On Wed, Oct 8, 2008 at 5:15 PM, Abram Demski <[EMAIL PROTECTED]> wrote:

> Given those three assumptions, plus the NARS formula for revision,
> there is (I think) only one possible formula relating the NARS
> variables 'f' and 'w' to the value of 'par': the probability density
> function p(par | w, f) = par^(w*f) * (1-par)^(w*(1-f)). Note: NARS
> truth values are more often (I think?) represented by the pair 'f'
> 'c', where 'c' is computed from 'w' by the formula c=w/(w+k), where k
> is a fixed constant. This is of little consequence at this point, and
> it was more intuitive to use 'f' and 'w' (at least for me).

At this stage, you are right. Since c and w fully determines each
other, in principle you can use either, and w is more intuitive.
However, in designing the truth-value functions, it is more convenient
to use c, a real number in [0, 1], than w, which has no upper bound.

> Here's the math. In NARS, the operation we're interested in is taking
> two pools of evidence, one concerning A=>X and the other concerning
> B=>X, and combining them to calculate the evidence they lend to A=>B.

Now things get tricky, In my derivation, in abduction/deduction the
evidence of a premise is not directly used as evidence for the
conclusion. Instead, it is the premise, as a summary of its own
evidences, that is used as evidence. That is, X is not a set, but an
individual. Consequently, the operation doesn't "taking two pools of
evidence" and somehow combine them into one pool (as in the revision
rule).

> So probabilistically, we want to determine the probability of the
> evidence for A=>X and B=>X given each possible 'par' value of A=>B.

According to the semantics of NARS, A=>X or B=>X, by itself, doesn't
provide evidence for A=>B.

Overall, it is a nice try, but given the difference in semantics
between probability theory and NARS, I'm still doubtful on how far you
can go in this direction.

Pei


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agi
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