Re: What is the 1st order formal system known as PA?

Rupert wrote:

> The answer to this, I suppose, is that when writers present the axioms
> informally they might throw around phrases like "is a number" which
> don't actually show up in the formalization. This is about all I can
> say unless you can point me to a specific writer.

Perhaps not from a renowned writer, but, for example, here is what you
get as the very first hit on a Google search for 'Peano arithmetic'
(the second link is just the one suggested at the page of the first
link):

http://mathworld.wolfram.com/PeanoArithmetic.html

http://mathworld.wolfram.com/PeanosAxioms.html

> > But Shoenfield on page 204 of 'Mathematical Logic' specifies Peano
> > arithmetic as axiomatized in first order logic with identity and
> > omitting leading universal quantifiers (cf. page 22 also):
> >
> > 0 0-place function symbol
> > S 1-place function symbol
> > + 2-place function symbol
> > * 2-place function symbol
> > < 2-place predicate symbol
> >
> > Sn not= 0
> >
> > Sn=Sk -> n=k
> >
> > n+0=n
> >
> > n+Sk=S(n+k)
> >
> > n*0=0
> >
> > n*Sk=(n*k)+n
> >
> > n not< 0
> >
> > n<Sk -> (n<k v n=k)
> >
> > (phi[0] & An(phi(n) -> phi[Sn])) -> An phi[n]
> >
> > So am I correct to take it that in current discussions, by 'Peano
> > arithmetic', people mean the theory axiomatized by the above axioms?
>
> Yes, except for the fact that < is usually taken as a defined symbol;
> see below.
>
> > (It could be a different set of axioms as long as it axiomatizes the
> > same theory.) Also, that each of those axioms is independent of the set
> > of the others? But, as to independence of primitive symbols, couldn't
> > we take '<' as defined by:
> >
> > Df. n < k <-> Ej(j not= 0 & n+j=k)
> >
>
> Yes.

Good, so I have a specific axiomatization I can reference. But here's a
different one from page 42 of Chang & Keisler's 'Model Theory':

Sn not= 0

Sn=Sk -> n=k

n+0=n

n+Sk=S(n+k)

n*0=0

n*Sk=(n*k)+n

(phi[0] & An(phi(n) -> phi[Sn])) -> An phi[n]

So Chang & Keisler don't include the seventh and eighth axioms of
Shoenfield. And I don't think those axioms are derivable, even with the
definition of '<', from the first six axioms along with axiom nine, are
they? So, without specification, I just don't know what people mean by
'Peano arithmetic'.

> > we could express
> > the axioms of a complete ordered field as a first order theory with
> > axioms and one axiom schema?

> You could do that; then it would have nonisomorphic models by
> Loewenheim-Skolem. This is because the first-order axiom schema doesn't
> capture the full power of the second-order axiom.
>
> In fact we consider nonstandard models for the theory you're talking
> about in nonstandard analysis.

Right, I don't know all the details of that, but I have a sense of the
outline. However, what I don't understand is that within set theory we
show that any two complete ordered fields are isomorphic with one
another. Yet, Lowenheim-Skolem tells us that there are non-isomorphic
models of the axioms, which axioms I picture in set theory as just
being the definition of 'a complete ordered field'. Where am I going
wrong in my picture of this?

Thanks,

MoeBlee

.

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