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.
When Sellars wrote his article on causal modalities and counterfactual conditionals in 1958, he was reacting to the arguments of Chisholm and Goodman. Sellars sets out to disarm those arguments. In doing so he develops a theory of conditionals that accommodates the troubling and puzzling examples. It is quite a decent theory of conditionals, when it is put in modern form, and could have held its ground in debates in the years before Stalnaker and Lewis changed the game.
The core of the problem page 1
Sellars’ representation of nature: the thing-kind framework page 2
Conditionals in this framework page 2
What conditionals are in this framework page 3
Sellars’ theory in modern form page 4
Principles that hold for Sellars’ conditionals page 5
Handling the examples about counterfactuals page 7
Nesting and iteration of conditionals page 8
1. The core of the problem
One point about conditionals was central and crucial. It had been assumed that the logic of conditionals must mimic the logic of valid arguments. That would mean that Weakening would hold: an argument will remain valid if extra premises are added. If A implies C then (A & B) implies C, and by that mimicry, “if A then C” must imply “If (A & B) then C”. But “this match would light if struck” does not imply “this match would light if moistened and struck”.
The crucial point about the failure of Weakening is that reasoning under supposition will fail for conditionals. For Weakening can be ‘proved’ as follows:
if A then C (Given)
A & B Supposition
A 2 conjunction elimination
C 1, 3 modus ponens
if A & B then C 2-4, conditional introduction
The rule of conditional introduction is the principle for discharging a supposition, which works fine for valid arguments. It fails when mimicked for conditionals.[1] Given this, there quickly follow other examples of reasoning with conditionals which are not in accord with earlier logics, such as C. I. Lewis’ modal logic of ‘strict implication’.
2. Sellars’ representation of nature: the thing-kind framework
To make his point, Sellars introduces what he takes to be the canonical form of the language of science, namely the language of the thing-kind framework. That is “the conceptual framework in terms of which we speak of what things do when actedupon in certain ways in certain kinds of circumstance” (1958: 225).
The basic pattern he describes as follows:
Suppose we have reason to believe that
𝜙-ing Ks (in circumstances C) causes them to 𝜓
(where K is a kind of thing – e .g., match). Then we have reason to believe of a particular thing of kind K, call it x, which is in C, that
x, would 𝜓, if it were 𝜙-ed.
And if it were 𝜙-ed and did 𝜓, and we were asked “Why did it 𝜓?” we would answer, “Because it was 𝜙-ed”; and if we were then asked, “Why did it 𝜓 when 𝜙-ed?” we would answer “Because it is a K.” If it were then pointed out that Ks don’t always 𝜓 when 𝜙-ed, we should counter with “They do if they are in C, as this one was.” (Sellars 1958: 248.)
The point is that the antecedent is an input (action or interaction) statement, the consequent an output statement, and neither input nor output statements describe circumstances (standing conditions). As an example I would take: “if this vase be thrown against the wall it will break”.
Sellars is not easy to read, but this article combined with the earlier “Concepts as involving laws and inconceivable without them” (1948), yields a clear and simple account of just what proposition is expressed by a conditional, on Sellars’ view.[2]
A given thing x of kind K will have a history, which is a trajectory in its state-space. What is characteristic of kind K is not only that specific state-space, but a selection from the set of logically possible histories: the family of histories such a thing can really have, its nomologically possible histories.
“In speaking of a family of possible worlds, what are we to understand by a “world”? Let us begin with the following: A world is a spatio-temporal structure of atomic states of affairs which exhibits uniformities of the sort we have in mind when we speak of the laws of nature.” (1948: 293)
This passage he immediately follows with the admonition to abandon the term “world”, and to speak of possible histories instead:
“Our basic framework is thus a family of possible histories, one of which is the actual and possible history.” (ibid.)
In that framework, kind K is characterized by a restricted family of possible histories: the histories which alone are possible for things of kind K.
3. Conditionals in this framework
Starting with the basic pattern in the quoted passage from Sellars, it is clear that “x, would 𝜓, if it were 𝜙-ed” is true in a situation where x’s kind and the circumstances are ‘just right’. It seems also that in the same conditions “x will 𝜓 , if it is 𝜙-ed” is true, and that this is in addition quite independent of whether the antecedent is true or false. But in a situation where the kind and circumstances are not ‘just right’ we need to think about other conditions. It may be that the conditional is just false, but it is also possible that the antecedent is impossible in those conditions, or the antecedent is possible but not in combination with the consequent. (I will give several examples below.) In that case, the conditional is not just a counterfactual, but a counterlegal.
Rather than ‘proof-texting’ with quotations, here is my summary for the special case in the 1958 passage quoted above:
Explication. To assert “x, would 𝜓, if it were 𝜙-ed” is to assert that the situation is felicitous for this conditional to be true.
The kind and circumstances are felicitous for that conditional exactly if there is a kind K to which x belongs, and there are circumstances C that x is presently in, which are together compossible with x being 𝜙-ed and 𝜓-ing, and it is the case that in all histories possible for any things of kind K in circumstances C, that they 𝜓 when 𝜙-ed.
I italicized “any” to show that on this understanding there is an underlying universal claim, overriding differences that may mark individuals of the same kind in the same circumstances.
The condition of compossibility is crucial to accommodate puzzling examples.
Example 1: There may be a kind of match which lights when moistened (with a special chemical that oxidizes rapidly when wet) and does not light if struck, for striking it removes the chemical. There is also the familiar kind of match which will light if struck only when dry. But there is no kind of match which will light when both moistened and struck. The antecedent is not impossible: take any match at all, and at will you can both moisten and strike it. But its combination with the consequent is impossible.
Example 2: The electric circuit in this room is such that the light will go on if either switch A or switch B, but not both, is in the down position. Currently both switches are in the up position. So both conditionals “if A is clicked the light will be on” and “If B is clicked the light will be on” are true, but the conditional “If A is clicked and B is clicked the light will be on” is false. Again, the antecedent is not impossible: you can click both switches. But the result will not be to turn on the light.
4. What conditionals are in this framework
In this Explication, the conditional in question is construed as an existential statement of fact: there is a kind K and circumstance C that characterize this situation such that …. An existential statement is in effect a disjunction, generalized to cover the case of infinitely many alternatives. So “If A then B” will amount to:
the case described in A amounts to a disjunction of combinations of kinds and circumstances {K(i) and C(i): i is in J}, each of which is compossible with A and B, and for which all histories that satisfy A also satisfy B
The implication relation signified by ‘all histories that satisfy … also satisfy …’ encodes the usual valid argument relationship.
So, on this view, a conditional is a disjunction involving a modality. Weakening does not hold for conditionals, due to the compossibility requirement.
We note one oddity, a difference from more familiar theories of conditionals, that has to do with modality. Suppose A logically implies B. In familiar theories of conditionals the conditional ‘if A then B’ is then automatically true. But there may be kinds K and circumstances C which are not compossible with A and B. In that case the conditional “if A then B” will not be logically valid. In fact, in that case this conditional is simply the disjunction of all kinds K(i) and circumstances C(i) compossible with A and B.
For example if something is scratched with a human nail then it is scratched with something human. But a steel vase cannot be scratched with a human nail. So then, on this Sellarsian account, it is not correct to assert, when looking at a steel vase in normal circumstances, that it would be scratched by something human if it were scratched by a human nail.
Yes, it is odd. But if someone were to assert that, while we are looking at a steel vase in normal circumstances, that would be very odd too! In this case, the counterfactual is a counterlegal conditional, and counterlegals pose a host of further difficulties for our intuitions (see e.g. Fisher 2017).
5. Sellars’ theory in modern form
Sellars 1948 and 1958 together spell out the thing-kind framework, with the elaboration into the possible histories of things of certain kinds in certain circumstances, but not in the precise form that we now require. Yet it is clear enough to provide a fairly straightforward way to put this in modern form.
If the main concepts are to be generalized, intuitively, then we can read “kind” and “circumstance” to stand for “whatever grounds the laws of nature” and “boundary conditions”. But I will keep the symbols K and C and the mnemonic “kind” and “circumstance” throughout.
A model will be a triple <H, P, F>, each element a non-empty set: I will call H the set of situations, P is a partition of H, and F, the family of propositions, is a set of subsets of H.
The elements of H, which correspond to Sellars’ possible histories, are also triples <K, C, 𝜋>. We think of possible situations as involving a single particular (whether a physical system or a whole world). This particular is classified as being of a certain kind K, as being in certain circumstances C, and as having a history 𝜋 (which could have various further attributes, not specifically spelled out here, such as being 𝜙-ed).
The family F may not be the family of all subsets of H. But we specify that it includes both H and the empty set Λ, and is closed under intersection, union, and relative complement. So F is a complete field (Boolean algebra) of sets. But moreover we specify that P is part of F, where we now define P as the family of all sets <K, C> with the definition:
<K, C> = {x in H: there is a factor 𝜋 such that x = <K, C, 𝜋>}.
In other words, a cell of this partition P is the set of all possible histories for an individual of kind K in circumstances C. How many cells does this partition have? How many kinds, and how many circumstances, are there that can go together in this way? We leave that open, so there are large models and small models, even models with just a single cell in that partition.
When are kind and circumstances in a given cell X of Pfelicitous for the conditional A →B? Compossibility is just being possible together. So what is required is that X ∩ A ∩B is not empty.
Abbreviation. “◊( X ∩ A ∩B)” for “( X ∩ A ∩B) ≠ Λ”
We can then define the operation that forms conditionals:
Definition. A → B = ∪{X in P: ◊( X ∩ A ∩B) and X ∩ A ⊆ B}
Abbreviation. To facilitate reading I will abbreviate “◊( X ∩ A ∩B)” to “◊” whenever the context allows that to be clear.
By this definition, every such conditional is a proposition, for if A and B are in F then A → B is a union of cells in partition P, all of which belong to F, which is closed under union. So F is closed under the operation →.
A conditional proposition is in general quite a large set of situations. It comprises all the situations in which the kind and circumstances are just right to make B follow from A. That is just what we saw above, in Sellars’ exposition, in terms of disjunctions.
Truth. If x is a situation and A is a proposition then A is true in x exactly if x is a member of A.
Notice that if x and y are both situations belonging to a given cell X of P then any conditional true in x is true in y. The special factor plays no role in the truth conditions for conditionals, though it may encode many other aspects of that situation, and figure in the truth conditions of sentences that involve no arrows.
What principles hold or fail in the logic of Sellars’ theory of conditionals?
6. Principles that hold for Sellars’ conditionals
Modus Ponens. A ∩ (A → B) ⊆ B
If A and A →B are both true in x then B is true in x.
For if the premises are true in situation x, then x is in A and x is in ∪{X in P: ◊ & X ∩ A ⊆ B}. But if x is in the union of the family {X in P: ◊ & X ∩ A ⊆ B} then x is in one of the cells, Xo, in that family. So x is in A and in Xo and X ∩ A ⊆ B}. there x is in B.
Impossibility. (A → ~ A) = Λ
For there is no cell X such that ◊( X ∩ A ∩ ~A) .
Definition. ◊(A, B) = ∪{X in P: ◊( X ∩ A ∩B)}
Entailment. If A ⊆ B then (A → B) = ◊(A, B)
Forif A ⊆ B then, for any cell X in P it is the case that ◊( X ∩ A ∩B) and X ∩ A ⊆ B if and only if ◊( X ∩ A ∩B).
This has the corollary that (A → A) = ◊(A, A).
The principle that A → B implies A → (A & B) does not seem to have a standard name; I’ll call it Carry-over. This has the corollaries that A → ~ A implies A → (A & ~A), and that ~A → A implies ~A → (A & ~A), which mark the cases in which A is impossible or necessary.
In terms of propositions:
Carry-over. (A → B) ⊆ [A →(A ∩ B)]
For suppose situation y is in the union of the cells in the family {X in P: ◊ & X ∩ A ⊆ B), and hence in a specific cell Xo in that family. Since Xo, A, B are compossible then so are Xo, A, A ∩ B. And since Xo ∩ A ⊆ B it is the case that Xo ∩ A ⊆ A ∩ B. So y is in [A → (A ∩ B)].
In normal modal logic the principle that if B1, … , Bn implies C then □B1, … , □Bn A implies □C, is sometimes called Goedel’s rule. So I’ll choose that to name the analogous principle for conditionals: If A, B1, … , Bn implies C then A → B1, … , A → Bn implies A → C. Notice that this is much weaker than the Classic or Intuitionist principle that if A, B ├ C then B ├ A → C.
In terms of propositions rather than sentences it becomes:
Goedel. If (B1 ∩ … ∩ Bn) ⊆ C then(A → B1) ∩ … ∩ (A → Bn) ⊆ (A → C)
Without real loss of generality I will provide the argument just for the case n = 2.
Suppose B1 ∩ B2 ⊆ C. Suppose also that (A → B1) ∩ (A → B2) is true in y. It follows that y is in a cell X such that ◊(X ∩ A ∩B1) and X ∩ A ⊆ B1, and that y is in a cell Y such that ◊( Y ∩ A ∩ B2) and Y ∩ A ⊆ B2.
Since y can only be in one cell, it must be a cell Xo such that ◊(Xo ∩ A ∩B1) and ◊( Xo ∩ A ∩ B2) and Xo ∩ A ⊆ B1and Xo ∩ A ⊆ B2. Therefore Xo ∩ A ⊆ C.
In addition, because ( Xo ∩ A ∩B1) is not empty it follows that ( Xo ∩ A) is not empty, and since that is part of C, it follows that ◊( Xo ∩ A ∩ C).
Therefore Xo, which contains y, is part of the union of the cells X such that ◊( Xo ∩ A ∩ C) and X ∩ A ⊆ C, which is A → C. Therefore A → C is true in y.
The Principle that “and” distributes over “if … then”, A → (B & C) ├ (A → B) & (A → C), and conversely, has its formulation in terms of propositions:
∩-Distribution 1. A → (B ∩C) ⊆ [(A → B) ∩ (A → C)]
If cell X is such that X ∩ A ⊆ B ∩ C then it is certainly such that X ∩ A ⊆ B and X ∩ A ⊆ C.
Moreover, if ◊( X ∩ A ∩ B ∩ C) then ◊( X ∩ A ∩ C).
∩-Distribution 2. [(A → B) ∩ (A → C)] ⊆ [A → (B ∩C)]
For suppose that situation y is in both [(A → B) and (A → C). Since y cannot be in more than one cell, y is in some cell Xo such that ◊( X ∩ A ∩ B) and ◊( X ∩ A ∩ C), and moreover X ∩ A ⊆ B and X ∩ A ⊆ C. The latter two facts entail that (X ∩ A) ⊆ (B ∩ C). The former two facts each entail that X ∩ A is not empty. But together with (X ∩ A) ⊆ (B ∩ C) that implies that ◊( X ∩ A ∩ B ∩ C).
Of the corresponding principle, that “or” distributes over “if .. then”, generally called Conditional Excluded Middle or CEX, one part holds but the other part fails, as we will see below.
CEX 1. (A → B) ∪ (A → C) ⊆ [A → (B ∪ C)]
For if X ∩ A ⊆ B then X ∩ A ⊆ (B∪ C). Note that ◊( X ∩ A ∩B) implies that ◊[ X ∩ A ∩ (B ∪ C)] . We may also note that CEX 1 follows from the simpler principle (again I don’t have a standard name, but it is a special case of Goedel):
Weakening on the right. If B ⊆ C then (A → B) ⊆ (A → C).
For if ◊( X ∩ A ∩B) and B ⊆ C then ◊( X ∩ A ∩ C).
7. Handling the examples about counterfactuals
The examples I gave above with matches and light switches can be represented by little models of this sort. So we know, for example, that Weakening fails for conditionals, in general.
Weakening. If A → C ⊆ [(A ∩ B) → C]. FAILS.
The simple reason is that ◊( X ∩ A ∩ C) does not guarantee that ◊( X ∩ A ∩ B ∩ C).
CEX2. [A → (B ∪ C)] ⊆ (A → B) ∪ (A → C). FAILS.
For example, C = ~B, then the part of A in any cell will be included in (B ∪ ~B) but there will in general be cells in which A is not included either in B or in ~B, but overlaps both.
Centering. A ∩ B ⊆ (A → B). FAILS.
Suppose that y is in A ∩ B. Then there is a cell Xo in P such that y is in Xo ∩ A ∩ B. Since P is a partition, the cells do not overlap, so y can be in the union of the cells {X in P: ◊ & X ∩ A ⊆ B} only if it is in one of those cells. But Xomay not be one of those, it may be a cell where A overlaps both B and ~B.
There is a traditional principle that is much discussed in this subject area, Import-Export (A & B) → C ├ A → (B → C), and its converse A → (B → C) ├ [(A & B) → C].
Counterexample. There is only one cell, X. A is non-empty, but A and B do not overlap, B is a non-empty part of C. So (B → C) = X. X is compossible with A and A is part of X, hence of (B → C). So it is also the case that X = [A → (B → C)].
Since B is part of C, (A ∩ B) is also part of C, but (A ∩ B) is empty, So (A ∩ B ∩ C) is empty. So X is not [(A ∩ B) → C], which is in fact the empty set, and does not have X as a part.
Import-Export 2. (A ∩ B) → C ⊆ [A → (B → C)]. FAILS.
Counterexample (see Diagram below). In this model there are exactly two cells, X0 and X1.
◊( Xo ∩ A ∩B ∩ C) and Xo ∩ A ∩ B ⊆ C, but Xo ∩ B is not part of C.
X1 ∩ A ∩ B is not empty but it is not part of C, and hence also X1 ∩ B is not part of C.
From these data we derive the following:
[(A ∩ B) → C] = X0
(B → C) = Λ
Neither (Xo ∩ A) nor (X1 ∩ A) is empty
[A → (B → C)] = Λ
Therefore (A ∩ B) → C is not included in [A → (B → C)].
Let’s also show how we can model the example of the two switches.
Example. The light will go on if either switch A or switch B, but not both, is in the down position. Currently both switches are in the up position. So both conditionals “if A is clicked the light will be on” and “If B is clicked the light will be on” are true, but the conditional “If A is clicked and B is clicked the light will be on” is false. Again, the antecedent is not impossible: you can click both switches, and the result will be that the light is still off.
The model is very simple. There is only one cell X. A and B are both non-empty, and each is part of C. So X = (A → C) = (B → C). But A and B do not overlap, so (A ∩ B) → C] = Λ.
8. Nesting and iteration of conditionals
Equating conditionals with suitably chosen disjunctions (finite or infinite), if accepted in general for all conditionals, has therefore yielded a specific family of logical principles to govern reasoning with conditionals.
The failure of Import-Export, in both directions, shows that nesting of conditionals in Sellars’ theory is not trivial. The nested conditional [A → (B → C)] is not in general equivalent to a statement in which there is no nesting of conditionals.
But this flexibility does not go very far. For Sellars, counterfactual conditionals are grounded on underlying lawlike strict conditionals, due to what is the case in all possible histories for things of a given kind. That is a differentiating feature, it is unlike anything along the lines of ‘nearest possible world’ relations, on which conditionals would be seen as grounded by Stalnaker and Lewis.
As a putatively difficult example, take
(*) If this vase would break if thrown against the wall then it would break if dropped on the floor
This has the form (A → C) → (B → C).
To begin, let us see how the antecedent can imply the consequent. There is a certain list of kinds of vases (glass, ceramic, steel) and circumstances (wall of brick, of wool, …). For some of these combinations of kind and circumstance it is the case that those vases will always, necessarily, break if thrown against the wall. Consider the antecedent conditional “If this vase were to be thrown against the wall then it would break”, and two cases:
the conditions are normal and this is a steel vase
the conditions are normal and this is a porcelain vase
In case 1. the antecedent conditional is a counterlegal. In the second case it is lawlike and true. But in that second case, the consequent “If this vase were to be dropped on the floor then it would break”, is also lawlike and true. We might put it this way: in neither case is this matter very iffy.
Reduction of a certain nesting. (A → C) → (B → C) = (A → C) ∩ (B → C)
For suppose first that y is in cell Xo and is in (A → C) → (B → C). Then ◊( X ∩ (A → C) ∩ (B→C)) and Xo ∩ (A → C) ⊆ (B → C).
But Xo ∩ (A → C) is non-empty only if Xo is a member of the family {X in P: ◊( X ∩ A ∩C) & X ∩ A ⊆ C}, for distinct cells do not overlap. Therefore (A → C) is true in y. Similarly, Xo ∩ (B → C) is non-empty only if Xo is a member of the family {X in P: ◊( X ∩ B ∩ C)& X ∩ B ⊆ C}, and so (B → C) is true in y as well.
Secondly, suppose y is in cell Xo and is in (A → C) ∩ (B → C). Then Xo belongs to both families {X in P: ◊( X ∩ A ∩ C) & X ∩ A ⊆ C} and {X in P: ◊( X ∩ B ∩ C) & X ∩ B ⊆ C}. Therefore the intersections of Xo with (A → C) and with (B → C) are not empty but both equal Xo itself. Hence also Xo ∩ (A → C) is part of (B → C), trivially, so y is in (A → C) → (B → C).
REFERENCES
Fisher, Tyrus (2017) “Counterlegal dependence and causation’s arrows: causal models for backtrackers and counterlegals”. Synthese 194: 4983-5003.
Fitch, Frederick (1952) Symbolic Logic. New York: Ronald Press.
Sellars, Wilfrid (1948) “Concepts as involving laws and inconceivable without them”. Philosophy of Science 15: 287-315.
Thomason, Richmond H. (1970) “A Fitch-style formulation of conditional logic”. Logique et Analyse Nouvelle Série 13: 397-412.
NOTES
[1] Fitch had modified reasoning under supposition for modal logic in 1952, and in 1970 Richmond Thomason would similarly modify reasoning under supposition for the logic of conditionals.
[2] It is a drawback for this account that it does not distinguish between present tense subjunctive and present tense indicative conditionals. For Sellars’ story will be the same, in this summary, if “would” and “were” are replaced by “will” and “is”. We may for now just note that the typical examples to distinguish subjunctive and indicative conditionals are not present tense, however, but of the “what would have been if” type.
My previous blog post on this subject was quite abstract. To help our imagination we need to have an example.
Result that we had. Let A→ be a Boolean algebra with additional operator →. Let P(A→) be the set of all probability measures on A→such that m(a → b) = m(b|a) when defined (“Stalnaker’s Thesis’ holds). Then:
Theorem. If for every non-zero element a of A→ there is a member m of P(A→) such that m(a) > 0 then P(A→) is not closed under conditionalization.
For the example we can adapt one from Paolo Santorio. A fair die is to be tossed, and the possible outcomes (possible worlds) are just the six different numbers that can come up. So the proposition “the outcome will be even” is just the set {2, 4, 6}. Now we consider the proposition:
Q. If the outcome is odd or six then, if the outcome is even it is six.
For the probability function m we choose the natural one: the probability of “the outcome will be even” is the proportion of {2, 4, 6} in {1, 2,3, 4, 5, 6}, that is, 1/2. And so forth. Lets use E to stand for “the outcome is even” and S for “the outcome is six”. So Q is [(~E ∪ S)→ (E → S)].
PLAN. What we will do is first to determine m(Q). Then we will look at the conditionalization m# of m on the antecedent (~E v S), and next on the conditionalization m## of m# on E. If everything goes well, so to speak, then the probability m(Q) will be the same as m##(S). If that is not so, we will have our example to show that conditionalization does not always preserve satisfaction of Stalnaker’s Thesis.
EXECUTION. Step One is to determine the probability m(Q). The antecedent of Q is (~E ∪ S), which is the proposition {1, 3, 5, 6}. What about the consequent, (E → S)?
Well, E → S is true in world 6, and definitely false in worlds 2 and 4. Where else will it be true or false?
Here we appeal to Stalker’s Thesis. The probability of (E →S) is the conditional probability of S given E, which is 1/3. So that proposition (E → S) must have exactly two worlds in it (2/6 = 1/3). Since it is true in 6, it must also be true in precisely one of {1, 3, 5}. Which it is does not affect the argument, so let it be 5. Then (E → S) = {5, 6}.
Now we can see that the probability of Q is therefore, by Stalnaker’s Thesis, the probability of {5,6} given {1, 3, 5, 6}, that is, 1/2. (Notice how often Q is false: if the outcome turns out to be 1 then the antecedent is true, but there is no reason why “if it is even it is six” would be true there, etc.)
Step Two is to conditionalize m on the antecedent (~E ∪ S), to produce probability function m#. If m# still satisfies Stalnaker’s Thesis then m#(E → S) = m(Q). Next we conditionalize m# on E, to produce probability function m##. Then, if things are still going well, m##(S) = m(Q).
Bad news! That is greater than m(Q) = 1/2. So things did not go well, and we conclude that conditionalization has taken us outside P(A→).
Why does that show that conditionalization has taken us outside P(A→)? Well suppose that m# obeyed Stalnaker’s Thesis. Then we can argue:
m##(S) = 1, so m#(S|E) = 1. Therefore m#(E → S) = 1 by Stalnaker’s Thesis. Hence m(E → S | ~E v S) = 1. Finally therefore m((~E v S) → (E → S)) = m(Q) = 1. But that is false, as we saw above m(Q) = 1/2.
So, given that m obeys the Thesis, its conditionalization m# does not.
Note. This also shows along the way that the Extended Stalnaker’s Thesis, that m(A → B|X) = m(B|A ∩ X) for all X, is untenable. (But this is probably just the 51st reason to say so.)
APPENDIX
Just to spell out what is meant by conditionalization, let’s note that it must be defined carefully to show that it is a matter of adding to any condition already present (and of course to allow that it is undefined if the result is a condition with probability zero).
So m(A|B) = m(A ∩ B)/m(B), defined iff m(B) > 0. Hence m(A) = m(A|K), where K is the tautology (unit element of the algebra).
Then the conditionalization m# of m on B is m(. | K ∩ B), and the conditionalization m## of m# on C is m#(. |K ∩ C) = m(. | K ∩ B ∩ C), and so forth. Calculation:
m##(X) = m#(X|C) = m#(X ∩ C)/m#(C) = m(X ∩ C |B) divided by m(C|B),
that is [m(X ∩ C ∩B)/m(B)] divided by [m(C ∩ B)/m(B)],
What exactly is so different and difficult about relevance logic?
I will illustrate with Urquart’s 1972 paper that I discussed in the previous post. I’ll assume my post was read, but won’t rely on it too much.
1. A parting of the ways: two concepts of validity
At first, we feel we are on familiar ground. The truth condition for conditionals Urquhart offers is this:
v(A → B, x) = T iff for all y, if v(A, y) = T then v(B, x ∪ y) = T
We see at once that Modus Ponens is valid. For if A → B is true at x, and A is true at x , then B is true at x ∪ x, for that is just x itself.
But used to the usual and familiar, we’ll have one puzzle immediately
This semilattice semantics yields as valid sentences precisely the theorems of the implicational fragment of R.
The first axiom of that logic is A → A. So, what about v(A → A, x)?
We might think that it must be T because if v(A, y) = T then v(A, x ∪ y) = T, because x ∪ y has in it all the information that x had and more. But that is not so! Urquhart points out emphatically that
the information in x ∪ y may not bestow T on A, because dragging in y may have dragged in irrelevant information.
So A → A is supposed to be a valid sentence although it does not receive T from all valuations?
Right! The criterion of validity is a different one:
a sentence is valid if there is an argument to it from no premises at all, from the empty set of premises.
(Being used to the usual and familiar, we would have thought that the two criteria would coincide …)
So Urquhart’ semilattice has an zero, 0, which is the empty piece of information. And A is valid if and only if v(A, 0) = T for all evaluations v.
And the join operation obeys the semi-lattice laws, so for example (x ∪ 0) = (x ∪ x) = x.
Now we can see that the first axiom is indeed valid. The condition for A → A to be valid is the tautology: if v(A, 0) = T then for all x, if v(A, x) = T then v(A, x ∪ 0) = T.
So within this approach:
That a sentence is valid does not imply that it is True in every possibility. Valid conditionals, specifically, are False in many possibilities.
And that is a significant departure from how possibilities are generally understood, however different they may be from possible worlds.
But it is right and required for relevance logic, where valid sentences are not logically equivalent. In general A → A does not imply, and is not implied by B → B, since there may be nothing relevant to B in what A is about.
2. Validity versus truth-preservation
What happens to validity of arguments? The first, and good, news is that Modus Ponens for → is not just validity-preservation but truth-preservation, in the good old way, as I mentioned above.
Btu after this, in relevance logic we will depart from the usual notion of valid argument. We can have instead:
The argument from A1, …, An to B is valid (symbolically, X =>> A) if and only if A1 →. →. An → B is a valid sentence.
That is different from our familiar valid-argument relation. Some characteristics are the same.
By Urquhart’s completeness theorem, the valid sentences are the theorems of the implicational fragment of R. This logic has, apart from the rule of Modus Ponens, the axioms:
A → A
(A →. A → B) → (A → B)
(A → B) → .(C → A) → (C → B)
[A → (B → C)] → [B → (A → C)]
By axiom 4), the order of the premises does not matter. Therefore =>> is still a relation from sets of sentences to sentences. So, for example, the argument from A and B to C is valid exactly if A → (B → C) is a valid sentence, which is equally the case if B → (A → C) is a valid sentence.
But there are crucial differences.
3. What it is to be a sub-structural logic
This consequence relation =>> does not obey the Structural Rules, and the consequence operator is not a closure operator.
Let X╞A mean that the argument from sentences X to sentence A is valid: the semantic entailment relation. The Structural Rules (which are the rules that can be stated without reference to specific features of the syntax) are these:
Identity if A is in X then X╞A
Weakening If X ⊆ Y and X╞A then Y╞A
Transitivity If X╞A for each member A of Y and Y╞ B then X╞ B
The corresponding semantic consequence operator is defined: Cn(X) = {A: X╞A}. If the Structural Rules hold then this is a closure operator.
In relevance logic, Weakening is seen as a culprit and interloper: extra premises may bring irrelevancies, and so destroy validity.
And the new argument-validity criterion above does not include Weakening. If A, …, N =>> B it does not follow that C, A, …, N =>> B.
Here is an extreme example, that actually throws some doubt on the motivating intuitions about the role of irrelevancy. Even A does not in general entail A → A in this sense. For:
v(A → (A → A), 0) = T only if:
for all y, if v(A, y) = T then v(A →A, y ∪ 0) = T,
…… then v(A → A, y) = T,
…….. then for all z, if v(A, z) = T then v(A, y ∪ z) = T
and that does not follow at all. For A’s being true at z and at y is no guarantee that A will be true at y ∪ z.
So even A → (A → A) is not a valid sentence form.
This can’t very well be because A has too much information in it, irrelevant to A → A.
Rather, the opposite: A → A has too much information in it to be concluded on the basis of A. We have to think of valid conditionals as not being ‘empty tautologies’ at all, but as carrying much information of their own.
4. Attempting to look at this algebraically
In subsequent work referring back to Urquhart’s paper the focus is on the ‘join’ operation, and the approach is called operational semantics. But the structures on which the models are based are still, unavoidably, semilattices.
The properties of the join operation are these: it is associative and commutative, but also idempotent (x ∪ x = x), and 0 is the identity (x ∪ 0 = x). So far this amounts to a semigroup. But there is a definable partial order: relation x ≤ y iff x ∪ y = y is a partial ordering:
x ∪ x = x, so x ≤ x
if x ∪ y = y and y ∪ z = z then x ∪ z = (x ∪ (y ∪ z) = (x ∪ y) ∪ z = y ∪ z = z ; so ≤ is transitive.
This partial order comes free, so to speak, and that makes it a semilattice.
Can we find some ordering directly related to truth or validity?
Relative to any specific evaluation v we can see a partial order R in the sentences defined by:
ARB iff v(A → B,0) = T
Then we see that:
R is idempotent: ARA
R is transitive: if v(A → B, 0) = T and v(B → C, 0) = T then v(A → C, 0) =T.
For suppose for all y, if v(A, y) = T then v(B, y ∪ 0) = T. Suppose also that all y, if v(B, y) = T then v(C, y ∪ 0) = T. Since y ∪ 0 = y, we see that for all y, if v(A, y) = T then v(C, y ∪ 0) = T.
So R s a partial order, defined in terms of truth-conditions, to be discerned on the sentences, relative to a valuation.
But trying to find a connection between this ordering of sentences relative to v, and the order in the semilattice, we are blocked. For example, define
If A is a sentence and v an evaluation then [[A]](v) = {x: v(A,x) = T} is the proposition that v assigns to A.
The proposition that v assigns to A will most often not have 0 in it, so it is not closed downward. Nor is it closed upward, for if x is in [[A]](v) it does not follow that x ∪ y is in [[A]](v). So the propositions, so defined, are neither the ideals nor the filters in the semilattice.
I have a feeling, Toto, that we are not in Kansas anymore …….
REFERENCES
Standefer, S. (2022). “Revisiting Semilattice Semantics”. In: Düntsch, I., Mares, E. (eds) Alasdair Urquhart on Nonclassical and Algebraic Logic and Complexity of Proofs. Outstanding Contributions to Logic, vol 22. Springer, Cham. https://doi.org/10.1007/978-3-030-71430-7_7
Urquhart, Alasdair (1972) “Semantics for Relevant Logics”. Journal of Symbolic Logic 37(1); 159-169.
A motivation for this, which will show up in Application 2, is to show that it is tenable to hold that in general, typically, conditionals are true only if they are certain. I do not propose this for conditionals in natural language. But I think it has merits in certain contexts in philosophy of physics, notably interpretation of the conditionals that appear in Einstein-Podolsky-Rosen and Bell Inequality arguments.
[1] The algebra page 1
[2] The language: first step in its interpretation page 1
[3] The algebra: filters and ideals page 2
[4] The language and algebra together: specifying a truth-filter page 2
[5] The language, admissible valuations, validity and the consequence relation page 3
APPLICATION 1: probability space models page 4
APPLICATION 2: algebraic logic of conditionals with probability page 5
NOTE: I will explain this approach informally, and just for the simple case in which we begin with a Boolean algebra.
The languages constructed will in general not be classical, but in this case validity of the classical sentential logic theorems will be preserved, even if other classical features are absent.
But this approach can be applied starting with some other sort of algebra.
[1] The algebra
Let us begin with a Boolean algebra A, with the operations ∩, ∪, -, relation ⊆, top K, and bottom Λ. From my choice of symbols you can see that I find it useful to think of it as an algebra of sets. That will be characteristic of some applications. But this play no role for now, it just helps the imagination.
I will use little letters p, q, r, … to stand for elements of A.
I have left open here whether there are other operations on this algebra, such as modal operators. Application 2 will be to a Boolean algebra with modal operator ==>.
[2] The language: first step in its interpretation
As far as the algebra is concerned, all elements have the same status. But we can introduce distinctions from outside, by choosing a language that can be interpreted in that algebra. When we do that each sentence E has a semantic value [[E]], which is an element of A, and we call it the proposition expressed by that sentence.
So let us introduce a language L. It has atomic sentences, the classical (‘Boolean’) connectives &, v, ~. It may have a lot more. The interpretation is such that
[[~E]] = -[[E]] (the Boolean complement)
[[E & D]] = [[E]] ∩ [[D]]
[[E v D]] = [[E]] ∪ [[D]]
and there will of course be more clauses if the language has more resources for generating complex sentences.
The atomic sentences, together with those three classical connectives, form a sub-language, which I will call Lat. This this is a quantifier-free, modal operator free, fragment of L. I tend to think of the members of Lat as the empirical sentences, the language of the data, but again, that is at this point only a mnemonic.
The set of propositions expressed by sentences in Lat I will call A0, that is {[[E]]: E is in Lat}, and it is clearly a Boolean algebra too, a sub-algebra of A. In general A will be much larger than A0.
[3] The algebra: filters and ideals
What about truth and falsity? I will take it that the true sentences in the language together form a theory, that is, a set closed under the language’s consequence relation — which clearly includes the consequence relation of classical sentential logic. I take it also that this theory is consistent, but do not assume that must be complete.
The algebraic counterpart of a theory is a filter: a set F of elements of A such that, if p ⊆ q and p is in F then so is q, and if r, q are both in F then so is (r ∩ q). A filter is proper exactly if it does not have Λ as a member. That corresponds to consistency.
The filter that consists of the propositions expressed by the members of a consistent theory is a proper filter. Obviously all filters contains K.
A set of elements of A is an ideal exactly if: if p ⊆ q and q is in G then so is p, and if r, q are both in G then so is (r ∪q). The ideal is proper if K is not in it. Obviously any ideal contains Λ.
Filter F has as counterpart an ideal G = {-p: p is in F}, where -p is the complement of p in A. This corresponds to what the theory rules out as false.
[4] The language and algebra together: specifying a truth-filter
Now we are ready to talk about assigning truth-values. Remember that the language L already has an interpretation [[.]] into the algebra of propositions A. What we need to do next then is to select the propositions that are true, and then assign value T to the sentences that express those propositions.
Well, I will show a way how we can do that; but there are many ways. I would like the ‘empirical sentences’ all to get a truth-value. In addition there may be a class of sentences that also should get truth-values, for some reason. They could be selected syntactically (in the way Lat is), or they could be selected as the ones that express a certain sort of proposition. The latter would be a new way of doing the job, so that is what I will outline.
Step 1 is to specify a proper filter T on A, which will be the set of propositions that we will specially specify as true, regardless of whether they belong to A0. Its corresponding ideal U is then the set of propositions that we will specially specify as false.
Step 2 is to specify a filter T0 on A0, as the set of true propositions which are values of ‘empirical sentences’, and indeed we want T0 to be a maximal proper filter on A0. Then its corresponding ideal U0 on A0 is a maximal proper ideal, and A0 is the union of T0 and U0. So every proposition in A0 is classified as true or false.
There is one important constraint on this step. Clearly we do not want any proposition to be selected as true in one step and false in the other step. So the constraint is:
Constraint on Truth Filtering. T0 does not overlap U.
It follows then also that U0 does not overlap T.
The final step is this: T* is the smallest filter that contains both T and T0. We designate T* as the set of true propositions in A. This is the truth-filter. Its corresponding ideal U* is the set of false propositions in A.
This is an unusual way of specifying truth conditions, not least because there will in general be propositions that belong neither to T* nor to U*: in general, bivalence fails.
We need to show that T* is a proper filter.
Lemma. For every proposition p in T* there is a proposition q in T and a proposition r in T0 such that q ∩ r ⊆ p.
It is easiest to prove this via the relation between filters and theories. Let Z be the least theory that contains theories X and Y: thus Z is the set of sentences implied by X ∪ Y. Implication, in our context, is finitary, so if A is in Z then there is a finite set of sentences belonging to X ∪ Y whose conjunction implies A.
Suppose now that T* is not proper. Then there is a proposition p such that both p and -p are in T*. They cannot both be in T nor both in T0. The Constraint on Truth Filtering implies that if p is in T0 then -p is not in T, so -p must a proposition that is not in either T or T0. Similarly if p is in T then -p cannot be in T0 so it must be in neither T nor T0. So we see that either p or -p belongs to neither T nor T0, but must be in the part of T* that is ‘implied’ by meets of elements taken from T and from T0.
By the Lemma there must be propositions q and r in T and T0 respectively such that (q ∩ r) ⊆ p, and also q’ and p’ in T and T0 respectively such that (q’ ∩ r’) ⊆ -p . But then there is a proposition s = (q ∩q’) in T and a proposition t = (r ∩ r’) in T0 such that (s ∩ t) ⊆ (p ∩ – p) = Λ.
In that case t ⊆ -s, while t is in T0 and -s belongs to U. And that is not possible, given the Constraint on Truth Filtering.
Therefore T* is a proper filter.
[5] The language, admissible valuations, validity and the consequence relation
Time to look into the logic in language L when the admissible assignments of truth-values are all of this sort!
What we have described informally now is the class of algebraic models of language L. The sentences E in L have as semantic values propositions [[E]] in A. A is a Boolean algebra with a designated filter T* and designated ideal U* = {-p: p is in T*}. An admissible valuation of L is a function v such that for all sentences E of L:
v(E) = T if and only if [[E]] is in T*
v(E) = F if and only if [[E]] is in U*.
This function is not defined on other sentences: those other sentences, if any, do not have a truth-value.
So an admissible valuation is in general a partial function on the set of sentences of L.
Validity
Boolean identities correspond to the theorems of classical sentential logic. If E is such a theorem then [[E]] = K, which belongs to every filter, and hence E is true.
This holds for any model of the sort we have described, so all theorems of classical sentential logic are valid.
Deductive consequence
E1, …, En imply F exactly if, in each such model, if [[E1]], …, [[En]] are all in T* then [[F]] is in T*.
In classical sentential logic E1, …, En imply F exactly if (E1 & … & En) ≡ (E1 & … & En & F) is a theorem. So then ([[E1]] ∩ …∩ [[En]]) = ([[E1]] ∩ …∩ [[En]] ∩ [[F]]).
It follows that if [[E1]], …, [[En]] are all in a given filter then so is [[F]].
Therefore all such classically valid direct inferences (such as Modus Ponens) are valid in L.
Natural deduction rules
Those which involve sub-arguments can be expected to fail. For example, (E v ~E) is valid, but it is possible that E lacks a truth-value, and so we would expect the Disjunctive Syllogism to fail.
We’ll see examples below.
APPLICATION 1: probability space models
The structure S = <K, F, P> is a probability space exactly if K is a non-empty set, F is a field of subsets of K (including K), and P is a probability function with domain F.
A field of sets is a Boolean algebra of sets. So we can proceed as above.
First there is a language LS, and if E is a sentence of LS then [[E]] is a measurable subset of K, that is to say, a set in F, a member of the domain of P. And as before we have a fragment LSat which is the closure of the set of atomic sentences under the Boolean connectives. The range of [[.]] restricted to LSat is a subfield — a Boolean subalgebra — F0 of F.
The set TS = {p is in F : P(p) = 1} is a proper filter. That is so because P( Λ) = 0, P(p) is less than or equal to P(q) if p ⊆ q, and P(p ∩ q) = 1 if and only if P(p) = P(q) = 1.
Similarly, there is a corresponding proper ideal US = {p is in F : P(p) = 0}.
Just as above, TS is the beginning, so to speak, of the set of true propositions. To determine an appropriate set of true propositions in F0 we begin with X = US ∩ F0 That is a proper ideal as well, within that subalgebra. Every such proper ideal can be extended (not uniquely) to a proper maximal ideal US0 on F0. This we choose as the set of false propositions in that subalgebra, and the corresponding maximal filter TS0 on F0 is the set of true propositions there.
And now, to complete the series of steps we are following, we define TS* to be the least filter on F which contains both TS and TS0. The general argument above applies mutatis mutandis to show that TS* is a proper filter — our truth filter in this setting.
Unless LSat is the whole of LS we will now have truth-value gaps: there will be non-empirical sentences that receive some probability intermediate between 0 and 1, and these are neither true nor false.
As before, there is no doubt that the axiomatic classical sentential logic is sound here. However there are natural deduction rules which are not admissible. For example, if something follows from each of P(p) = 1 and P(q) = 1 it may still not follow from P(p v q) = 1. For example, if we are going to toss a coin then Probability(Heads) = 1 entails that the coin is biased, and Probability(Tails)= 1 also entails that the coin is biased. But Probability( Heads or Tails) =1 is true also if the coin is fair.
APPLICATION 2: algebraic logic of conditionals with probability
This is an example of a probability space model, in which the algebra is a Boolean algebra with a binary modal operator ==>. It begins with a ‘ready to wear’, off the shelf, construction, which I’ll describe. And then I will apply the recipe developed above to give a picture of a language in which conditionals, typically, are true only if they have probability 1, and and false only if they have probability 0.
I am referring to the logic CE, which is like Stalnaker’s logic of conditionals, but weaker (van Fraassen 1975; see also my preceding blogs on probabilities of conditionals).
The language has the Boolean connectives plus binary connective –>. A structure M = <K,F, s> is a model of CE exactly if K is a non-empty set (the worlds), F is a field of subsets of K (the propositions), and s, the selection function, is a function which maps K x F into the subsets of K, with these properties:
s(x,A) ⊆ A
if x is in A then s(x,A) = {x}
s(x, A) has at most one member
s(x, A) = Λ only if A = Λ
The truth conditions for &, v, ~ are as usual, and for –> it is:
A –> B is true in world x if and only if s(x,A) ⊆ B
equally: [[A –> B]] = {x is in K: s(x, [[A]]) ⊆ [[B]]}
and we can see that there is therefore an operator on F , for which I’ll use the symbol ==>:
[[A –>B]] = [[A]] ==> [[B]].
This differs from Stalnaker’s semantics only in not imposing the further restriction on the selection function that it must derive from an ordering. We may intuitively refer to s(x, A) as the world nearest to x that is in [[A]], but this “nearest” metaphor has no content here.
When this language is thus interpreted in model M, the propositions form a Boolean algebra with operator ==>, which has the properties:
(I) [p ==> (q ∪ c)] = [(p ==> q) ∪ (p ==> c)]
(ii) [p==> (q ∩ c)] = [(p ==> q) ∩ (p ==> c)]
(iii) [p ∩ (p ==> q)] = (p ∩ q)
(iv) (p ==> p) = K ( “necessity” )
(v) (p ==> -p) = Λ unless p = Λ (“impossibility”)
Let us call this a CE algebra.
A probability model for CE is a structure <K, F, s, P> such that <K, F, s >is a model for CE and P is probability function with domain F such that for all p, q in F
P(p ==> q) = P(q | p) when defined
This condition is generally called Stalnaker’s Thesis (or more recently, just “the Thesis”). Stalnaker’s logic of conditionals could not be nontrivially combined with this thesis but CE could. As it happens, CE has a rich family of probability models.
Thus, if <K, F, s, P> is a probability model for CE then S = < K, F, ==>, P> is a probability space model in the sense of the previous section, with some extra structure.
Now we can proceed precisely as in the preceding section to define a truth filter T* on the algebra of propositions. As empirical statements we take the closure of the set of atomic sentences under just the Boolean connectives, that is the sentences in which there are no occurrences of –>. The image of this language fragment by the map [[.]] is the relevant, privileged Boolean subalgebra F0 of F in which every proposition is classified as true or false, as a first step.
In addition the propositions which have probability 1 are true. And finally, anything implied by true propositions is true — all this understood as coming about as shown in the preceding section. Thus all theorems of CE are valid, and inference by modus ponens is valid.
As to sentences of form (A –> B), they are typically true only if P(A | B) = 1. I say “typically” because we cannot rule out that the proposition [[A]] ==> [[B]] is a member of F0. For the model of CE could be a model of a stronger theory, perhaps one that entails (implausibly!) that “if it is lit then it burns” is the meaning of “it is flammable”. But typically that will not be the case, so typically (A –>B) will be classified as true only if P([[B]] | [[A]]) = 1.
REFERENCES
van Fraassen, B. C. (1976) “Probabilities of Conditionals”, in W. Harper and C.A. Hooker (eds.) Foundations of Probability and Statistics, Volume l. Reidel: 261-308.
At a conference at Notre Dame in 1987 Paul Teller “made an issue” as he wrote later of “the fallacious form of argument I called the ‘Candy Bar Principle’:
from ‘If I were hungry I would eat some candy bar’ conclude ‘There is some candy bar which I would eat if I were hungry’.”
And Henry Stapp, whom Teller had criticized mentioned this in his presentation: “Paul Teller has suggested that any proposed proof of the kind I am setting forth must contain such a logical error ….”
I do not want to enter into this controversy, if only because there were so many arguments swirling around Stapp’s proposed proofs. Instead I want to examine the question:
is the Candy Bar inference a fallacy?
Let’s formulate it for just a finite case: there are three candy bars, A, B, and N. The first two are in this room and the third is next door. I shall refer to the following form of argument as a Candy Bar inference:
If I choose a candy bar it will be either A or B
therefore,
If I choose a candy bar it will be A, or, if I choose a candy bar it will be B
and I will symbolize this as follows:
C –> (A v B), therefore (C –> A) v (C –> B)
This has a bit of a history of course: it was submitted as valid in Robert Stalnaker’s original theory of conditionals and was rejected by David Lewis in his theory. Lewis showed that Stalnaker’s theory was inadequate, and blamed this principle. But we should quickly add that the problems Lewis raised also disappeared if this principle were kept while another one, shared by Stalnaker and Lewis, was rejected. This is just by the way, for now I will leave all of this aside.
How shall we go about testing the Candy Bar inference?
I imagine that the first intuitive reaction is something like this:
Imagine that I decide to choose a candy bar in this room. Then it will definitely be either A or B that I choose. But there is nothing definite about which one it will be.
I could close my eyes and choose at random.
Very fine! But unfortunately that is not an argument against the Candy Bar inference, but rather against the following different inference:
It is certain that if I choose, then I will choose either A or B,
therefore
Either it is certain that if I choose I will choose A, or, it is certain that if I choose I will choose B
That is not at all the same, for we cannot equate ‘It is certain that if X then Y’ with ‘if X then Y’. As an example, contrast the confident assertion “If the temperature drops it will rain tomorrow” with “It is certain that if the temperature drops it will rain tomorrow”. The former will be borne out, the prediction will be verified, if in fact the temperature drops and it rains the next day — but this is not enough to show that the latter assertion was true.
So the intuitive reaction does not settle the matter. How else can we test the Candy Bar inference?
Can we test it empirically? Suppose two people, Bob and Alice of course, are asked to predict what I will do, and write on pieces of paper, respectively, “if Bas chooses a candy bar in this room, he will choose A” and “if Bas chooses a candy bar in this room, he will choose B”. Surely we will say:
we know that if Bas chooses a candy bar in this room, he will choose A or B.
So if he does, either Bob or Alice will turn out to have been right.
And then, if Bas chooses A, we will say “Bob was right”.
That is also an intuitive reaction, which appears to favor the Candy Bar inference. But again, it does not really establish much. For it says nothing at all about which of these conditionals, if any, would be true if Bas does not choose a candy bar. That is the problem with any sort of empirical test: it deals only with facts and does not have access to what would have happened instead of what did happen.
Well there is another empirical approach, not directly to any facts about the choice and the candy bars, but to how reasonable, practical people would let this situation figure in their decision making.
So now we present Alice and Bob with this situation and we ask them to make bets. These are conditional bets, they will be Gentlemen’s Wagers, which means that they get their money back if Bas does not choose.
Alice first asks herself: how likely is he to choose a bar from this room, as opposed to from next door (where, you remember, there is bar N) Suppose she takes that to have probability 3/4. She accepts a bet that Bas will choose A or B, if he chooses at all, with payoff 1 and price 0.75. Her expectation value is 0, it is just a fair bet.
Meanwhile Bob agrees with her probability judgment, but is placing two bets, one that if Bas chooses he will choose A, and one that if Bas chooses he will choose B. These he thinks equally probable, so for a payoff of 1 he agrees to price 3/8 for each. His expectation value is 1/4(0) + 3/8(1) + 3/8(1) minus what he paid, hence 0: this too is just a fair bet.
Thus Alice and Bob pay the same to be in a fair betting situation, where the payoff prices are the same, though one was, in effect, addressing the premise and the other the conclusion of the Candy Bar inference. So, as far as rational betting behavior is concerned then, again, there is no difference between the two statements.
Betting, however, as we well now by now, is also only a crude measuring instrument for what matters. The fact that these are Gentlemen’s Wagers, as they pretty well have to be, once again means that we are really only dealing with the scenario in which the antecedent is true. The counterfactual aspect is beyond our ken.
To be clear: counterfactual conditionals are metaphysical statements, if they are statements about what is the case, at all. They are not empirical statements, and this makes the question about the validity of the Candy Bar inference a metaphysical question.
There is quite a lot of every-day metaphysics entrenched at the surface of our ordinary discourse. Think for instance of what Nancy Cartwright calls this-worldly causality, with examples like the rock breaking the window and the cat lapping up the milk.
Traditional principles about conditionals, just as much as traditional principles about causality, may guide our model building. And then nature may or may not fit our models …
So the question is not closed, the relation to what is empirically accessible may be more subtle than I managed to get to here. To be continued ….
REFERENCES
The Notre Dame Conference in question had its proceedings published as Philosophical Consequences of Quantum Theory: Reflections on Bell’s Theorem (ed. J. T. Cushing and E. McMullin; University of Notre Dame Press 1989).
My quotes from Teller and Stapp are from pages 210 and 166 respectively.
plays no role in the truth conditions for conditionals, though it may encode many other aspects of that situation, and figure in the truth conditions of sentences that involve no arrows.
Bas van Fraassen's philosophy blog 