I believe my disagreement was specifically about this (quoted):
> In principle a computer can correctly recognize 2 apples as
> being 2 apples without knowing anything about addition and
> without being able to recognize 1 apple.
I have a hard time seeing this. If I am not mistaken again, there is no definition of the symbol 2 that does not include addition. The proof above doesn't really prove this either as Peano's axiom rely on the definition of "successor". Successor definition: "a+1 is the successor of a".
A computer can be given an explicit map between the symbols S(0), S(S(0)), s(S(S(0))), ... and 1,2,3..., so to identify S(S(S(S(S(S(0)))))) with 6 - it wouldn't need to know anything at all about + or about 1.
Edit: also consider the following quote from Wittgenstein's Philosophical Investigations and whether the person described needs to have any concept of 'addition' in order to correctly use numbers: "Now think of the following use of language: I send someone shopping. I give him a slip marked 'five red apples'. He takes the slip to the shopkeeper, who opens the drawer marked 'apples', then he looks up the word 'red' in a table and finds a colour sample opposite it; then he says the series of cardinal numbers--I assume that he knows them by heart--up to the word 'five' and for each number he takes an apple of the same colour as the sample out of the drawer.--It is in this and simlar ways that one operates with words--"But how does he know where and how he is to look up the word 'red' and what he is to do with the word 'five'?" ---Well, I assume that he 'acts' as I have described. Explanations come to an end somewhere.--But what is the meaning of the word 'five'? --No such thing was in question here, only how the word 'five' is used."
You cannot calculate S^(x+1)(0) before the result of S^x(0) is known.
To calculate S^6(0) you start by calculating S(0) ----> It's not possible to "identify S^6(0) with 6" if you haven't calculated S(0) first, because you cannot know S^6(0) at this point.
From the definition of +: S(0)+S(0)=S(S(0)+0) Again using the definition of +: S(0)+0=S(0). And we get: S(0)+S(0)=S(S(0))
Q.E.D.
There is also a famous proof by Whitehead and Russell on page 379 of Principia Mathematica: http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhist...