
Re: Axiomization of Number Theory
Posted:
Aug 4, 2003 12:18 AM


> >Andrew Boucher <Helene.Boucher@wanadoo.fr> wrote: > >>"n is a perfect number" can be translated as "Sum{x : x  n} = n" where > >>"Sum(P)" refers to the x in (hope I got this right): > >> > >>"[m][R][S] ( Seq(R,m,P) & Seq0(S,m) & S'm = x & S'0 = 0 & (i)(i < m => S'(i+1) = S'i + R'(i+1)) )". > >> > >>Here: > >>[..] means "there exists .." > >>Seq(R,m,P) means "R is a sequence from {x : 1 <= x <= m} onto P" > >>Seq0(S,m) means "S is a sequence from {x : 0 <= x <= m}" > >>S'z refers to the y for which Sz,y
> First, secondorder Peano Arithmetic is not in the "Predicate Calculus," > which is firstorder logic.
IMHO the terms are "1st Order Predicate Calculus" (or "(1st Order) Predicate Calculus") and "2nd Order Predicate Calculus". I used "Predicate Calculus" to mean both, but I won't worry about it if you won't.
> Secondly, you have to accept abbreviations, > or be willing to unbundle the abbreviating terms. So e.g. if you don't > want to accept > "Sum{x : x  n} = n" > then I would write > "[m][R][S] ( Seq(R,m,P) & Seq0(S,m) & S'm = n & S'0 = 0 & (i)(i < m => S'(i+1) = S'i + R'(i+1)) )" > > for "n is a perfect number," where someone (you can't and I won't) has > to unbundle the Seq wffs. If "<" is not part of the Peano Arithmetic > under consideration, it has to be rewritten as well (say in terms of > addition).
Of course it is not a question of my "accepting" it. The question is, are you going to give a PA wff that represents "N is a perfect number", or are you going to keep giving constructs and definitions that are not a PA wff?
Which of the following do you agree with?
1. PA wffs consist of relations, parentheses, quantifiers ("for all" and "there exists"), variables, ^ (and), v (or), => (implication  if you want to use it), <=> (equivalence  if you want to use it) and ~ (negation) only.
2. PA rules of inference act only on what I call "PA wffs" in # 1.
3. Your syntax for "N is a perfect number" uses constructs that are not "PA wffs" as defined by # 1.
4. You should either give a PA wff for "N is a perfect number" or admit that, as far as you know, PA cannot represent "N is a perfect number".
[ Aside: Notice that I cleverly avoided getting sucked into a discussion of your syntax and semantics (a red herring), other than to note that it is not a PA wff. :p ]
Charlie Volkstorf Cambridge, MA

