🚀 go-pugleaf

On May 29, 11:10 pm, "R. Srinivasan" <sradh...@in.ibm.com> wrote:

> 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.

You want me to start by taking Aristotelian logic as a basis? And how
basic in Aristotelian logic do you want me to confine to? Please say
exactly and specifically the principles you want me to start with. And
then since you want me to clear all the other knowledge I have about
mathematical logic, please provide for filling again by introducing
your alternative concepts one at a time in a systematic way.

Whatever I know about mathematical logic, I learned by following each
step at a time, systematically. First I learned the truth tables for
the basic connectives, which made sense to me. Then I learned other
basics about propositional logic, which made sense to me. Then on to
more and more complex material. Each module of the subject matter made
sense to me, step by step, as I went along. If you want people to
learn your alternative, then you need to be able to provide them with
that kind of systematic presentation.

> 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.

Okay, of course, I understand all of that, except you say NAFL has the
equivalence of ~(P&Q) with ~Pv~Q but that NAFL requires proving ~Pv~Q
only by either proving ~P or proving ~Q. So I don't know how you
define your proof system so that  ~(P&Q) is equivalent with ~Pv~Q yet
we can't infer ~Pv~Q from ~(P&Q).

MoeBlee