Free Choice Impossibility Results

Abstract

Free Choice is the principle that possibly p or q implies and is implied by possibly p and possibly q. A variety of recent attempts to validate Free Choice rely on a nonclassical semantics for disjunction, where the meaning of p or q is not a set of possible worlds. This paper begins with a battery of impossibility results, showing that some kind of nonclassical semantics for disjunction is required in order to validate Free Choice. The paper then provides a positive account of Free Choice, by identifying a family of dynamic semantics for disjunction that can validate the inference. On all such theories, the meaning of p or q has two parts. First, p or q requires that our information is consistent with each of p and q. Second, p or q narrows down our information by eliminating some worlds. It turns out that this second component of or is well behaved: there is a strongest such meaning that p or q can express, consistent with validating Free Choice. The strongest such meaning is the classical one, on which p or q eliminates any world where both p and q are false. In this way, the classical meaning of disjunction turns out to be intimately related to the validity of Free Choice.

Appendix

Definition 7

If a set of possibility operators is partitionable and all satisfy Free Choice, then disjunction is not propositional.

• 1. {w^′∣wRw^′},{v^′∣vR^′v^′}, and {u^′∣uR^″u^′} are not empty.

• 2. {w^′∣wRw^′} = {v^′∣vR^′v^′}∪{u^′∣uR^″u^′}

• 3. {v^′∣vR^′v^′}∩{u^′∣uR^″u^′} = ∅

• 4. ∃A,B ∈L : ⟦ A⟧ = {v^′∣vR^′v^′} & ⟦ B⟧ = {u^′∣uR^″u^′}.

Fact 1

If a set of possibility operators is partitionable and all satisfy Free Choice, then disjunction is not propositional.

Proof

Suppose ♢_A, ♢_B, and ♢_C are partitionable and satisfy Free Choice. Let B and C denote {v^′∣w^′R_Bv^′}, and {v^″∣w^″R_Cv^″} respectively, where A = B ∪ C. Suppose disjunction is propositional, with ∗ as the witnessing operator. Let A and B be true at exactly B and C respectively. Then by the validity of Free Choice for ♢_A, ♢_A(A ∨B) is true at w just in case ♢_AA and ♢_AB are true at w. ⟦ A ∨B⟧ = B ∗ C. So B ∗ C is consistent with B ∪ C just in case B is consistent with B ∪ C and C is consistent with B ∪ C. However, we also know that B ∗ C is inconsistent with C if B is inconsistent with C, since ♢_C(A ∨B) is true at w^″ only if ♢_CA is true at w^″. Since B is inconsistent with C, it follows that B ∗ C is inconsistent with C. Similarly, we know that B ∗ C is inconsistent with B if C is inconsistent with B, since ♢_B(A ∨B) is true at w^′ only if ♢_BB is true at w^′. Since C is inconsistent with B, it follows that B ∗ C is inconsistent with B. But if B ∗ C is inconsistent with both B and with C, then B ∗ C is inconsistent with B ∪ C. Contradiction. □

Fact 2

If any conditional possibility operator ♢_i(⋅)(⋅) satisfies Conditional Free Choice, then ∨ is not propositional.

Proof

Let ♢(⋅)(⋅) be the witnessing operator with accessibility relation R. Now take some arbitrary world w such that ♢(⋅)(⋅) has a non-trivial accessibility relation at that world, such that there are at least two worlds v and u where wRv and wRu. Now I will assume that there are two propositions A and B that partition {v∣wRv}, and are the semantic value of some sentences A and B. Now let ⊤ be a claim true at every possible world. By Conditional Free Choice, w ∈⟦♢(⊤)(A ∨B)⟧ just in case w ∈⟦♢(⊤)(A)⟧ and w ∈⟦♢(⊤)(B)⟧. In this semantic framework, the last condition in turn requires that ∃v : wRv & v ∈ A ∗ B iff ∃v : wRv & v ∈ A and ∃v : wRv & v ∈ B. By the construction of R, we know that this condition is satisfied. But now consider the conditionals ♢(A)(A ∨B) and ♢(B)(A ∨B). If ♢ satisfies Conditional Free Choice, then we know that w ∈⟦♢(A)(A ∨B)⟧ only if w ∈⟦♢(A)(B)⟧. This in turn implies that ∃v : wRv & v ∈ A & v ∈ A ∗ B only if ∃v : wRv & v ∈ A & v ∈ B. But there is no such v in both A and B, since these propositions are inconsistent. So there is no accessible v where both A and A ∗ B hold. Likewise, w ∈⟦♢(B)(A ∨B)⟧ only if w ∈⟦♢(B)(A)⟧; which in turn implies that ∃v : wRv & v ∈ B & v ∈ A ∗ B only if ∃v : wRv & v ∈ A & v ∈ B. Since this last condition cannot obtain, we know that there is no accessible v where both B and A ∗ B. But every accessible v world is either an A or a B world. So A ∗ B can’t hold at any accessible world in the first place. □

Definition 10 (Credal Free Choice)

There is some operator ∗ such that for any rational Pr,A,B:

Pr(A ∗ B) > 0 iff Pr(A) > 0 and Pr(B) > 0.

Fact 3

If Credal Free Choice holds, then Pr(A) = 0 or Pr(A) = 1 for every rational Pr,A.

Proof

Pr(A∗(W−A)) = Pr(A∗(W−A))×Pr(A∗(W−A)∣A)+Pr(W−A)×Pr(A∗(W−A)∣W−A). Pr(A∣A) = 1, which means Pr(W − A∣A) = 0, which implies Pr(A ∗ (W − A)∣A) = 0 by Credal Free Choice. By similar reasoning, Pr(A ∗ (W − A)∣W − A) = 0. So Pr(A ∗ (W − A)) = 0. So by Credal Free Choice again, Pr(A) = 0 or Pr(W − A) = 0. So Pr(A) is 0 or 1. □

Fact 4

If Free Choice is valid, then A ∨B is not distributive for arbitrary atomic A and B.

Proof

Consider some arbitrary atomic p and β. Let s be some context which contains some p worlds, some β worlds, and no worlds where both p and β are false. Since s[p]≠∅ and s[β]≠∅, we know by Dynamic Free Choice that s[p ∨ β]≠∅. Now let s_¬p be {w ∈ s∣w(p) = 0}, and let s_¬β be {w ∈ s∣w(β) = 0}. By Dynamic Free Choice, s_¬p[p ∨ β] = ∅, since s_¬p[p] = ∅. Similarly, s_¬β[p ∨ β] = ∅. Now suppose p ∨ β is distributive. Then for any w ∈ s_¬p, {w}[p ∨ β] = ∅. Similarly, for any w ∈ s_¬β, {w}[p ∨ β] = ∅. But s = s_¬p ∪ s_¬β. So for any w ∈ s, {w}[p ∨ β] = ∅. So, supposing p ∨ β is distributive, s[p ∨ β] = ∅. This is inconsistent; so p ∨ β is not distributive. □

Fact 5

Suppose that entailment is transitive, that ♢ is upwards monotonic, that the T axiom is valid, and that the 4 axiom is valid. Then FC I is valid iff Disjunction-Possibility Link is valid.

Proof

By the T axiom, A ∨B implies ♢(A ∨B). By Free Choice, this last implies ♢A ∧♢B. So by the transitivity of entailment A ∨B implies ♢A ∧♢B. Now suppose A ∨B⊧♢A. Then ♢(A ∨B)⊧♢♢A by upwards monotonicity, which by the 4 axiom implies ♢A. □

Definition 16

s[A ∨_tB] = s[♢A][♢B]

Fact 6

If ∗ validates Free Choice, then ∗ is at least as strong as ∨_t.

Proof

By the T axiom for ♢, . Suppose FC 1 is valid for ∗; then and . This is sufficient to show that , and hence that □

Definition 18

∗ is factorizable iff there is some distributive operator ⋈ such that [A ∗B] = s[A⋈B][A ∨_tB].

Definition 19

s[A ∨_uB] = s[A] ∪ s[B]

Definition 20

s[A ∨_utB] = s[A ∨_uB][A ∨_tB].

Fact 7

if ∗ is factorizable and satisfies Free Choice, and if A⊮B and B⊮A, then .

Proof

Suppose that s[A ∨_uB][A ∨_tB] = s. We must show that s⊧A ∗B. We can suppose that s is not ∅, since the proof would be trivial in that case. Since s[A ∨_uB][A ∨_tB] = s, we know that (i) s[A]≠∅ and s[B]≠∅, and (ii) s is made up exclusively of worlds w where either {w}⊧A or {w}⊧B. From (i), we know that the possibility tests required by ∗ are satisfied. To complete our proof, we can use (ii) to show that s[A⋈B] = s.

Since ⋈ is distributive, we know that s[A⋈B] = {w ∈ s∣{w}⊧A⋈B}. So to prove that s[A⋈B] = s, we can take an arbitrary w ∈ s and show that {w}⊧A⋈B. From (ii) we know that either {w}⊧A or {w}⊧B. We can show that in either case {w}⊧A⋈B.

So suppose the former. Either {w}⊧B or not. Suppose the former. Then since {w}⊧A and {w}⊧B, we have that {w}⊧♢A and {w}⊧♢B. So since ∗ satisfies Free Choice, we know that {w}⊧♢(A ∗B). Since {w} is maximal, this means that {w}⊧A ∗B and hence {w}⊧A⋈B.

But now suppose instead that {w}⊧A and {w}⊮B. Now let v be such that {v}⊧B and {v}⊮A. Here, we rely on the assumption that A and B are not ordered by strength. First, this means that {w,v}⊧♢A and {w,v}⊧♢B. So by Free Choice {w,v}⊧♢(A ∗B). But this means that {w,v}[A⋈B]≠∅. This in turn implies that {w,v}[A⋈B] = {w,v}. For otherwise one of w or v would be removed, in which case one of ♢A or ♢B would not be supported by {w,v}[A⋈B], and so {w,v}⊮♢(A ∗B). So since {w,v}⊧A⋈B and ⋈ is distributive, {w}⊧A⋈B. □

Definition 23

∗ is quasi-conjunctive relative to R iff for any A and B, if A and B are expressible via R, then A ∗ B = A ∩ B.

Fact 8

If Wide Free Choice is valid for ♢_i and disjunction is propositional, then disjunction is quasi-conjunctive relative to R_i.

Proof

Suppose Wide Free Choice is valid for ♢_i and disjunction is propositional, expressing the operation ∗. Now suppose that A and B are expressible via R_i. That is, suppose ⟦♢_iA⟧ = A and suppose ⟦♢_iB⟧ = B, for some A and B. By Wide Free Choice, ♢_iA ∨♢_iB implies each of ♢_iA and ♢_iB, and vice versa. But this means that ⟦♢i A ∨♢i B⟧ = ⟦♢_iA⟧∩⟦♢_iB⟧. Since ⟦♢i A ∨♢i B⟧ = A ∗ B, this means that A ∗ B = A ∩ B. □

Fact 9

Wide Free Choice is valid for ♢_i and ∨_t iff for any w,v,u: if wR_iv then uR_iv.

Proof

Suppose Wide Free Choice is valid for ♢_i and ∨_t. Then whenever s⊧♢♢_iA and s⊧♢♢_iB, we have that s⊧♢_iA and s⊧♢_iB. This means that if there is some w ∈ s where wR_iv and v ∈ A, we are guaranteed that for every w ∈ s there is an accessible v world where A. This means that if wR_iv, then uR_iv for any u. Otherwise, s could contain just w and u, and then {v} would be a counterexampling choice of A. Conversely, suppose that for any w,v,u: if wR_iv then uR_iv. Now suppose s⊧♢♢_iA and s⊧♢♢_iB. Then ∃w ∈ s : {v∣wR_iv}[A]≠∅. Now let u be an arbitrary world in s. By the constraint above, we know that {v^′∣uR_iv^′}[A]≠∅, since {v∣wR_iv} = {v^′∣uR_iv^′}. So s⊧♢_iA. □

Fact 10

Free Choice, Dual Prohibition, Contraposition, and Transitivity imply Explosion.

Proof

By Free Choice, . So by Contraposition . By Dual Prohibition, . So by Transitivity, and hence by Contraposition again . □

Fact 11

If Free Choice and Dual Prohibition are valid, then Explosion is valid.

Proof

Suppose s⊧♢A. Suppose that s is non-empty (otherwise the proof is trivial). We can now show that s⊧♢B. First, s[A]≠∅. Since Dual Prohibition is valid, this means that s[A ∨B]≠∅. But since Free Choice is valid, this last implies that s[B]≠∅. So s⊧♢B. □