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


MoeBlee wrote:
> David C. Ullrich wrote:
>
> > it's awesomely obvious that "S" can have
> > many different interpretations.
>
> I understand that point very well. But I am unclear about a few things
> regarding PA. In some instances I'm not sure what people mean when they
> refer to 'first order PA'. Is the predicate 'is a number' always a
> predicate of the language and do the axioms always include the common
> clauses 'if n is a number'?
>

Not usually, no. Usually there is no predicate "is a number".

> If there is a 1-place predicate 'is a number' in the language, and the
> axioms include such clauses as 'if n is a number', then am I correct
> that the predicate symbol 'is a number' can be assigned by a structure
> to a proper subset of the domain? In other words, it might be that only
> a proper subset of the domain of interpretation is such that 0 and the
> successor operation provide for a Peano system. Right? So, for example,
> as I mentioned in another post, there can be models of the theory in
> which S is not 1-1 on the entire domain. Is that correct?
>

It probably would be correct for the hypothetical version of PA you're
describing. But not for the usual versions, no.

> Another point I am unclear on is addition and multiplication. Can we
> define these in Peano arithmetic?

Yes, they are primitive symbols.

> Don't we have to add axioms for them?

Yes.

> To use recursion to define them, we have to step out into a broader
> theory such as set theory, right?

In second-order PA, we can give recursive definitions of addition and
multiplication.

But in first-order PA, we take addition and multiplication as primitive
symbols and the equations in their recursive definitions are taken as
axioms.

> Or we could take the addition
> operation symbol and the multiplication operation symbol as primitive
> and then add the usual four axioms for them (two axioms for each
> symbol).

Yes, that's what we do.

> Do I have that right?
>
> So axiomatizations that include axioms for addition, multiplication, or
> ordering are not PA,

Yes, they are.

> but rather they are axiomatizations of different
> theories (ones that are incomplete in the sense that not all sentences
> true in the standard model are provable from the axioms axioms). Am I
> correct or am I missing something?
>

First-order PA is both deductively incomplete (there are sentences S
such that neither S nor its negation is derivable from the axioms) and
descriptively incomplete (there are nonisomorphic models for it).

Second-order PA is deductively incomplete, but is descriptively
complete (there is only one model up to isomorphism). This is possible
because second-order logic is semantically incomplete (not every
sentence which is true in all models is provable).

> Also, in set theory we prove that any two Peano systems are isomorphic.
> And, if I'm not mistaken, we prove that any two completely ordered
> fields are isomorphic. Yet, Lowenheim-Skolem tells us that for the
> axioms of Peano arithmetic there are non-isomorphic models, and for the
> axioms for a complete ordered field, there are non-isomorphic models.

The axioms for a complete ordered field are second-order.
Loewenheim-Skolem applies to first-order theories.

No first-order theory with any infinite models can be descriptively
complete. But a second-order theory can be.

> But I'm having a hard time picturing how to reconcile these facts. In
> set theory they're isomorphic, but in the meta-theory for first-order
> logic, they're not isomorphic. But the meta-theory for first-order
> logic is, we can suppose, set theory. So I am having a hard time
> picturing why things come out differently. What am I missing?
>
> Those are a lot of questions, I know, so I appreciate whatever help
> with any of them that anyone would lend.
>
> Thanks,
>
> MoeBlee

.

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