused, I had learned the Peano Axioms as well in Tao's Analysis, and in there, it seemed so only 4 or so axioms were there and from that we derive all the numbers and properties of them – tryst with freedom May 14 '22 at 08:30
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    Once again (assuming first-order logic), you learned a finite bunch of axioms and a single axiom schema (induction) which packages up infinitely many axioms. "Whenever $\phi$ is a formula, the following is an axiom: $\phi(0) \wedge (\forall n, \phi(n) \to \phi(n+1)) \to \forall n, \phi(n)$" is not an axiom. – Patrick Stevens May 14 '22 at 08:31