Re: Pa without 0 and successorfunction

On Aug 13, 1:33 am, translogi <wilem...@xxxxxxxxxxxxxx> wrote:
On Aug 11, 9:30 am, Rupert <rupertmccal...@xxxxxxxxx> wrote:

On Aug 9, 6:35 pm, translogi <wilem...@xxxxxxxxxxxxxx> wrote:

Boolos "the logic of provability" says that PA can be axiomised
without:
zero and the successor functions as primitive symbols.

Unfortunedly he doesn't tell how to do so.

Can somebody enlighten me here?

I would think you could replace the two primitive symbols 0 and s with
one primitive symbol <. Perhaps that is what he has in mind. Can you
provide the context of what he said? That might give us some clues.

Boolos " logic of provability" pag 17

"(Since zero can be proved in PA to be the unique number k such that j
+k=j for all j and the successor of i can be proved to be the unique k
such that j x k = j x i +j for all j 0 and s could have been
dispensed with, but it simplifies matters concideerably for them to be
taken as primitives)"

so if i am correct zero can be defined as:
Ek ((k=0 <--> Aj ((j+k) =j) & Ax Aj (((j+x) =j) -> (x=k))

and the successor as

AiEk (Si=k <--> Aj ((j x k)= ((j x i) +j))

But if you add these as definitions doesn't PA becomes hopelesly
circulair?

also how do you define a term then (for syntax)
0 and S are not primitive symbol anymore so they should not be used as
defining symbols for the syntax.

What you do is you add axioms which assert that there is a unique
number k such that j+k=j for all j, and for all i there is a unique
number k such that j x k = j x i + j. Then you translate all the
axioms of PA into the language containing only addition and
multiplication using Boolos' definitions of zero and successor. This
yields a theory in which PA can be interpreted.
.

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