From: Osher Doctorow on
From Osher Doctorow

From the physical importance of A "enwrapping" B (embedding B in
itself), which involves AB (the intersection of A and B), we are
stimulated to return to the expression:

1) P ' (A-->B) - P(A-->B)

since when B is embedded in A, that is to say P(B C A) = 1, then (1)
is 0 because P(AB) = P(B). The complete expansion of (1) is:

2) P ' (A-->B) - P(A-->B)= P(B) - P(AB) = P(A ' B)

and this leads us to evaluate the analogous:

3) P ' (A ' --> B) - P(A ' --> B) = P(AB)

The "Embedding-Expansion-Repulsion Theorem" is then the following ((4)
and (5) below):

4) P ' (A-->A ' ) - P(A-->A ' ) = P(A-->A ' ) if P(A ' ) < = P(A).

5) P ' (A ' --> A) - P(A ' --> A) = P(A ' --> A) if P(A) < = P(A ' ).

The proof is easy, and readers can try to prove it before I give the
proof.

Notice that P(A-->A ' ) is Expansion/Repulsion from the previous
posts, and P(A ' --> A) is Contraction/Attraction from the previous
posts. So Embedding is clearly related to Expansion and Contraction
by (4) and (5). The Embedding part is perhaps more obvious from (2)
and (3), noting that (3) represents embedding and (2) represents
"counter-embedding".

Osher Doctorow


From: Osher Doctorow on
From Osher Doctorow

(4) is proven by (2), and (5) is proven analogously. Recall also
that:

1) P(A-->B) = 1 + P(AB) - P(A)
2) P ' (A-->B) = 1 + P(B) - P(A) if P(B) < = P(A)
3) P(A-->A ' ) = P(A ' ) = 1 - P(A)
4) P(A ' --> A) = P(A)
5) P ' (A-->A ' ) = 1 + P(A ' ) - P(A) (if P(A ' ) < = P(A)) = 1 + (1
- P(A) - P(A)) = 2 - 2P(A) = 2(1 - P(A))

6) P ' (A ' --> A) = 1 + P(A) - P(A ' ) = 1 + P(A) - [1 - P(A)] =
2P(A)

Substituting these as appropriate into the equations to be proven
yields the desired results. For example, 2(1 - P(A)) - (1 - P(A)) =
1 - P(A), and 2P(A) - P(A) = P(A).

Osher Doctorow