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


MoeBlee wrote:
> 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
>

This is an informal presentation of Peano's axioms, probably modelled
on the presentation Peano himself originally gave. It's different to a
formal presentation.

> > > 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?

Yes, they are.

> 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?

> 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?
>

You seem to be assuming that there is a set of first-order axioms such
that any model of them is a complete ordered field. This is not the
case.

> Thanks,
>
> MoeBlee

.

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