Re: Programming is the Engineering Discipline of the Science that is Mathematics

vc wrote:
> Recall that the OP claimed that PT is a logic
> 'generalization' in the sense that 'probability depends
> only on the constituent probabilities'. He failed to
> prove the bizzare assertion and refused/or was unable to
> solve the trivial puzzle that disproves his statement.

No idiot. That is not the sense in which I claimed "one view" of probability theory is as a generalization of logic. The "bizzare" assertion was part of an elaboration in the refutation of your idiotic claim

  "PT 'connectives' (+/*) are not truth functional." -- vc

In which you confused addition and multiplication with connectives and failed to realize that PT uses the SAME connectives as logic.

> However, below, instead of talking about such
> 'generalization, he attempts to demonstrate 'reducing'
> probabilistic statements to their logical truth valued
> counterparts by conjuring up, in vain, the spirit of
> conditional probability:

Mindless drivel. "conjuring"? "spirit"? Absolutely senseless.

> > Erwin, what vc was referring to is that
> >
> > P(AB) = P(A|B)P(B) -or-
> > P(AB) = P(B|A)P(A)
> >
>
> Note, that I said nothing about conditional
> probabilities.

No but you claimed "P(p1 and p2) is not equal P(p1)*P(p1) in general" which is of course pointing to the possibility of p1 and p2 being dependent which is of course equivalent to a conditional statement P(p1|p2) = P(p1).

> I merely requested to compute the P(A and
> B) probability in terms of P(A) and P(B) which was
> promised by the OP (see above).

Which the proof you called "mindless playing with formulas" provides. How about this, since you believe I'm wrong why don't YOU provide a proof? I would love to see what you consider is /not/ "mindless playing with formulas"

> > where | means given and AB is short for "A and B". This
> > is called the product rule. Something that vc seems not
> > to know (given his questions in the other post) is that
> > in the limit of true (0) and false (1) the conditional
> > probability product rule reduces to the logical
> > conjunction truth table. Here is the proof
> >
> > g : P(A) = 0
> > p : P(AB) = P(B|A)P(A)
> > u : P(AB) = 0
>
> Unfortunately, it's no proof but just mindless playing
> with formulas. The conditional probability is *defined*
> as
>
> P(B|A) def P(A and B)/P(A)
>
> the requirement for such definition being that P(A) <>0,
> naturally. The definition can be found in any
> introductory PT textbook.

LMAO. "mindless playing with formulas"? You vc are a mindless idiot who has an extremely limited understanding of probability theory. You clearly know nothing about the Cox formulation or his derivation of the product rule IN THE FORM I gave above as p) from functional requirements ALONE. Your total knowledge is apparently limited to a (small) part of the Kolmogorov formulation. In Cox's formulation there are NO such <>0 requirements. You are a VI. Go educate yourself and stop spouting nonsense. Go pick up your copy of Jaynes' book, turn to Chapter 2 and start reading. After you get through the derivation of an equation called "the product rule" (equation 2.18 in my copy), stare at it, think about the derivation, realize that you are wrong, then come back here and apologize, and finally STFU.

> Even, if P(A) were <> 0, the step 'p' is invalid since
> P(B|A) is unknown, only P(A) and P(B) are given and P(A
> and B) has to be computed. It's, like, secondary school
> algebra.

It's called the product rule idiot. It is ALWAYS valid regardless of the values. Just like modus ponens or any other /rule/ is always valid. It's, like, duh, the same reason 2 x = x + x is valid even though x is unknown. You are turning out to be an excellent example of VI.

> > g : P(A) = 1
> > g : P(B) = 1
> > s : P(~B) = 0
> > m : P(A) = P(AB) + P(A~B)
> > p : P(A) = P(AB) + P(A|~B)P(~B)
> > u : P(A) = P(AB)
> > c : P(AB) = P(A)
> > u : P(AB) = 1
>
> This is even funnier. First, we do not know what P(A|B) is
> (see above)

It doesn't matter idiot. It's a real number between 0 and 1 and it's multiplied by 0. Which means? Duh, the product is 0 REGARDLESS of it's value. Do you understand this? It's like elementary school arithmetic.

> and second the question is what kind of events might A and
> B be if the probability of either is one ? What about P(A
> or B) given the respective probabilities are one ? Is it
> two by any chance ? (I asked the same question in another
> message).

It doesn't matter what A and B are idiot. I explained this in another post AND answered your ignorant disjunction question.

> > thus
> >
> > P(A) : P(B) : P(AB)
> > 0 : 0 : 0
> > 0 : 1 : 0
> > 1 : 0 : 0
> > 1 : 1 : 1
> >
>
> Unfortunately, there can be no 'thus'.

Unfortunately, you are acting like a total VI moron at this point. Read Jaynes' book or Cox's paper, comprehend them, educate yourself. After that come here and apologize for being a VI. Before that STFU.

• Keith --