Re: primitive recursive: obsolete?

MoeBlee wrote:
On May 9, 10:27 pm, "Nam D. Nguyen" <namducngu...@xxxxxxx> wrote:
Aatu Koskensilta wrote:

Solely on basis of our basic understanding of the naturals as what
one obtains from 0 by repeatedly applying the "add one"-operation
So what happens to the prime numbers and the multiplication-operation?

Both the predicate 'is prime' and multiplication are primitive
recursive.

But what does that have to do with my critique on AK's *neglecting*
to mention multiplication, in his characterizing the naturals above?

Naturally, what's perceived as the natural numbers is more than just
a model of Presburger Arithmetic!

Huh? Who said anything about the natural numbers as being nothing more
than that which is modeled by Presburger artihmetic?

His characterization of the natural numbers above said that.

the principle of induction and the principle of definition by primitive
recursive induction are equally compelling.
As you've noted above the two are equivalent; if one is compelling, so would
be the other. But "compelling" in what sense?

In the sense of thinking what is the case about natural numbers.

This is the reason for taking primitive recursive
arithmetic as the canonical formalisation for finitism; if we further
allow that properties definable from primitive recursive properties by
means of the usual logical operations of first-order logic --
including unrestricted quantification --, in effect accepting the
totality of the naturals as something determinate, we get PA.
But at what cost are you willing to pay for accepting such "totality of
the naturals as something determinate"? For example, are you willing
to accept the distinct possibility that GC is arithmetically true, but
can't be arithmetically proven,

What do you mean by "arithmetically proven"?

It means the formula is a theorem in a formal system that is
"as strong as arithmetic".

or can't be proven in any theory
"as strong as arithmetic"?

What do you mean by 'the theory of arithmetic'?

Since that's not what I said, I can't answer your question.


There is the theory of
all true sentences of arithmetic in the language of PA.

Are you sure you'd know exactly what that theory is? For example,
what are some examples of its *non-trivial* theorems? [Hint: one
can actually define "true" sentences even for an inconsistent theory!]

Eiher GC or ~GC is provable in that theory, though it is not a recursively
axiomatized theory.

You wouldn't know "that theory" enough to assert this kind of provability
"fact"!

Meanwhile, if GC is true then there is a
recursively axiomatized theory in which GC is provable and true and if
GC is false then there is a recursively axiomatized theory in which
~GC is provable and true. We can devise such theories in two seconds,
just by taking either GC or ~GC as axioms.

You had "taking either GC or ~GC as axioms" but "axioms" is plural!
And I don't think you meant {GC} and {~GC} since they're not considered
as "arithmetic theories"! So, I don't really understand what you said above!

Though, we have no choice
at this moment as to accepting - we must accep that at this moment -
that we don't know which one we would add.

Wouldn't it be simpler to just say that if ~GC is not provable in Q,
it's impossible to prove GC in Q.


And suppose you're willing to pay such a
steep price, what would that even mean to FOL reasoning?

I have no idea in what way you think this challenges, in particular,
first order reasoning nor what reasoning system for mathematics you
thus propose.

This challenges the notion that the set of FOL rules of inference is adequate
enough to make it possible to solve all the provability problems. The shortcoming
of FOL is that it doesn't even formally admit this shortcoming.




MoeBlee

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