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From: Osher Doctorow on 3 Jul 2010 03:47
From Osher Doctorow

It is fairly easy to prove, as I have done earlier in this thread,
that:

1) P(A-->B) + P(B-->A) = 1 + P(A<-->B)
2) P(A-->B) - P(B-->A) = P(B) - P(A)

To prove (1), simply consider P{(A-->B) U (B-->A)} which is easily
shown to be 1 since (A-->B) U (B-->A) = A ' U B U B ' U A. Then use
P(C U D) = P(C) + P(D) - P(CD) for any (random) sets C, D. The proof
of (2) is easy by direct substitution of the definitions.

We can write in LISP-like notation:

3) P(A-->B) = 1 + P(AB) - P(A) = (1, B, A)
4) P(B-->A) = 1 + P(AB) - P(B) = (1, A, B)

Define the Operators:

5) REVERSE(P(A-->B)) = P(B-->A)
6) CORREL(P(A-->B)) = P(A<-->B)

Then we can write (1) as:

7) P(A-->B) + REVERSE(P(A-->B)) = 1 + CORREL(P(A-->B))

We can write (2) as:

8) P(A-->B) - REVERSE(P(A-->B)) = P ' (A-->B) - 1

at least formally. Whether P(B) < = P(A) in (8) is required will be
left for another time.

Osher Doctorow
From: Osher Doctorow on 3 Jul 2010 04:01
From Osher Doctorow

In LISP-like notation, we can write (1) above as:

1) P(A-->B) + P(B-->A) = (1, B, A) + (1, A, B) = 1 + CORREL(1, B, A)

Similarly, for (2):

2) P(A-->B) - P(B-->A) = (1, B, A) - (1, A, B) = P ' (A-->B) - 1

at least formally. Notice that we do not add lists like (1, B, A) +
(1, A, B) componentwise, and similarly we do not subtract them
componentwise. Their sums and differences, however, are related to
important Operators and to +/- 1. If we want to use {1, B, A} to
mean P ' (A-->B) = 1 + P(B) - P(A), then in (2) P ' (A-->B) - 1 can be
written {1, B, A} - 1 formally.

Osher Doctorow
From: Osher Doctorow on 3 Jul 2010 08:05
Let us write:

1) [0, B, A] = CORREL(1, B, A) = P(AB) + P(A ' B ' )
2) [1, B, A] = 1 + CORREL(1, B, A) = 1 + P(AB) + P(A ' B ' )

Then (1) of the previous post becomes:

3) P(A-->B) + P(B-->A) = (1, B, A) + (1, A, B) = [1, B, A]

Next, from the paragraph below (2) of the previous post, we have:

4) {1, B, A} = P ' (A-->B)

Let us write:

5) {0, B, A} = P ' (A-->B) - 1 = {1, B, A} - 1

Then (2) of the previous post becomes:

6) P(A-->B) - P(B-->A) = (1, B, A) - (1, A, B) = {0, B, A}

Notice in (6) that the subtraction of the 1s become 0 with a change in
list symbols, while in (3) the addition of 1s becomes 1 with a change
in list symbols.

We have 3 list symbols now, representing P(A-->B), P ' (A-->B), and
P(A<-->B), the 3 types of Probable Causation/Influence and/or Probable
Correlation, +/- 1, namely (A, B, C), {A, B, C}, and [A, B, C], which
potentially can describe quark triples, the genetic code triple
codons, etc.

Osher Doctorow
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