Online Papers - Brian Weatherson

Dogmatism and Intuitionistic Probability 567
If p is an ⊢I L-thesis, then p ∧ r ⊣⊢ r , so P r3 ( p ∧ r ) = P r3 (r ), so P r3 ( p|r ) = 1, as
required for (P1).
If p ⊢I L q then p ∧ r ⊢I L q ∧ r . So P r3 ( p ∧ r ) ≤ P r3 (q ∧ r ). So P r3 ( p∧r ) (q∧r )
P r3 (r ) ≤ P r3 . P r3 (r )
That is, P r 3 ( p|r ) ≤ P r 3 (q|r ), as required for (P2).
Finally, we need to show that P r 3 ( p|r ) + P r 3 (q|r ) = P r 3 ( p ∨ q|r ) + P r 3 ( p ∧
q|r ), as follows, making repeated use of the fact that P r 3 is an unconditional ⊢ I L -
probability function, so we can assume it satisfies (P3), and that we can substitute
intuitionistic equivalences inside P r 3 .
P r3 ( p|r ) + P r3 (q|r ) = P r3 ( p ∧ r )
P r3 (r ) + P r3 (q ∧ r )
P r3 (r )
= P r3 ( p ∧ r ) + P r (q ∧ r )
P r3 (r )
= P r3 (( p ∧ r ) ∨ (q ∧ r )) + P r3 (( p ∧ r ) ∧ (q ∧ r ))
P r3 (r )
= P r3 ( p ∨ q) ∧ r ) + P r3 (( p ∧ q) ∧ r )
P r3 (r )
= P r3 ( p ∨ q) ∧ r )
+
P r3 (r )
P r3 (( p ∧ q) ∧ r )
P r3 (r )
= P r 3 ( p ∨ q|r ) + P r 3 ( p ∧ q|r ) as required
Now if r ⊢ I L p, then r ∧ p I L ⊣⊢ I L p, so P r 3 (r ∧ p) = P r 3 ( p), so P r 3 ( p|r ) = 1,
as required for (P5).
Finally, we show that P r 3 satisfies (P6).
P r3 ( p ∧ q|r ) = P r3 ( p ∧ q ∧ r )
P r3 (r )
= P r3 ( p ∧ q ∧ r ) P r3 (q ∧ r )
P r3 (q ∧ r ) P r3 (r )
= P r 3 ( p|q ∧ r )P r 3 (q|r ) as required
Theorem 6 Let x be any real in (0,1). Then there is a probability function
C r that (a) is a coherent credence function for someone whose credence
that classical logic is correct is x, and (b) satisfies each of the following
inequalities:
P r (Ap → p|Ap) > P r (Ap → p)
P r (¬Ap ∨ p|Ap) > P r (¬Ap ∨ p)
P r (¬(Ap ∧ ¬ p)|Ap) > P r (¬(Ap ∧ ¬ p))