3 ms·

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:

• cousin_it 10y ago
  It's =>, not ==. We need to show that (4) is contained in the union of (2) and (3).
  • mh-cx 10y ago
    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 10y ago
      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 10y ago
        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 10y ago
          They are arrows, logical implications :-)
          • mh-cx 10y ago
            Argh. Right. After more thinking equal or greater than wouldn't make much sense.