Re: Arithmetic with (only) exponentiation

On 26 Apr 2006 19:06:05 -0700, george wrote:

Bertie Reed:
or to sum up: "blah blah blah I can't here you".

I didn't make that spelling-error; you did.
And if you are not being understood,
the burden is on you to make yourself clear.

I am not sure which spelling error you mean.

I heard you just fine; I just didn't like what I heard.
And what you post in response is not a refutation
of anything I said.


No, it was a clarification of an earlier post that you did not seem to
understand. I showed that from the function x^y, one can define 0 and 1. I
also noted that this was true for the function 2^x. This refuted the
implicit assertion in

"No, no, no!...How are you even getting "2" into this vocabulary if you
don't have a successor function AS WELL as an exponentiation function?"

Proving that you can use N as a domain of a model
for something is boring and trivial; if T is a first-
order theory over a countable language THEN OF COURSE
N can be a domain of a model for it. That is NOT the issue here.

True, if T has an infinite model. True also that this is not the issue.


The original treatment spoke in terms of definig 2^(whatever),
not of x^y.

No, it spoke, perfectly correctly, about multiplication and arithmetic
being definable from 2^x.

As I noted, the treatment I gave works for 2^x function too*.

The actual question here is, simply, what axioms do you need,
if you are going to do this?

This is your question, perhaps. Do what?

The origional question was

"Does anyone know the status of arithmetic with just
exponentiation? My guess is that it would not turn out to be decidable,
but I was hoping this matter might be settled and someone could clue me
in."

Whilst this is not entirely clear, the meaning that I (and A N Niel and
quazi) understood was the following:

"Is the theory of N with a function for exponentiation (either binary or
unary) decidable?"

The answer was given by A N Niel, and clarified by me because you did not
seem to understand it.

The secondary question is, can you do it without = ?

Do what?

PA is doable without = at all, let alone without = in the logic.

I am afraid you have completely lost me.

Bertie
.