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Re: Axiomization of Number Theory
Posted:
Aug 4, 2003 12:18 AM
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> >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, second-order Peano Arithmetic is not in the "Predicate Calculus,"
> which is first-order 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
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