🚀 go-pugleaf

On May 29, 11:25 pm, MoeBlee <jazzm...@hotmail.com> wrote:
> 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.
>
All right. I see that I will have to get started again and at least
try to get you in a frame of mind to understand NAFL. You really have
to do a "memory clear" operation and start with basic laws of logic,
i.e., approximately from about 350 BC, when they were first stated by
Aristotle. I will make another attempt when time permits, and start
with the ideas mentioned in the thread that I referred to.

But anyway, here is a general framework for what you might think is a
counter-example to pairwise consistency implying consistency.

Suppose T is a theory with the sentences P and Q both being
undecidable in T. Suppose T proves ~(P&Q), but let us say that ~(P&Q)
is not a tautology (which allows P and Q to be consistent with each
other)., So you can see that P, Q and ~(P&Q) together with the axioms
of T are pairwise consistent. But T+P+Q+~(P&Q) is inconsistent.

This argument does not work in NAFL. Because ~(P&Q) is equivalent to
(~Pv~Q) in NAFL and if a consistent NAFL theory T proves this
disjunction, it must prove either ~P or ~Q (i.e., both P and Q cannot
be undecidable in T as assumed). Even in intuionistic logic, if  a
theory proves a disjunction it must prove one of the disjuncts. But
intuitionistic logic does not accept the classical- (and NAFL-)
equivalence of ~(P&Q) with ~Pv~Q. So pairwise consistency does not
imply consistency in either classical or intuitionistic logic for
different reasons. But NAFL is different in this respect and I assert
that NAFL correctly represents finitary reasoning.

Regards, RS