Re: Proposal: 6NF

vc wrote:
> > A much simpler example. Let {0, 1, 2, 3} be a set of four integers
> > with addition modulo 4. Then, none of its subsets, except {0} and
> > {0, 2}, retains the addition mod 4 operation which makes the idea of
> > 'subtype as subset' utterly silly, [....].
This example is perfectly valid.
Jan Hidders belongs to a category of people that believe that computing paradigms such typing make *de facto* good mathematics. He is a fool who should be ignored.

> You keep on making the same mistake. The expression a +[mod 4] b has a
> well defined result if a and b are from any subset of {0, 1, 2, 3}.
NO you have not made any mistake, he is...(Once more) Now he is trying to convince you that you'are making a mistake...Total lack of intellectual honnesty. That's what happens when people get exposed by sound proofs: they get pissed...

Ask him to express the *well defined result*?

Some more of Jan's bullshit...

Consider domain of Integers D1 {-1, 0, 1} and derived domain of positive integers PD1{0, 1}. Consider SQRT operator.

SQRT does not apply on D1 because of negative values. (except we do COMPLEX numbers, which Jan's finds practical to ignore because it simply proves all the bullshit he produced as being wrong)

SQRT applies on ALL values of PD1 to produce REALS, INTEGERS or COMPLEX so what is the well defined value?

If the mathematicians that are supposed to *enlighten* him did not explain the nondeterministic aspect of domains, they simply must be as dum as he is. What an idiot, what a moron. Quit wasting time with that time black hole.. Received on Thu Oct 19 2006 - 15:40:27 CEST