I'm stuck solving problem 14 in Chapter 5 of Jeffrey's Logic of Decision.
The first part of the problem says:
Show that in presence of prob is nonnegative (prob X≥0) and prob is normalized (T=1),
(5-1) (c) if XY=F, then prob (X∨Y) = prob X + prob Y implies
(5-1) (h) if prob XY=0, then prob (X∨Y) = prob X + prob Y but not viceversa.
(Hint: what if prob assigned the value 1 to all propositions?).
I don't understand because Jeffrey says (page 76) that prob F=0. So both 5-1 (c) and (h) would be identical.
The other part of the problem is that I wouldn't know the procedure to prove this implication. A truth table seems cumbersome and inadequate. I suppose there's a better way.
Would appreciate any help regarding both the difference between XY=F and XY=0 or the procedure to prove the implication.
Thanks in advance.