Online Papers - Brian Weatherson

Dogmatism and Intuitionistic Probability 564
Ap → p ⊣⊢ ((Ap → p) ∧ Ap) ∨ ((Ap → p) ∧ ¬Ap) (31.1)
P r (Ap → p) = P r (((Ap → p) ∧ Ap) ∨ ((Ap → p) ∧ ¬Ap)) 1, P2 ∗
(31.2)
P r ((Ap → p) ∧ Ap) ∨ ((Ap → p) ∧ ¬Ap)) =
P r ((Ap → p) ∧ Ap) + P r ((Ap → p) ∧ ¬Ap)
P3 ∗
(31.3)
P r ((Ap → p) ∧ Ap) = P r (Ap)P r (Ap → p|Ap) P6 (31.4)
P r ((Ap → p) ∧ ¬Ap) = P r (¬Ap)P r (Ap → p|¬Ap) P6 (31.5)
P r (Ap → p) =
P r (Ap)P r (Ap → p|Ap) + P r (¬Ap)P r (Ap → p|¬Ap)
2, 4, 5 (31.6)
¬Ap ⊢ Ap → p (31.7)
P r (Ap → p|¬Ap) = 1 7, P5 (31.8)
P r (Ap → p|Ap) ≤ 1 P1*, P4 (31.9)
P r (Ap → p) ≥
P r (Ap)P r (Ap → p|Ap) + P r (¬Ap)P r (Ap → p|Ap)
6, 8, 9 (31.10)
⊢ Ap ∨ ¬Ap (31.11)
P r (Ap ∨ ¬Ap) = 1 11, P1 (31.12)
P r (Ap) + P r (¬Ap) = 1 12, P3 ∗
(31.13)
P r (Ap → p) ≥ P r (Ap → p|Ap) 10, 13 (31.14)
Note (14) is an equality iff (9) is an equality or P r (¬Ap) = 0.
Theorem 2
If P r is a classical probability function, then
• P r (¬(Ap ∧ ¬ p)|Ap) ≤ P r (¬(Ap ∧ ¬ p)); and
• P r (¬Ap ∨ p|Ap) ≤ P r (¬Ap ∨ p).
Proof: Assume P r is a classical probability function, and ⊢ the classical consequence
relation.
Ap → p ⊣⊢ ¬(Ap ∧ ¬ p) (31.1)
P r (Ap → p) = P r (¬(Ap ∧ ¬ p)) 1, P2 ∗
(31.2)
P r (Ap → p|Ap) = P r (¬(Ap ∧ ¬ p)|Ap) 1, P4, P5 (31.3)
P r (Ap → p) ≥ P r (Ap → p|Ap) Theorem 1 (31.4)
P r (¬(Ap ∧ ¬ p)|Ap) ≥ P r (¬(Ap ∧ ¬ p)) 2,3,4 (31.5)
Ap → p ⊣⊢ ¬Ap ∨ p (31.6)
P r (Ap → p) = P r (¬Ap ∨ p) 6, P2 ∗
(31.7)
P r (Ap → p|Ap) = P r (¬Ap ∨ p|Ap) 6, P4, P5 (31.8)
P r (¬Ap ∨ p|Ap) ≥ P r (¬Ap ∨ p) 4,7,8 (31.9)