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.

  • Was the question copied correctly? I'm not sure what "presence of" would mean here, and it seems like prob can't be its object.
    – Dan Hicks
    Commented Dec 26, 2018 at 15:58
  • Yes, I checked. It says "Show that in the presence of (5-1) (a,b),...".
    – martin
    Commented Dec 26, 2018 at 16:05

1 Answer 1


See page 76 :

For the impossible proposition, the proposition which is false in all cases, we shall use the special symbol F and set

prob F=0.

Thus, to say that XY=F means to say that XY is never true.

Axiom 5-1(c) says :

prob is additive: if XY = F, then prob (X v Y) = prob X + prob Y.

See page 81; using Ax.5-1(c), Jeffrey proves (5-1)(g) :

prob (A v B) = prob A + prob B - prob AB.

Thus, if we assume prob AB=0, using (5-1)(g) [that derives from (5-1)(c)] we conclude that :

prob (A v B) = prob A + prob B.

The "Hint : what if prob assigned the value 1 to all propositions?" allows us to check the difference between (5-1)(c) and (5-1)(h).

Consider : (5-1)(d). We have AA' = F [I've used A' for the denial of A] and we can set X = A and Y = A' in (5-1)(h) [now we have it instead of (5-1)c)] to get ...

And here we are stuck; if all propositions have prob = 1, then also prob AA' = 1 and thus from the fact that AA' = F we cannot deduce that prob AA' = 0 and thus we cannot apply the antecedent of (5-1)(h), i.e. prob AA'=0, to derive the fundamental :

prob A' = 1 - prob A.

In fact, if all propositions have prob =1, then : prob A = prob A'=1.

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