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

Bob Badour wrote:


Are you seriously suggesting that true and false are trivial and
uninteresting? Should we all pack up and go home?


I am suggesting that a meaningful PT statement reduction to a
propositional logic statement is trivial and uninteresting. Also,
there is a problem with the conditional probability not being equal
the probability of the conditional (see Lewis's result) which makes
conditional probabily untranslatable to modus ponens in principle.

Going in the opposite direction, generalization, PT is not truth
functional , that is the probability of a compound statement is not
determined solely by its components probabilities (see my trivial
puzzle). Also, importantly, it appears impossible to find an axiom
system for any known probabilistic logic that would be sound and
complete (except some special cases). Obviously, lack of such axiom
system makes a formal derivation (a hallmark of any logic) impossible.

Apparently, despite obvious similarities and profound connections,
both had better be used what they are best at and attempts to merge
them do not seem very productive (see abundant literature on
probabilistic logics).


What Jaynes did in his derivation of the sum/product rules has got
nothing to do with your mindless playing with formulas. See the
argument from authority in my previous messages.

Your argument from authority was flawed. I will reply in the other thread.

The argument from authority was a quote from Jaynes' book , not mine.


that should have been P(p1|p2) != P(p1).

That's assuming that P(p1|p2) even makes sense. More general
formulation of such independence is just P(p1 and p2) = P(p1)* P(p2).

The formulation is neither more general nor less general. It is, in
fact, a simple substitution of the equation describing independence:

(1) P(p1|p2) = P(p1)

Again, the substitution is possible only when P(p2) > 0. (See the
Jaynes book).


into the formula for conditional probability:

(2) P(p1 and p2) = P(p1|p2)*P(p2)

Substitute (1) into (2) gives P(p1 and p2) = P(p1)*P(p2)

.

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