🚀 go-pugleaf

On May 29, 12:06 am, "R. Srinivasan" <sradh...@in.ibm.com> wrote:
> 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."

I can't make any sense of it.

> 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.

Okay; that's just as in ordinary mathematical logic.

> 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.

Oh, "it follows from basic NAFL principles". Yes, that sure expains
it.

Anyway, meanwhile, counterexample in ordinary mathematical logic:

ZFC set theory is consistent & ZF+~C is consistent. But there is no
theory in which both C and ~C are 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.

Okay, some kind of logic in which there are models of contradictions.

> This is a fundamental
> requirement of NAFL and to understand this you need to follow the
> ideas posted in the thread mentioned earlier.

You posts there make no sense to me.

> 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.

I can't make sense of any of that.

> 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.

That also I can make no sense of.

> > > 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.

I know what PA is, but I can't make sense of your postings.

MoeBlee