1. Introduction

Our ability to embed indicative conditionals in larger sentences is a reason to analyze them as expressing some sort of genuine logical connective. However, embedding conditionals under negations provides some of the best evidence against the most popular analysis, the material conditional. In this article, I will show how negations of conditionals also help formulate the main challenges for stronger theories of the indicative conditional. Although the basic challenge primarily focuses on simple conditionals, it leads to a result about iterated conditionals: they do not obey modus ponens.

2. Troubles with Negating Conditionals

Consider this adaption of Grice's (Reference Grice[1967] 1989) example of an election in which the candidates are Heath, Thorpe, and Wilson. Suppose one election watcher gets information that Thorpe will lose and says, “If Wilson loses, then Heath will win.” A second person responds, “That's not the case.” The second speaker's response is puzzling on the material conditional analysis, where the conditional is equivalent to the disjunction “Either Wilson or Heath wins.” For that would commit the second speaker to affirming that Wilson will lose, even though he might well be unsure of that. Grice, crediting this sort of observation to Bromberger, tries to save the material conditional analysis of the indicative by rejecting the claim that disagreement always amounts to affirming the truth-functional negation. Indeed, he suggests that denying a conditional often amounts to affirming the conditional with its consequent negated.

A more plausible strategy, if it can be carried out, is to reject the material conditional analysis in favor of some other theory that gets the right interaction with negation. Any such proposal faces a difficulty of its own. Suppose the second person got the information that Heath will lose and affirms, “If Wilson loses then Heath will lose.” Even though it sounds like the two speakers disagree, it seems that neither speaker was wrong if it turns out that Wilson wins. Grice would have it that this means that both speakers spoke truly, supporting the material conditional analysis. Gibbard (Reference Gibbard, Harper, Stalnaker and Pearce1981) concludes that Ramsey (Reference Ramsey[1929] 1990) was right in claiming that conditionals do not express proper objects of beliefs but, rather, dispositions to revise one's beliefs.

Ramsey's basic idea was that in evaluating whether to accept a conditional, an agent should check whether he accepts the consequent upon revising his beliefs to accept the antecedent. Several philosophers—especially Stalnaker (Reference Stalnaker, Harper, Stalnaker and Pearce[1968] 1981, Reference Stalnaker[1970] 1981)—have claimed that one can endorse this positive suggestion by Ramsey, while abandoning his negative conclusion that conditionals are not objects of belief. In the next section I shall present a glimpse of a way of articulating such a theory. Then I shall show how the theory runs into problems, not by considering apparent disagreements between different persons but, rather, by examining a similar problem that arises by considering the potential belief states of a single agent.

3. The Ramsey Test

As a step toward making things more precise, consider a language containing a conditional connective $> $, as well as negation $\neg $, conjunction $\wedge $, and $\supset $ for the material conditional. The letters $A,\,B,\,C$ range over the sentences not containing $> $. Greek $\phi ,\,\psi $ range over all sentences. The constant $\perp $ stands for some truth-functional contradiction.

A theory $K$ is a set of sentences closed under the language's consequence relation: $K=Cn( K) $. The Expansion of $K$ by $\phi $, written $K+\phi $, is $Cn( K\cup \{ \phi \} ) $. The exact nature of the consequence relation $Cn$ is to be determined by the structure imposed on the $> $ connective by the Ramsey Test. However, it is assumed that $Cn$ is an extension of classical consequence. For instance, it obeys modus ponens and the deduction theorem for the material conditional, so $\psi \in K+\phi $ if and only if (iff) $\phi \supset \psi \in K$. Also, it obeys Monotonicity: if $\phi $ entails $\psi $, then $K+\psi \subseteq K+\phi $. If $\{ \phi ,\:\neg \phi \} \subseteq K$ then $K$ is inconsistent and contains every sentence. If $\neg \phi \not\in K$ then both $K$ and $K+\phi $ are consistent, and $\phi $ is said to be compatible with $K$.

We use theories to represent the belief states of an idealized agent, who accepts all the logical consequences of his beliefs. Such an agent has a policy for revising his beliefs on the basis of new information (observations or suppositions; we do not here decide on the exact interpretation). This is captured by a Revision function $*$ which takes a prior belief state $K$ into a posterior belief state $K*A$ on the basis of the input $A$. (Note that although conditionals are in belief states, we are not here considering them as inputs to revision, although one might do so.)

The Ramsey Test is formulated in terms of the Revision operation. Here is a version that is restricted to simple conditionals (those containing no conditional parts):

RT. If $\neg A\not\in K$, then

This formulation is restricted to conditionals that are “noncounterfactual,” in that their antecedents are compatible with belief. (This is not to say that we don’t use indicatives whose antecedents are rejected or that the Ramsey Test shouldn’t be extended to cover them.) Additionally, RT is rather minimal in that it gives only sufficient conditions for accepting conditionals and their negations. If the first (“positive”) clause were strengthened to a necessary condition, then accepting $A> \neg C$ would suffice for accepting $\neg ( A> C) $, which is very plausible. If the “negative” clause were strengthened to a necessary condition, accepting $\neg ( A> C) $ would suffice for accepting $A> \neg C$, which is a subject of controversy in the study of counterfactual subjunctive conditionals.Footnote ¹

Although there is a case to be made for richer formulations of the Ramsey Test, this version suffices for the purposes of this paper. The Ramsey Test constrains the logical behavior of the conditional by means of properties of the revision function. Thanks to the clause for negations, RT is incompatible with the material conditional as a reasonable interpretation of $A> C$. Because the negation of a material conditional entails its antecedent, interpreting RT as about the material conditional gives the unacceptable result that one is in suspense about something only if revising by it leads to being in suspense about everything.Footnote ² While it looks like progress is being made, the Ramsey Test turns out to conflict with seemingly plausible principles of belief revision.

4. Triviality Results for Simple Conditionals

We say that a theory $K$ is Nontrivial in case there are sentences $A,\: C$ such that $A\wedge C$, $A\wedge \neg C$, $\neg A$ are each compatible with $K$. (Equivalently, $K$ is compatible with pairwise incompatible $B,\: C,\: D$; let $A=B\vee C$.) The Ramsey Test is clearly trivial if it only holds in belief states that are more opinionated than this.

We will discuss some principles of Revision that lead to the triviality of the Ramsey Test, but a good deal of the work in proving the theorem can be done without those details. Consider this version of a principle from Stalnaker Reference Stalnaker[1975] 1981, for accepting $> $ conditionals on the basis of material conditionals.

Direct Argument. If $\neg A\not\in K$, then

The heart of the triviality proof for RT is contained in the following lemma, which doesn’t say anything about the Ramsey Test.

We want a way of representing the belief states of an agent who uses a conditional in accord with the Ramsey Test. Let a belief revision model $\langle \mathbf{K},\: *\rangle $ consist of a set $\mathbf{K}$ of theories paired with a Revision function $*$, which is closed under RT and Revision: if $K\in \mathbf{K}$, then $K$ obeys RT, and if $\neg A\not\in K$, then $K*A\in \mathbf{K}$. Given a proper specification of the Revision function, the closure condition ensures that if $\mathbf{K}$ contains a theory representing an agent's prior belief state, then it contains all the potential posterior belief states that might come from it.

What is the nature of the Revision function? There is a popular theory of belief revision according to which Revision simply goes by Expansion when the input is consistent with prior belief: $K*A=K+A$, if $\neg A\not\in K$. This is defended in Gärdenfors Reference Gärdenfors1988 as the minimal change to be made to $K$ to accept $A$. If we adopt this rule, then RT is consistent only in trivial cases. But this assumption about Revision is stronger than what is needed for the triviality result. It turns out to be instructive to explore a couple sets of weaker assumptions, in order to facilitate comparison with some influential triviality proofs and to get a better understanding of the tenability conditions for the Ramsey Test.

Since it is assumed that belief revision models are closed under Revisions, if the latter go by Expansion then $\mathbf{K}$ is closed under Expansions: if $K\in \mathbf{K}$ and $\neg A\not\in K$, then $K+A\in \mathbf{K}$. The rule that Revision goes by Expansion can be split in two:

Inclusion says one should not leap to any conclusions that do not follow from drawing the consequences of adding $A$ to one's old beliefs. Preservation says one should accept $A$ without losing any old beliefs. We only need a weaker version of Preservation, just applying to beliefs not containing conditionals.

The proof we give below requires either Inclusion, or closure under Expansions, but not both.

5. Inclusion, Iteration, and Indexicality

6. Related Triviality Theorems

Two of the most influential triviality proofs about the Ramsey Test are those of Lewis (Reference Lewis[1976] 1981) and Gärdenfors (Reference Gärdenfors1986, Reference Gärdenfors1988). These proofs proceed from different enough assumptions that one might wonder how much they really have in common. It turns out that we can use the Main Theorem of this paper to generalize both of them, as well as a theorem by Harper (Reference Harper, Harper and Hooker1976) which has not received as much attention.

Gärdenfors doesn’t work with a clause for accepting negations of conditionals but, rather, with belief-contravening revisions. Gärdenfors's proof relies on the claim that $\mathbf{K}$ is closed under Expansion along with the following:

An interesting feature of Gärdenfors's proof, which he attributes to Alchourrón and Makinson, is that G1 entails a property of Revisions that suffices for a triviality proof without conditionals. The triviality of G1 follows from that as a corollary. It turns out that the only requirements of Revision are G2 and Basic Preservation.Footnote ⁶

However, by focusing on the Ramsey Test rather than Revision in the abstract, we may use Gärdenfors's G1 and G2 to give a variation of both cases of the Main Theorem of this paper. Indeed, we may dispense with G2 and belief-contravening revisions and use case $( E) $ of the Main Theorem to derive Gärdenfors's result about G1. To do so, assume that $\mathbf{K}$ is closed under Expansions $K+\phi $ by conditionals. From G1 holding for all $K\in \mathbf{K}$, it follows that

From G2 also holding in all members of $\mathbf{K}$, it follows that

(Suppose for reductio that H2 is false: $\neg A\not\in K$ but $\neg ( A> \:\perp ) \not\in K$. Hence $A$ and $K+( A> \:\perp ) $ are each consistent. By closure under Expansions, G1 applies to $K+A> \:\perp $, so $\perp \in ( K+A> \:\perp ) *A$, which contradicts G2.) Harper Reference Harper, Harper and Hooker1976 proves a version of the Main Theorem basically using H1, H2, and G1.Footnote ⁷ But if G1, H1, and H2 hold in all $K\in \mathbf{K}$, so does RT, just given the closure of $\mathbf{K}$ under Revisions. Hence the Main Theorem entails these results and distinguishes the two cases: closure under Expansions and Revision obeying Inclusion.

Ramsey's original test was a probabilistic one. A probability $P$ is a function from sentences into real numbers obeying the laws that $P( \phi ) \geq 0$, $P( \phi \vee \neg \phi ) =1$, and $P( \phi \vee \psi ) =P( \phi ) +P( \psi ) $ if $\phi ,\:\psi $ are incompatible. A conditional probability is equal to a ratio of probabilities $P( \psi /\phi ) =P( \phi \wedge \psi ) /P( \phi ) $, if $P( \phi ) > 0$.

Let the operation $P_{\phi }( \psi ) $ be the posterior probability assigned to $\psi $ upon revising prior $P$ by $\phi $. In its probabilistic form, the Ramsey Test tells us that $P( A> C) =P_{A}( C) $, if $P( A) > 0$. The standard rule for probabilistic revision says that it obeys

Combined with the rule that $P_{A}$ goes by conditionalization, the Ramsey Test gives us an equation often named after Adams's Reference Adams1975 development of Ramsey's idea.

From the law of probability that $P( \psi /\phi ) +P( \neg \psi /\phi ) =1$, it follows from AT that $P( A> \neg C) =1$ iff $P( A> C) =0$. Hence Ramsey's remark that denying a conditional is equivalent to denying its consequent.

Lewis (Reference Lewis[1976] 1981) proves that except in trivial cases, AT is not closed under conditionalization. Although the proof that Lewis gives involves the full use of AT (in its last step), it turns out that Lewis's theorem only requires the extreme cases: if $P( A) > 0$ then $P( A> C) =1$ if $P( C/A) =1$, and $P( A> C) =0$ if $P( C/A) =0$.Footnote ⁸ From just that weak version of AT and the laws of probability, Lewis's result can be obtained from this paper's Main Theorem, in the following way.

Let $\mathbf{P}$ be a set of probability assignments satisfying the hypothesis of Lewis's theorem (closure under both AT and conditionalization by sentences not containing conditionals). The main task is to define a belief revision model $\langle \mathbf{K},\: *\rangle $ in terms of $\mathbf{P}$ in such a way as to satisfy the hypothesis of the Main Theorem of this paper. This is easy to do, in a way that captures the qualitative features we need from Lewis's quantitative model.

First, define $\mathbf{K}$ as the smallest set where for every $P\in \mathbf{P}$, there is a $K\in \mathbf{K}$ such that $K=\{ \phi \thinspace{:}\thinspace P( \phi ) =1\} $. We show that defining $\mathbf{K}$ in this way entails that it is closed under Expansions: if $K\in \mathbf{K}$ and $\neg A\not\in K$, then $K+A\in \mathbf{K}$. For each $P\in \mathbf{P}$, consider the corresponding $K$. (This ensures that we cover each member of $\mathbf{K}$, since it is the smallest set meeting the definition.) If $\neg A\not\in K$, then $P( \neg A) \neq 1$, and so $P( A) > 0$. Since $\mathbf{P}$ is closed under conditionalization, it contains $P_{A}$, where $P_{A}( \psi ) =1$ iff $P( \psi /A) =1$. Moreover, $\mathbf{K}$ is defined so as to contain $H=\{ \psi \thinspace{:}\thinspace P_{A}( \psi ) =1\} $. By probability, $P( \psi /A) =1$ iff $P( A\supset \psi ) =1$. So $H=\{ \psi \thinspace{:}\thinspace A\supset \psi \in K\} $, which is $K+A$ by the definition of Expansions.

Next, we can define $K*A=K+A$ when $\neg A\not\in K$. From this we have that $\mathbf{K}$ is closed under Revisions, and clearly $*$ satisfies Basic Preservation. Since $\mathbf{P}$ is closed under AT, it follows that $\mathbf{K}$ is closed under RT. Finally, suppose for reductio there is some $P\in \mathbf{P}$ that is nontrivial in Lewis's sense: there are $A,\: C$ such that $P( A\wedge C) $, $P( A\wedge \neg C) $, $P( \neg A) $ are all positive. That supposition entails that there is some $K\in \mathbf{K}$ that is compatible with $A\wedge C$, $A\wedge \neg C$, and $\neg A$. But the Main Theorem tells us that this is impossible, which proves Lewis's result.Footnote ⁹

Adams himself takes the triviality results to show that Ramsey was right: a conditional expresses a conditional probability rather than any truth conditional connective. But as in the qualitative case, we may maintain the probabilistic Ramsey Test by suitable restrictions. We may hold on to our standard theory of revision when no conditionals are involved, $P_{A}( C) =P( C/A) $. So we can keep $P( A> C) =P( C/A) $, even though we must deny that in general $P_{A}( \psi ) =P( \psi /A) $. Exploring the details of the best way to do this is outside the scope of this paper.Footnote ¹⁰

7. Conclusion

In the tradition inspired by Goodman (Reference Goodman[1947] 1955), counterfactual conditionals have provided the strongest impetus to search for a theory other than the truth-functional one. But indicative conditionals pose their own significant challenges to the material conditional. The Ramsey Test promises insight, but the triviality results present an obstacle to using principles of belief revision to constrain the logic of the conditional. Indeed, the variety of triviality results seems to present several obstacles.

I have argued that several of the most important triviality results have more in common than is at first apparent, thanks to a triviality proof based on a shared formulation of the Ramsey Test. A consequence of the proof is that in order to escape triviality without making conditionals indexical, iterated conditionals fail to obey modus ponens. The phenomenon about iterated conditionals also holds on any reasonable Ramsey Test theory that avoids triviality by making conditionals indexical to belief states.