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

Date: Mon, 12 Jun 2006 01:12:22 GMT
Message-ID: <Wb3jg.20869\$A26.479965@ursa-nb00s0.nbnet.nb.ca>

vc wrote:

```> Keith H Duggar wrote:
> [ A lot of irrelevant chaff  skipped]
>
```

>>>1. PT is a logic generalization.
>>
>>No, I the OP claimed quote:
>>
>> "interestingly, one view of logic is as a specialization
>> of conditional probability theory. One that deals only
>> with certainty (1) and impossibility (0) rather than a
>> range of probability." -- KHD
>>
>>In other words to paraphrase myself and Cox:
>>
>> "interestingly, one view of probability theory, the Cox
>> formulation, is as a generalization of logic. One that
>> deals with a range of probability corresponding to a
>> degree of rational belief bounded in the extremes by
>> certainty (1) and impossibility (0)."
>>
>>
>>>My response was that PT is not truth functional in the
>>>sense the propositional logic is namely that the compound
>>>sentence truth, in logic, is determined by its
>>>constituent's truth values
>>
>>
>> "PT cannot be 'a generalization of logic' because PT
>> 'connectives' (+/*) are not truth functional." -- vc
>>
>>Which, I tried to explain to you is wrong for two reasons.
>>First, (+/*) are NOT connectives in PT they are the real
>>operators addition and multiplication. Second, PT uses the
>>SAME connectives as logic. The connectives haven't changed
>>they are still truth-functional as well as probability-
>>functional.
```>
> OK,  good, so you continue to claim that the  probability of P(A and B)
> is determined solely by  P(A) and P(B) just as you did before:
>
> "When you apply the connectives to a probability-valued
> statements you get probability-valued statements whose
> probability depends only on the constituent probabilities.
> "

```

I don't see how you draw the inference you draw. Clearly, P(B|A) and P(A|B) are constituent probabilities as well.

Show how to derive P(A and B) given P(A) and P(B). If you > can show that, you can claim that "probability depends only on the > constituent probabilities". Are you unable to do that ?

P(A and B) = P(B|A)P(A)
P(A and B) = P(A|B)P(B)

To derive P(A and B) one must know either P(B|A) or P(A|B) just as one must know the angle between two sides to use the cosine law. You fail to demonstrate anything useful or meaningful by your challenge.

>>>What kind of sentences do you have in mind whose
>>>probability is one?
>>
>>
>>What are you talking about? It's not hard. It's trivial and
>>irrelevant! For proving probability theorems or that
>>probability is a generalization of logic it does not matter
>>what A, B, C etc stand for! "My first name is Keith" "My
>>last name is Duggar" "My first name is Keith or my last name
>>is Duggar". Does that satisfy you? I hope so because it is
>>TOTALLY irrelevant (and somewhat VI honestly).
>>
>>[disjunction nagging]
>>
>>
>>>2. In some case, namely when probabilities are 0 and 1,
>>>the probabilistic statements 'reduce' to logical
>>>statements. I asked to provide two or more statements
>>>whose probability would be one and show what the
>>>probability of the disjunction of such statements might
>>>be. There has been no answer. Are you unable to answer the
>>>question ?

```>
> What you've provided is not a meaningful example,  but just two
> mutually irrelevant true propositions (similarly to introducing
> irrelevancy with penguins and the leaking roof in the Jaynes book).  To
> provide a meaningful example of reducing probabilities, take for
> example two relevant statements A='It will rain today' and 'The roof
> will leak' (see the same book), assign priors and show how P(A and B)
> can be equal to one.
>
> Also,  your 'basic proof' is quite meaningless (as I pointed out in
> another message):
>
>
```

>>One last time I will provide one of these basic proofs. And
>>I will provide more than you ask for, that is below is a
>>proof for the complete reduction to logic in the limit of
>>certainty (1) and impossibility (0).

[light bulb goes on]

Keith already specified the argument applies to a limit.

>>P(A) = 1
>>P(A or B) = P(~(~A and ~B))
>>P(A or B) = 1 - P(~A and ~B)
>>P(A or B) = 1 - P(~B|~A)P(~A)

```>
>
> Since P(~A) equals zero,  the above statement does not make sense.

```

But Keith stated "in the limit of". Thus, one could read the first line of what he wrote as

lim P(A) as P(A) -> 1

which makes the other factor

lim P(~A) as P(~A) -> 0.

Thus, in the limit as P(A) approaches 1, P(A or B) = 1 - P(~B|~A)(1-P(A)) = 1

If he rewrote the proof using explicit limit notation, would you still object to his proof? Received on Sun Jun 11 2006 - 20:12:22 CDT

Original text of this message

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