Ancestral arithmetic

[Newberry]

It is possible to define the ancestor relation anc(x,y) meaning that x
is an ancestor of y. For example x is an ancestor of Sx. We then can add
the following axiom to PA

∀x(x = 0 ∨ anc(0,x))

meaning every number is either zero or one of its successors. A couple
of other axioms are needed to precisely define anc(x,y) from the
successor operator S. And we obtain a theory, which is CATEGORICAL, i.e.
there are no non-standard models. It is nevertheless alleged that this
theory has the same power as PA.

I do not understand that. Correct me if I am wrong, but it seems to me
for example that if there are no non-standard numbers then
∃x(Prf(x, ⌜P⌝) → P (A)
That leads to a contradiction:
(1) ∃x(Prf(x, ⌜G⌝) Assumption
(2) G by A
(3) ~∃x(Prf(x, ⌜G⌝) RAA, (2) is self-contradictory
(4) G (3) is equivalent to G
Inconsistency!
Does anybody have any insight into this?

[Khong Dong]

As long as you don't "touch" the (definition of) primes, as long as you concentrate purely on induction via 'S', then it doesn't matter: you're just talking about some kind of Presburger arithmetic and not the "natural numbers"!

If you want to deal with the "natural numbers", "standard model", it's the non-inductive/non-recursive well-ordering of primes that's the matter -- not the succession denoted by 'S'.

[Ross A. Finlayson]

Isn't it of "arbitrary order", thus outside schema?

[Pierre Asselin]

Newberry <newbe...@gmail.com> wrote:
> It is possible to define the ancestor relation anc(x,y) meaning that x
> is an ancestor of y. For example x is an ancestor of Sx. We then can add
> the following axiom to PA

> ???x(x = 0 ??? anc(0,x))

> meaning every number is either zero or one of its successors. A couple
> of other axioms are needed to precisely define anc(x,y) from the
> successor operator S.

Hmm, the transitive closure of S ... Okay, there exists a y that
codes a finite-sequence starting with 0 and ending with x, with each
member but the last followed by its S().

> And we obtain a theory, which is CATEGORICAL, i.e.
> there are no non-standard models.

How so ? The y could be nonstandard.

Besides, you can prove forall x anc(0,x) -> anc(0,S(x)) so by induction
forall x anc(0,x).

If you have a different formal definition of anc(), please sketch it.

--
pa at panix dot com

[FredJeffries]

If I may be allowed momentarily to step above my pay grade, I have no idea of the answer to the original inquiry, but it presumably refers to Peter Smith's 'Ancestral Logic/Arithmetic'. See

https://www.logicmatters.net/2006/05/07/ancestral-logic/

Smith, Peter. “Ancestral Arithmetic and Isaacson’s Thesis.” Analysis 68, no. 1 (2008): 1–10.
https://www.logicmatters.net/resources/pdfs/Isaacson_paper.pdf

[Ross A. Finlayson]

You might be surprised how many people are way above their, "pay grade".

Quoting your refernce:

"It’s pretty obvious that the semantic consequence relation
for such an ‘ancestral logic’ won’t be compact, so the logic isn’t axiomatizable"

"... which is an generalized induction schema. "

"... if we treat the ancestral operator as a logical constant
with a fixed interpretation — this is a categorical theory
whose only model is the intended one (up to isomorphism).
But while semantically strong it is deductively weak.
It is conservative over PA."


Another way to look at that is exercises in Skolem,
in lieu of exercises in Cohen. (For compactness,
and a model of compact naturals.)

Until Peter Smith backpedals about "material implication", well,
until logic overall reflects on the mistakes of "material implication",
it's a-waffling.

[Newberry]

Kind of. I actually got it from his book "Goedel's Theorems."

I thought about the question, and I am guessing that maybe ancestral
arithmetic is not semantically complete. It is the same case with 2-nd
order arithmetic. It is negation incomplete, categorical and
semantically incomplete.