Comment by woopsn

If I had to write again I might say "same theorems about natural numbers" and capitalize ROUGHLY. It is a conversation, what exactly I am weaseling around (not just nonstandard model theoretic issues), and I take your caveat about consistency strength - with that said would you still call it misleading? Why is it that eg x+y=y+x for x y given takes exponential length proof in Robinson compared to PA? For the reason stated, which is true in a very broad sense.

>> If for all n we can prove P(n), then `∀n P(n)` should be an acceptable proposition

> But how can you prove that P(n) for all n without induction?

Just to be clear to all readers, the axiom of COUNTABLE choice is uncontroversial. Nobody is disturbed by induction.

The issue it that when you allow UNCOUNTABLE choice - choices being made for all real numbers (in a non-algorithmic way, I believe - so not a simple formula) - there are some unpleasant consequences.

> But how can you prove that P(n) for all n without induction? Maybe I misinterpret what you're saying, or I'm naive about something in formal languages, but if we can prove P(n) for all n. then `∀n P(n)` just looks like a trivial transcription of the conclusion into different symbols.

Of course it is likely that an interesting result about all positive integers, that is "really" about positive integers, is proved by induction, but you certainly don't need induction to prove P(n): n = 1.n, or, more boringly, P(n): 0 = 0. (These statements are obviously un-interesting, both in the human sense of the word and in the sense that they are just statements about semi-rings, of which the non-negative integers are an example.)

My understanding is that the difference between "For every positive integer n, I can prove P(n)" and "I can prove ∀n.P(n)" is that the former only guarantees that we can come up with some terrible ad hoc proof for P(1), some different terrible ad hoc proof for P(2), and so on. How could I be sure I have all these infinitely many different terrible ad hoc proofs without induction? I dunno, but that's all that the first statement guarantees. Whereas the second statement, in the context of computability, guarantees that there is some computable function that takes a positive integer n and produces a proof of P(n); that is, there is some sort of guaranteed uniformity to the proofs.

I think it may be easier to picture if connected with math_comment_21's analogy in https://news.ycombinator.com/item?id=44269153: the analogous statements in the category of topological spaces (I think one actually has to work about topos, but I don't know enough about topos theory to speak in that language) about a map f : X \to Y are "every element of Y has a pre-image under f in X" versus "I can continuously select, for each element of Y, a pre-image of it under f in X", i.e., "there is a continuous pre-inverse g : Y \to X of f."

Rejecting induction could be quite useful if you want to be very precise about the implications of your constructions wrt. computational complexity. This is of course only a mildly strengthened variant of the usual arguments for constructivism.