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vc wrote:
> Bob Badour wrote:
>
>>Are you seriously suggesting that true and false are trivial and >>uninteresting? Should we all pack up and go home?
Then why did you bring it up in the first place?
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.
Modus ponens requires a conditional probability of 1 and says nothing about situations with other conditional probabilities which is why the conditional "If A then B" says nothing about B when A is false.
For modus ponens, one must start with the premise "If A then B", which is synonymous with P(B|A) = 1. To have a sound argument, one must not only have a valid argument, but the premise "If A then B" must be true.
It think your argument might cause more problems for modus tollens than for modus ponens.
> 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).
I fail to see what an indeterminate problem demonstrates. If I give you the lengths of two sides of a triangle and ask you for the length of the third side, you cannot answer unless you know the angle where the given sides meet.
Such a challenge would neither disprove pythagoras nor disprove the law of cosines. Neither would it disprove that the law of cosines generalizes the pythagorean theorem.
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.
That precludes it from deductive logic but not from inductive logic.
> 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).
Are you suggesting that they need merging? Classical mechanics is a special case of relativity. Do they need merging? The pythagorean theorem is a special case of the cosine law. Do they need merging? Deductive logic is a special case of inductive logic. Do they need merging?
Each of the special cases has properties that do not generalize and each of the generalizations considers factors irrelevant to the special case.
The real question is whether inductive logic is appropriate for data management, and I do not think it is.
With a deductive formalism, one knows that the answers one gets are as valid as the operations and as sound as the premises. With inductive formalisms, one has no such knowledge.
Inductive logic, on the other hand, seems to have utility in science for gauging the validity of an hypothesis when one lacks any clearly contradictory evidence.
>>>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 source of the quote is irrelevant because the flaw itself was a lack of relevance. Keith never relied on the meaning of any meaningless value. The fact that the value of x might be indeterminate is unimportant to the conclusion that x times zero is zero because the conclusion holds for all x.
>>>>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)
Okay, the formula for independence only applies when P(p2) > 0. And how, exactly, does that support your assertion that one formulation is more or less general than the other?
>>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)Received on Sun Jun 11 2006 - 17:18:31 CDT