Acceptance and Certainty, Doxastic Modals, and Indicative Conditionals

Abstract

I give a semantics for a logic with two pairs of doxastic modals and an indicative conditional connective that all nest without restriction. Sentences are evaluated as accepted, rejected, or neither. Certainty is the necessity-like modality of acceptance. Inferences may proceed from premises that are certain, or merely accepted, or a mix of both. This semantic setup yields some striking results. Notably, the existence of inferences that preserve certainty but not acceptance very directly implies both failure of modus ponens for the indicative conditional in the logic of acceptance and failure of the deduction theorem for the material conditional in the logic of certainty. The latter failure dissolves, in the logic of certainty, the much - discussed tension between modus ponens and the law of import-export.

Appendices

Appendix A

Semantics in which sentences are evaluated at sequences of worlds, as in the lower level of the present semantics, have been used to try to uphold the claimed identity Pr(A ⇒ B) = Pr(B∣A), often called Adams’s Thesis. I think that this program ought to be abandoned. The primary reason is not any sort of triviality result but, rather, a conflict between two claims. The first, an implication of Adams’s Thesis, I will call Adams’s Corollary, or AC.

AC If A and B are probabilistically independent, and Pr(A) > 0, then Pr(A ⇒ B) = Pr(B).

The second claim I’ll call Edgington’s Thesis, or ET, because it is the general principle behind one of Edgington’s criticisms of McGee’s treatment of conditionals ([8], p. 202, par. (e) and n. 32). Here is the principle.

ET If A and B are probabilistically independent, and 0 < Pr(A) < 1, then Pr((A ⇒ B) ∧ (¬A ⇒ B)) = Pr(B).

Anyone who accepts Adams’s Thesis will accept AC. But ET is just as plausible. It says that when A and B are uncorrelated, the probability of “B, whether or not A” is simply the probability of B. Consider two successive coin flips, with results not yet revealed. What’s the probability of “The second flip landed heads, whether or not the first one did,” if not 1/2? (How often does a bet on that sentence win, if not half the time? As a bit of news, how much surprise should the sentence elicit, compared to “The second flip landed heads”?)¹

The problem, now, is that even if AC and ET are formally compatible, together they lead to substantively absurd conclusions. Specifically, given AC and ET, and probabilistically independent p and q, we can reason as follows.

1. By AC, Pr(p ⇒ q) = Pr(q).

2. By ET, Pr((p ⇒ q) ∧ (¬p ⇒ q)) = Pr(q).

Because (p ⇒ q) ∧ (¬p ⇒ q) entails ((p ⇒ q) ∧ (¬p ⇒¬q)) ∨ ((p ⇒ q) ∧ (¬p ⇒ q)), which entails p ⇒ q, 1 and 2 jointly imply:

3. Pr(((p ⇒ q) ∧ (¬p ⇒¬q)) ∨ ((p ⇒ q) ∧ (¬p ⇒ q))) = Pr(q)

Now consider the identity Pr(X) + Pr(Y ) = Pr(X ∧ Y ) + Pr(X ∨ Y ), with X = (p ⇒ q) ∧ (¬p ⇒¬q) and Y = (p ⇒ q) ∧ (¬p ⇒ q). Since 2 and 3 jointly imply that Pr(Y ) = Pr(X ∨ Y ), it follows that Pr(X) = Pr(X ∧ Y ). Dropping a duplicate p ⇒ q conjunct in X ∧ Y yields

4. Pr((p ⇒ q) ∧ (¬p ⇒¬q)) = Pr((p ⇒ q) ∧ (¬p ⇒¬q) ∧ (¬p ⇒ q))

Because (p ⇒ q) ∧ (¬p ⇒¬q) ∧ (¬p ⇒ q) entails (¬p ⇒¬q) ∧ (¬p ⇒ q),

5. Pr((p ⇒ q) ∧ (¬p ⇒¬q)) ≤Pr((¬p ⇒¬q) ∧ (¬p ⇒ q))

This is a reductio. We have found that, for instance, “The two coin flips matched” is no more probable than “If the first flip landed tails, the second landed both tails and heads.”

Given that AC and ET are comparably plausible but we can’t embrace both, is there a reason to choose one over the other? Yes. Unlike AC, ET doesn’t give rise to triviality problems. We know this because ET is consistent with treating ⇒ as the material conditional. So the triviality results are relevant, but they are only needed to break a tie between two comparably attractive claims that can’t both be correct.²

Appendix B

Below are brief proofs of key claims made in the main discussion. The following lemma will be useful.

Lemma Given a model containing sequence u, a model can be constructed in which u counts as full, and u is proper in this model just in case it is proper in the original model, and any sentence X is accepted at u in the new model just in case X is accepted at u in the original one.

By construction: Let the new model’s state contain exactly the worlds in u, ordered just as they are in the original model’s state. This makes u full. And for every atomic sentence A, let each world be an A-world in the new model just in case it is an A-world in the original one. Any sequence in the new model, including u, will be proper there just in case it is proper in the original one. Therefore, every sentence can and will have the same acceptance status at every sequence, including u, in the new model as at the same sequence in the original model. $\blacksquare$

The logic of acceptance

If the inference schema X₁,X₂,... / Y is a substitution-instance of a schema that is valid in S5, with either $\square$ or $\boxdot$ (but not both) as the necessity modal, then the schema X₁,X₂,... / Y preserves acceptance.

By contraposition: Let schema X₁,X₂,... / Y be a substitution-instance of a schema in the language of S5 that uses either $\square$ or $\boxdot$ (but not both) as the modal, and suppose that X₁,X₂,... / Y fails to preserve acceptance. Then there is a model where X₁,X₂,... are accepted but Y is not. This model must contain a full, proper sequence, u, where X₁,X₂,... are accepted but Y is not. Identify the appropriate set of permutations of u: the set of proper permutations if the modal in the S5-schema is $\boxdot$, the set of all permutations if the modal is $\square$. Use that set to construct a corresponding S5 model (i.e., one for sentences in the language of S5), one world for each permutation. Let any sentence that was substituted for in the S5-schema be true at a world just in case the sentence that was substituted for it is accepted at the corresponding permutation. The S5 model will be a counterexample to the S5 schema of which X₁,X₂,... / Y is a substitution-instance. $\blacksquare$

$$ \textbf{DT} \Rightarrow \quad \text{If} \square {{A}}, \square {X} \vDash {Y}, \text{then} \square {A} \vDash {X} \Rightarrow {Y} $$

By contraposition: Consider a model where $\square$A is accepted but X ⇒ Y is not. This model must contain a full, proper sequence, u, where $\square$A is accepted but X ⇒ Y is not. Then v, the most-similar reduction of u where $\square$X is accepted, exists and is proper, and Y is not accepted there. Since u must contain only A-worlds, $\square$A is accepted at v, as well. By the Lemma, a model can be constructed where v counts as full and proper, and where $\square$A and $\square$X are accepted at v but Y is not. Then $\square$A and $\square$X are accepted at all permutations of v, including proper ones, and so $\square$A and $\square$X are accepted in the new model. But Y is not accepted there, because it is not accepted at v. $\blacksquare$

$$ \mathrm{A}, \mathrm{A} \Rightarrow \mathrm{B} \vDash \mathrm{B} $$

By contraposition: Consider a model where A is accepted but B is not. This model must contain a full, proper sequence, u, where A is accepted and B is not, hence where the first world is an A-world but not a B-world. Since this world will also be the first world in the most-similar reduction of u where $\square$A is accepted, A ⇒ B cannot be accepted at u. Therefore, A ⇒ B cannot be accepted in the model. $\blacksquare$

The logic of certainty

If $A, X \vDash _{C} Y$, then $A \vDash _{C} X \Rightarrow Y$.

By DT⇒, if $\square$A, $\square$$X \vDash$ $\square$Y, then $\square$$A \vDash X \Rightarrow$ $\square$Y. It therefore suffices to show that $\square$$A \vDash X \Rightarrow$ $\square$Y implies $\square$$A \vDash$ $\square$(X ⇒ Y ). Proof, by contraposition: Consider a model where $\square$A is accepted but $\square$(X ⇒ Y ) is not. This model must contain a full sequence, u, where $\square$A is accepted but X ⇒ Y is not. Then v, the most-similar reduction of u where $\square$X is accepted, exists, and Y is not accepted there. Since u must contain only A-worlds, $\square$A is accepted at v, as well. By the Lemma, a model can be constructed in which v is full, such that $\square$A and $\square$X are accepted at v but Y is not. Sequence v has a permutation, x, that is full and proper in the new model. At x, $\square$A and $\square$X are accepted but $\square$Y is not. Since x is its own most-similar reduction where $\square$X is accepted, X ⇒ $\square$Y is not accepted at x. In this new model, then, $\square$A is accepted but X ⇒ $\square$Y is not. $\blacksquare$

$$ {X},{X} \Rightarrow {Y} \vDash_{C} {Y} $$

Since $\square$X,X ⇒ $\square$$Y \vDash$ $\square$Y, it suffices to show that $\square$$(X \Rightarrow Y) \vDash X \Rightarrow$ $\square$Y. By contraposition: Consider a model where X ⇒ $\square$Y is not accepted. This model must contain a full, proper sequence, u, where X ⇒ $\square$Y is not accepted. Then v, the most-similar reduction of u where $\square$X is accepted, exists, and $\square$Y is not accepted there. Sequence v must have a permutation, x, where $\square$X is accepted but Y is not. Now, u has at least one permutation, y, that starts with x at the head. And x must be the most-similar reduction of y where $\square$X is accepted, because any competitor sequence would have to start with x and add more worlds, but then a permutation of that competitor, not v, would have been the most-similar reduction of u where $\square$X is accepted. Since Y is not accepted at x, X ⇒ Y is not accepted at y. But since u has a permutation, y, where X ⇒ Y is not accepted, $\square$(X ⇒ Y ) is not accepted at u and therefore is not accepted in the model. $\blacksquare$

$$ \vDash_{C} ((A \wedge X) \Rightarrow Y) \equiv (A \Rightarrow (X \Rightarrow Y)) $$

Direct proof: Given a model containing full, proper sequence u, let v be the most-similar reduction of u where $\square$A is accepted. (If v does not exist, then the biconditional is accepted in the model.) S, the set of reductions of u where $\square$(A ∧ X) is accepted, is also the set of reductions of v where $\square$X is accepted, and in S, the sequence most similar to u is also the sequence most similar to v, since for any two sequences in S, the first world in u over whose inclusion they differ is also the first world in v over whose inclusion they differ. Therefore, evaluating (A ∧ X) ⇒ Y for acceptance at u will always produce the same result as evaluating A ⇒ (X ⇒ Y ) for acceptance at u, and so the biconditional must be accepted at u. Since u is arbitrary, the biconditional must be accepted at all full, proper sequences and therefore accepted in the model. $\blacksquare$