Any proof he could construct in NA could also be constructed in PA. (If this is not so, finding (for example) finite X and Y such that X^Y is infinite in NA does not provide a contradiction to PA.)

Therefore, he doesn’t need to bother working in NA to achieve his goal – he can work in PA instead. In fact, if PA is inconsistent, it might be that he must work in PA to prove it. It may be that the only “easy to find” statement P in PA for which both P and ~P are proveable in PA all suffer from the deficiency that neither P nor ~P is proveable in NA. Then his approach will fail.

The only advantage he might find in working in NA is that working with a restricted set of axioms refines his intuition. However, it’s not in the least a logical necessary, and has other disadvantages such as the one I mentioned.

“If I want to prove that for all X,Y, X^Y is finite, you are right. However, for any specific X,Y, there exists a computer that will run the program. That’s the computer he should use. If he run the program on the wrong computer, how is it my fault if it fails to terminate? How can it invalidate the proof?” The problem is, for sufficiently large X and Y, X^Y may be too big for any computer that you can practically make to finish the proof that X^Y is finite within a human lifetime. So you’ll never know directly that the proof will terminate, unless you give a proof somehow.

“granted, he could add an “induction doesn’t hold” axiom to his system,” But he doesn’t want to add it an as axiom, he wants to prove it as a theorem, thereby invalidating PA.

Best of luck to him :-) However, if he succeeds in showing, within NA, that PA is inconsistent, then
1) he could construct the same proof within PA, with equal validity and importance. Why, then, cripple himself from the outset with a weaker logical system, where (for example) induction holds but is unproveable?
2) his proof that PA is inconsistent might arise because of an inconsistence in NA. Granted, this would also imply that PA is really inconsistent, but it wouldn’t mean NA is the “correct” system to use.

“I’ve already presented a proof to them. I told him if he didn’t believe me he should run the program and he’ll see the proof. He just needs a powerful enough computer.”…. The problem is, regardlesss of how powerful the computer is, for sufficiently large X and Y the program may run longer than a human lifetime, so you’ll never find out whether it terminates. So then in order to prove that X^Y is finite, you need some way to prove that the program will terminate in a finite amount of time. How will you do that?

If I want to prove that for all X,Y, X^Y is finite, you are right. However, for any specific X,Y, there exists a computer that will run the program. That’s the computer he should use. If he run the program on the wrong computer, how is it my fault if it fails to terminate? How can it invalidate the proof?

He cannot reasonably do so, because your proof that induction holds in his system is based on induction in PA, which he sees no reason to accept

granted, he could add an “induction doesn’t hold” axiom to his system, and get an interesting system where, for example,
* “There exists a finite number that cannot be expressed as the sum of 4 or fewer squares”
* “0 is not that number”
* “1 is not that number”
* “2 is not that number”
* “3 is not that number”
* (etc)
are all true.

By the way, have you taken a look at the book by Nelson I linked to earlier, Predicative Arithmetic? You might find it interesting.

I believe I would find it interesting :-) However, I haven’t had time to look at it yet :-(

“However, if you are correct, he surely has no objection to my 2^billion-1 line proof that 2^billion is finite?” He would not have any objection to a 2^billion – 1 line proof if you actually wrote such a proof down. What he won’t accept is you just claiming that there exists such a proof, because how can you know that there exists such a proof unless you’ve either written it down or you know that 2^billion – 1 is a counting number?

“Since I now know that the principal of induction holds in NA, I can use it to derive proveable statements in NA, without necessarily constructing the proofs. As an example, I could prove that N(2^billion is finite), since (you insist) I can prove that N(2^0 is finite), and in his talk he shows that N(2^x is finite implies 2^Sx is finite). Therefore, in fact, N(2^x is finite) for all x, even if not N(2^x is finite for all x).” Right, if you work in PA you can easily show that exponentiation is total in Nelson’s system of predicative arithmetic. But the whole goal of Nelson’s work is to prove PA inconsistent, and thus prove that exponentiation is not total.

“I’ve already presented a proof to them. I told him if he didn’t believe me he should run the program and he’ll see the proof. He just needs a powerful enough computer.” The problem is, regardlesss of how powerful the computer is, for sufficiently large X and Y the program may run longer than a human lifetime, so you’ll never find out whether it terminates. So then in order to prove that X^Y is finite, you need some way to prove that the program will terminate in a finite amount of time. How will you do that?

“If he won’t accept my proof encoded as a computer program, why should I accept his alleged “proof” that 1 is finite encoded as smudges of ink on a piece of dead tree?” It doesn’t matter how the proof is written down. The proof can be written down on a computer or any other medium. What Nelson won’t accept is a mere algorithm which purports to write down the proof, without either seeing the algorithm terminate or having some way of knowing that the algorithm will terminate in a finite amount of time.

“I hereby assert that 1 is infinite in NA, and his “proof” is infinitely long. I reject it.” As I said, provable finiteness is not something that is needed to check the validity of a proof in Nelson’s view. So your bizarre assertion that a one-line proof is infinitely long doesn’t change anything.

“I’ll back down and concede the finiteness of 1 if he will concede the finiteness of 2^billion, which he can reasonably do in light of my inductive proof that induction holds in NA.” He cannot reasonably do so, because your proof that induction holds in his system is based on induction in PA, which he sees no reason to accept.

By the way, have you taken a look at the book by Nelson I linked to earlier, Predicative Arithmetic? You might find it interesting.

But the problem is, how do you know that your proposed program won’t go on an infinite loop forever?

I’ve already presented a proof to them. I told him if he didn’t believe me he should run the program and he’ll see the proof. He just needs a powerful enough computer.

If he won’t accept my proof encoded as a computer program, why should I accept his alleged “proof” that 1 is finite encoded as smudges of ink on a piece of dead tree?

I hereby assert that 1 is infinite in NA, and his “proof” is infinitely long. I reject it. I’ll back down and concede the finiteness of 1 if he will concede the finiteness of 2^billion, which he can reasonably do in light of my inductive proof that induction holds in NA.

I, who am happy to work in PA, can prove the following. Let P(x) be a proposition.
* If N(P(0)) and
* N(for all x, P(x) implies P(Sx)), then
* for all x, N(P(x))

So, in fact, induction holds in NA, even if we can’t actually prove this within NA.

Since I now know that the principal of induction holds in NA, I can use it to derive proveable statements in NA, without necessarily constructing the proofs. As an example, I could prove that N(2^billion is finite), since (you insist) I can prove that N(2^0 is finite), and in his talk he shows that N(2^x is finite implies 2^Sx is finite). Therefore, in fact, N(2^x is finite) for all x, even if not N(2^x is finite for all x).

However, if you are correct, he surely has no objection to my 2^billion-1 line proof that 2^billion is finite?

“If the contention is that this can happen, I’d ask for a proof that it does, in fact, happen – that there does in fact exist X,Y such that X^Y is infinite. If possible, a constructive proof, please.” If Nelson succeeded in showing in his system that X^Y is infinite even though X and Y are finite, then he would have proven that Peano arithmetic is inconsistent. Nelson is actually trying to do just that (he calls it his “Modified Hilbert Program”), because he DOES believe that exponentiation is not total, but so far he has been unsuccessful, despite the false alarm a few months ago.

“If the contention is merely that it cannot be proven one way or the other that X^Y is finite for all X and Y, I’d say “fine”. That’s a much weaker claim.” For better or worse, Nelson is indeed making the stronger claim.

“I’d also ask for their axioms. If they include
A1) 1 is finite
A0) if x is finite, so is Sx
then I’d complain that there is a proof that X^Y is finite, and show how to construct the proof. I’d perhaps even write the program for writing down the proof, and ask them to run the program and check the proof if they don’t believe me.” But the problem is, how do you know that your proposed program won’t go on an infinite loop forever?

“If the axioms are
A1) 0 is finite
A0) if x is finite, so is Sx
I’d ask how they know 1 is finite. If they write me down a 1 line proof and say it’s obviously a proof, I’d claim equal rights to use intuition, and prove that the axiom of induction holds, even if it’s unproveable.” They’re not claiming that the 1-line proof is a proof on the grounds of intuition. It is simply not the case that Nelson is insisting that proofs have to have a provably finite number of lines. But in any case, this is a side issue, so if you want to avoid getting distracted by the “1 is finite” question, then you can change the axioms to start from “1 is finite”, it really doesn’t make a difference for the purpose of Nelson’s logic. Nelson would still insist that you can’t just baselessly assert that there exists a 2^billion – 1 line proof that 2^billion is a counting number, because you would have to first show that 2^billion – 1 is a counting number.

And you can’t just say “The finiteness of 2^billion can be justified by the finiteness of 2^billion – 1, which can be justified by the finiteness of 2^billion – (1+1), which can be justified by the finiteness of 2^billion – (1+1+1), etc.” The problem with that is that perhaps there is an infinite sequence 2^billion, 2^billion – 1, 2^billion – (1+1), … that never reaches 1. Now you or I may find that absurd, because you can’t have an infinite descending chain of natural numbers, but if 2^billion were really infinite then the sequence wouldn’t be a descending chain of natural numbers.

What is your basis for this claim that there exists a proof of the finiteness of 2^billion? How can you refute Nelson’s view that there we can keep writing proofs, in order, that “1 is finite”, “1+1″ is finite, “1+1+1 is finite”, etc., but in that infinite sequence we will never get to the statement “1+1+1+…+1 is finite” where you have 2^billion 1’s, because 2^billion is infinite?

I can use the principle of induction. I know that the principle of induction is not in his list of axioms. However, I’m pretty sure that I can prove (using induction) that it holds in his system. You will (partially correctly) point out that this would be a circular proof. You are only partially correct, since I can choose to do these proofs using a different logical system, say, PA for example.

You may still insist this is circular, even if I don’t do the proof within Nelson’s system. However, it is no more circular than accepting a proof with 1 step to prove that 1 is finite in his system, surely? You accept it the proof only after accepting the finiteness of ’1′, which you do for other reasons (either because you proved it in some other system, or because it’s “intuitively obvious”)

“How would you refute Nelson’s contention that you can have a typewriter with a finite symbol set writing on a finite piece of paper, and yet there being an infinite number of possible pages you can type? I find that point of view so shocking, so off the wall that I would love to come up with a good counter argument to it”

If the contention is that this can happen, I’d ask for a proof that it does, in fact, happen – that there does in fact exist X,Y such that X^Y is infinite. If possible, a constructive proof, please.

If the contention is merely that it cannot be proven one way or the other that X^Y is finite for all X and Y, I’d say “fine”. That’s a much weaker claim.

I’d also ask for their axioms. If they include
A1) 1 is finite
A0) if x is finite, so is Sx
then I’d complain that there is a proof that X^Y is finite, and show how to construct the proof. I’d perhaps even write the program for writing down the proof, and ask them to run the program and check the proof if they don’t believe me.

If the axioms are
A1) 0 is finite
A0) if x is finite, so is Sx
I’d ask how they know 1 is finite. If they write me down a 1 line proof and say it’s obviously a proof, I’d claim equal rights to use intuition, and prove that the axiom of induction holds, even if it’s unproveable. Or complain that if proofs become valid just because a mathematician says so, that a whole lot of rubbish suddenly becomes true, and could they please write down an axiomatic system describing the mathematician.

If they can’t either
* acknowledge the proof of 1′s finiteness is circular, and should be rejected along with my proof that induction holds, OR
* allow me to use induction, in exchange for me allowing them to do 1 line proofs, OR
* axiomatise the mathematician and his or her intuitions, OR
* acknowledge there’s no solid logical basis to the whole thing and it’s just fanciful,
then I know not to take them seriously.

Is mathematical truth to be determined with reference to a particular species of mammal on a particular planet?

Nobody is remotely claiming this. We’re not arguing about mathematical truth; we are arguing about what we can prove. And of course the question of what we can prove is a question about what can be done with a particular collection of proof techniques that happen to be available to a particular species on a particular planet, because we are a particular species on a particular planet.