🚀 go-pugleaf

On May 28, 2:13 am, MoeBlee <jazzm...@hotmail.com> wrote:
> On May 25, 12:44 am, "R. Srinivasan" <sradh...@in.ibm.com> wrote:
>
> > E.g consider the NAFL version of Peano Arithmetic, call this
> > NPA.
>
> What exactly is NPA? For that matter, what exactly is an NAFL theory?
>
First let me know if you understood the basic ideas in the thread that
I had posted earlier, titled something like "FOL/intuitionistic logic
versus NAFL. Part 1. Failure of non-contradiction."

We need to start from basics, i.e., the fundamental laws of logic,
before discussing how a NAFL theory can be constructed.
>
> > It is true that NPA has infinitely many axioms, which may be
> > generated by some rule. Here is a proposition about NPA:
>
> > NPA is consistent if and only if its axioms are pairwise consistent
> > (1)
>
> How do you infer that?
>
It is very straightforward. Take a consistent NAFL theory T and let P
and Q be sentences (in the language of T) that are undecidable in T.
Then T+P and T+Q are consistent theories and there must exist a NAFL
model of T in which P must be true and another NAFL model of T in
which Q must be true. It follows, from basic NAFL principles, that
there must also exist a NAFL model of T in which P and Q are both
true, i.e., P&Q is true. This must hold even if T+P proves ~Q, or If T
+Q proves ~P; in these cases, the NAFL model is a non-classical one in
which the law of non-contradiction fails. This is a fundamental
requirement of NAFL and to understand this you need to follow the
ideas posted in the thread mentioned earlier.

The basic idea here is that T can only reject P&Q by the reasoning "If
P (Q), then ~Q (~P)". But since the truth of P (Q) is necessarily
axiomatic with respect to T, this reasoning only holds in the NAFL
theory T+P (T+~Q). I.e., the "If P...." part in the above sentence is
an axiomatic declaration of P by the human being who already has the
axioms of T in mind. So the human being has the theory T+P in mind
after the "If P..." part. It follows that there is no way for T to
refute P&Q.

Now if we know that P and Q are consistent with each other, the above-
mentioned model will be a classical one, from which the consistency of
T+P+Q will follow. By induction, we may conclude that if the axioms of
a NAFL theory are pairwise consistent, then T is consistent.

Now take a theory like classical PA. To refute what I have said above,
you will need the kind of "coding" employed by Godel. I.e., you will
have to show that PA encodes a theory like Z in which it is
demonstable that pairwise consistency does not imply consistency. But
this encoding is rejected in NAFL as infinitary.
>
>
> > Now (1) will go through in NAFL, but not in classical logic. To see
> > that (1) is true, we only need to know the rules by which the
> > infinitely many axioms of NPA are generated.
>
> What are those rules?
>
>
The axioms of NPA (the NAFL version of PA) are constructed in the same
way in which those of classical PA are constructed. But to describe
NPA in detail requires effort and you need to demonstrate some
understanding of the ideas posted in the thread mentioned earlier.

Regards, RS