1

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 Dec 26 '18 at 15:58
  • Yes, I checked. It says "Show that in the presence of (5-1) (a,b),...". – martin Dec 26 '18 at 16:05
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.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.