[BrandonSmithJ]

I don't quite follow your proof on the last two steps; would you mind explaining? You say it's by logical operations, but I'm missing something because I see:

(A && B && C) || (A && D && C) == (A && B) || (D && C)

[cousin_it]

It's =>, not ==. We need to show that (4) is contained in the union of (2) and (3).

[mh-cx]

I think you misunderstood his question. I have the same problem.

How do you get from this line:

    (A1Y and B2Y and B3Y) or (A1Y and A2Y and B3Y) 

to this

    (A1Y and B2Y) or (A2Y and B3Y)

BrandonSmithJ just did some replacement to make it easier to read. Maybe he shouldn't have used the same letters to avoid confusion.

Let's write it differently: Why are these terms equivalent?

   (u && v && w) || (u && x && w)
   (u && v)  || (x && w)

[cousin_it]

They are not equivalent. The former term logically implies the latter term. That means the set of situations described by the former term is a subset of the situations described by the latter term. That means the probability of the former set should be less or equal than the probability of the latter set.

[mh-cx]

Ah, makes sense now. Thanks!

Now I also get your previous comment. I've read your => as arrows instead of "equal or greater".

[cousin_it]

They are arrows, logical implications :-)

[mh-cx]

Argh. Right. After more thinking equal or greater than wouldn't make much sense.