1. Background
Stalnaker’s Thesis that the probability of a conditional is the conditional probability, of the consequent given the antecedent, ran quickly into serious trouble, in the first instance (famously) by David Lewis.
When I took issue with David Lewis’s triviality results, Robert Stalnaker wrote me a letter in 1974 (Stalnaker 1976). Stalnaker showed that my critique of Lewis did not save his Thesis when applied to his (Stalnaker’s) own logic of conditionals (logic C2).
Stalnaker proved, without relying on Lewis’ special assumptions:
If the logic of conditionals is C2, and for all statements A and B, P(A → B) = P(B|A) when defined, then there are at most two disjoint propositions with probability > 0.
At first blush this proof must raise a problem of a result I had presented, namely:
Theorem. Any antecedently given probability measure on a countable field of sets can be extended into a model structure with probability, in which Stalnaker’s Thesis holds, while the field of sets is extended into a probability algebra.
This theorem does not hold for a language of which the logic is Stalnaker’s C2. Rather, it can be presented equivalently as a result for a language that has the same syntax as C2, but has a weaker logic, that I called CE.
While Stalnaker acknowledged that his proof was specifically for C2, and did not claim that it applied to CE, neither he nor I showed then just how the difference between the two logics resolves the apparent tension.
Here I will show just how Stalnaker’s triviality argument does not hold for CE, with a simple counterexample.
2. Stalnaker’s Lemma
Stalnaker’s argument relies on C2 at the following point, stated without proof, which I will call his Lemma.
Definition. C = A v (~A & (A → ~B))
Lemma. ~C entails C → ~(A & ~B)
We may note in passing that these formulas can be simplified using principles that hold in both C2 and CE, for sentences A and B that are neither tautologies nor contradictions. Although I won’t rely on this below, let’s just note that C is then equivalent to [A v (A → ~B)] and ~C to [~A & (A → B)].
3. The CE counter-example to the Lemma
I will show that this Lemma has a counter-example in the finite partial model of CE that I constructed in the post “Probabilities of Conditionals: (1) Finite et-ups” (March 29, 2021).
The propositions are sets of possible outcomes of a tossed fair die, named just by the numbers of spots that are on the upper face. To begin we take propositions
p = {1, 3, 5} “the outcome is odd”
q = {1, 2, 3} “the outcome is low”
The probability of (p → q) will be P(q|p) = P(1, 3)/P(1, 3, 5) = 2/3. That is the clue to the construction of the selection function s(x, p) for worlds x = 1, 2, 3, 4, 5, 6.
In this model the choices are these. First of all if x is in p then s(x, p) = x. For the other three worlds we choose:
s(2, p) = 1, s(4, p) = 3, s(6, p) = 5
Thus (p → q) is true in 1 and 3, which belong to (p ∩ q), and also in 2 and 4, but not in 5 or 6.
Hence (p → q) = {1, 3, 2, 4}, “if the outcome is odd then it is low”, which has probability 2/3 as required.
Similarly we see that (p → ~q) = {5, 6}.
To test Stalnaker’s Lemma we define:
c = p ∪ (~p ∩ (p → ~q))
= {1, 3, 5} ∪ ({2,4, 6} ∩ {5, 6})
= {1,3, 5} ∪ {6}
= {1,3,5, 6} “the outcome is odd or 6” or “the outcome is neither 2 nor 4”
~c = {2, 4} “the outcome is 2 or 4” (the premise of the Lemma)
Now proposition c has four members, and that means that in the construction of the model we need to go to Stage 2. There the original 6 world model is embedded in a 60 world model, with each possible outcome x replaced by ten worlds x(1), …, x(10). These are the same as x, except that the selection function can be extended so as to evaluate new conditionals. The previously determined choices for the selection function carry over. For example, s(4(i), p) = 3(i), so (p → q) is true in each world 4(i), for i = 1, …, 10.
We refer to the set {x(1), …, x(10)} as [x]. So in this stage,
c = [1] ∪ [3] ∪ [5] ∪ [6]
The conclusion of the Lemma is:
c → ~(p ∩ ~q} = c → ~[([1] ∪ [3] ∪ [5]) ∩ ([4] ∪ [5] ∪ [6])]
= c → ~[5] “If the outcome is either odd or 6 then it is not 5”
What must s(x, c) be? The way to determine that is to realize again that each member of c must have probability ¼ conditional on c. Probability ¼ equals 15/60 so for example (c → {1}) must have 15 members.
Since [1] is part of c, we must set s(1(1), c) = 1(1), and so forth, through s(1(10), c) = 1(10). Similarly for the other members of c.
To finish the construction we need to get up to 15, so we must choose five worlds y not in [1] such that s(y, c) = 1. Similarly for the rest. To do so is fairly straightforward, because we can divide up the members of [2] and [4] into four bunches of five worlds each:
S(2(i), c) = 1(i) for i = 1, .., 5
S(2(j), c) = 3 (j) for j = 6, .., 10
S(4(i), c) = 5(i) for i = 1, .., 5
S(4(j), c) = 6 (j) for j = 6, .., 10
Now each conditional c → [x] is defined for each of the 60 worlds, and has probability ¼ for x = 1, 3, 5, 6.
The Lemma now amounts to this, in this model:
~c implies c → ~{[5]}
or, explicitly,
[2] ∪ [4] ⊑ [[1] ∪ [3] ∪ [5] ∪ [6]] → ~[5]
For a counter-example we look at a specific world in which ~c is true, namely world 4(1). Above we see that s(4(1), c) = 5(1). Therefore in that world the conditional c → {5(1)} is true, and hence also c → [5], which is contrary to the conclusion of the Lemma.
4. Conclusion
To recap: in this finite partial model of CE the examined instance of Stalnaker’s Lemma amounts to:
Premise. The outcome is either 2 or 4
Conclusion. If the outcome is neither 2 nor 4 then it is not 5 either
And the counter-example is that in this tossed coin model, there is a certain world in which the outcome is 4, but the relevant true conditional there is that if the outcome is not 2 or 4 then it is 5.
Of course, given that the Lemma holds in C2, this partial model of CE is not a counter-example to Stalnaker’s argument as it applies to his logic C2 or its extensions. It just removes the apparent threat to CE.
REFERENCES
Stalnaker, Robert (1976) “Stalnaker to van Fraassen”. Pp. 302-306 in W. L. Harper and C. A. Hooker (eds.) Foundations of Probability Theory, Statistical Inference, and Statistical Theories of Science. Dordrecht: Reidel.
Bas van Fraassen's philosophy blog 