Observation 8. If (Alt3) replaces ( $\rightarrow $ Rej), then for any $\varphi $ and $\psi $ and any coherent m: either $m\vdash {\text {\rm Acc}} (\varphi \rightarrow \psi )$ or $m\vdash {\text {\rm Acc}} (\varphi \rightarrow \neg \psi )$ .

Proof. Assume that $m\not \vdash {\text {Acc}} (\varphi \rightarrow \psi )$ . So $m\,\circledast {\text {Acc}}\ \varphi \not \vdash {\text {Acc}}\ \psi $ . By Strong Success and Weak Redundancy, $m\,\circledast {\text {Acc}}\ \varphi \,\circledast {\text {Acc}}\ \varphi \not \vdash {\text {Acc}}\ \psi $ . So, by Alt3, $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}} \neg (\varphi \rightarrow \psi )$ . By Monotonicity, $m\,\circledast {\text {Acc}}\ \varphi \,{\oplus }{\text {Acc}}\ \psi \vdash {\text {Acc}} \neg (\varphi \rightarrow \psi )$ . By Success $m\,\circledast {\text {Acc}}\ \varphi \,{\oplus }{\text {Acc}}\ \psi \vdash {\text {Acc}}\ \psi $ . By Strong Success, $m\,\circledast {\text {Acc}}\ \varphi \,{\oplus }{\text {Acc}} \psi \Vdash {\text {Acc}}\ \varphi $ , so, by Weak Redundancy, $m\,\circledast {\text {Acc}}\ \varphi \,{\oplus }{\text {Acc}} \psi = m\,\circledast {\text {Acc}}\ \varphi \,{\oplus }{\text {Acc}}\ \psi \,\circledast {\text {Acc}}\ \varphi $ . So $m\,\circledast {\text {Acc}}\ \varphi \,{\oplus }{\text {Acc}}\ \psi \,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ and so $m\,\circledast {\text {Acc}}\ \varphi \,{\oplus }{\text {Acc}}\ \psi \vdash {\text {Acc}} (\varphi \rightarrow \psi )$ . As we also have $m\,\circledast {\text {Acc}}\ \varphi \,{\oplus }{\text {Acc}}\ \psi \vdash {\text {Acc}} \neg (\varphi \rightarrow \psi )$ , $m\,\circledast {\text {Acc}}\ \varphi \,{\oplus } {\text {Acc}}\ \psi $ is incoherent. So by Classicality $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}} \neg \psi $ . So $m\vdash {\text {Acc}}( \varphi \rightarrow \neg \psi )$ .□

Proof of Theorem 1.

The consistency part of the proof is omitted. For the language $L_R$ it is largely trivial. For the full language L note that Modus Ponens, Classicality, CN, K and LLE do not make use of any particular feature of $\!/\!\!/\!$ other than that it is a function of the proposition it takes as an argument, while Import-Export and rMP and rCS only involve factual antecedents.

The completeness part has a standard structure (it proceeds by showing how a ‘canonical’ ICL model can be constructed and then shows that if $\Gamma \not \models _{\text {ICL}} \varphi $ we can find a point of evaluation in the canonical model where the sentences of $\Gamma $ are true but $\varphi $ is false). Note throughout that due to CN, K and LLE, $\rightarrow $ is a normal conditional modality and I will rely on this in a number of steps.

Let $W_c$ be the set of maximal ICL-consistent sets of factual sentences. Let $V_c(p) = \{w\in W_c: p\in w\}$ and let $\mathcal {B}_c$ be the set we get by closing the set $\{V_c(p)\, :\, p \text { atomic}\}$ under complements, finite intersections and finite unions. When $A\in \mathcal {B}_c$ let $S(A)$ denote some canonically chosen factual sentence such that $F({|S(A)|}) = A$ . Let $M_c$ be the model $(W_c, \mathcal {B}_c, V_c)$ .

Let $\Sigma $ denote a maximal ICL-consistent set on the whole language $L_R$ (so not only on factual sentences). Define, for any $A\in \mathcal {B}_c$ such that $S(A)\rightarrow \perp \, \not \in \Sigma $ (Where $\perp $ is an arbitrary factual contradiction):

$$ \begin{align*} f_\Sigma(A) = \{\varphi: S(A)\rightarrow \varphi\in \Sigma \text{ and } \varphi \text{ is factual}\}. \end{align*} $$

Due to LLE it doesn’t matter which sentence $S(A)$ picks out (so $f_\Sigma (A)$ is well-defined When $S(A)\rightarrow {\perp }\not \in \Sigma $ . Furthermore, $f_\Sigma (A)$ is an element of $W_c$ (for due to CEM and the fact that $S(A)\rightarrow \perp \,\not \in \Sigma $ , $f_\Sigma (A)$ will be a maximal consistent set of factual sentences, i.e. an element of $W_c$ ).

Let ${\scriptstyle \mathrm {MB}}_f$ be the range of $f_\Sigma $ , the set $\{f_\Sigma (A)\,:\, A\in \mathcal {B}_c \text { and } S(A)\rightarrow {\perp }\not \in \Sigma \}$ .

Continuing the proof of the completeness theorem. Assume that $\Gamma \not \models _{\text {ICL}} \varphi $ . One can show (using standard techniques) that $\Gamma \cup \{\neg \varphi \}$ can be extended to a maximal ICL-consistent set $\Sigma $ . By Lemma B.3, $f_\Sigma \models _{M_c} \psi $ for all $\psi $ in $\Gamma $ , furthermore, $f_\Sigma \not \models _{M_c} \varphi $ .

Proof of Theorem 2.

First show that the expressivist consequence relation $\models _{\mathfrak {E}}$ contains $\models _{\text {ICL}}$ . The proof of the opposite direction follows from Theorem 5 which is proved below.

Take any expressivist model $\mathcal {E}$ and any coherent mental state m: one needs to show that $m\vdash $ is closed under the rules and axioms of $\models _{\text {ICL}}$ .

MP: Assume that $m\vdash {\text {Acc}}\ \varphi $ and $m\vdash {\text {Acc}}\ \varphi \supset \psi $ . Then $m\,{\oplus } {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ ; by Redundancy $m\vdash {\text {Acc}}\ \psi $ .

CL: As the binary boolean connectives are classically interdefinable, it is enough to show that every instance of the following axiomatisation of classical logic is accepted: (1) $\varphi \supset (\psi \supset \varphi )$ , (2) $(\varphi \supset (\psi \supset \chi )) \supset ((\varphi \rightarrow \psi )\supset (\varphi \supset \chi ))$ , (3) $(\neg \psi \supset \neg \varphi ) \supset (\varphi \rightarrow \psi )$ .

Recall throughout that by Explosion if a mental state is incoherent, then it supports every attitude.

(1) $m\,{\oplus } {\text {Acc}}\ \varphi \,{\oplus } {\text {Acc}}\ \psi \vdash {\text {Acc}}\ \varphi $ by Success and Monotonicity, so $m\vdash {\text {Acc}}\ \varphi \supset (\psi \supset \varphi )$ for all m.

(2) $m \,{\oplus } {\text {Acc}}\ \varphi \supset (\psi \supset \chi ) \,{\oplus } {\text {Acc}}\ \varphi \supset \psi \,{\oplus } {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \varphi \supset \psi $ by Success and Monotonicity. By Success and modus ponens, $m \,{\oplus } {\text {Acc}}\ \varphi \supset (\psi \supset \chi ) \,{\oplus } {\text {Acc}}\ \varphi \supset \psi \,{\oplus } {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ . By Success and Monotonicity, $m \,{\oplus } {\text {Acc}}\ \varphi \supset (\psi \supset \chi ) \,{\oplus } {\text {Acc}}\ \varphi \supset \psi \,{\oplus } {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \varphi \supset (\psi \supset \chi )$ . By modus ponens $m \,{\oplus } {\text {Acc}}\ \varphi \supset (\psi \supset \chi ) \,{\oplus } {\text {Acc}}\ \varphi \supset \psi \,{\oplus } {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi \supset \chi $ . By modus ponens $m \,{\oplus } {\text {Acc}}\ \varphi \supset (\psi \supset \chi ) \,{\oplus } {\text {Acc}}\ \varphi \supset \psi \,{\oplus } {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \chi $ . So $m\vdash {\text {Acc}} (\varphi \supset (\psi \supset \chi )) \supset ((\varphi \rightarrow \psi )\supset (\varphi \supset \chi ))$ .

For the last axiom, note that ${\text {Acc}}\ \psi $ and ${\text {Acc}} \neg \psi $ are complementary attitudes. For recall that $m\,{\oplus } {\text {Acc}}\ \psi = m\,{\oplus } {\text {Rej}} \neg \psi $ for any m and so, as ${\text {Rej}} \neg \psi $ and ${\text {Acc}} \neg \psi $ are complementary (Classicality), so are ${\text {Acc}}\ \psi $ and ${\text {Acc}} \neg \psi $ .

(3) $m\,{\oplus } {\text {Acc}} \neg \psi \supset \neg \varphi \,{\oplus } {\text {Acc}}\ \varphi \,{\oplus } {\text {Acc}} \neg \psi \vdash {\text {Acc}}\ \varphi $ by Success and Monotonicity while $m\,{\oplus } {\text {Acc}} \neg \psi \supset \neg \varphi \,{\oplus } {\text {Acc}}\ \varphi \,{\oplus } {\text {Acc}} \neg \psi \vdash {\text {Acc}} \neg \varphi $ by Success, Monotonicity and modus ponens. So $m\,{\oplus } {\text {Acc}} \neg \psi \supset \neg \varphi \,{\oplus } {\text {Acc}}\ \varphi \,{\oplus } {\text {Acc}} \neg \psi \vdash {\text {Rej}}\ \varphi $ . By Classicality, $m\,{\oplus } {\text {Acc}} \neg \psi \supset \neg \varphi \,{\oplus } {\text {Acc}}\ \varphi \,{\oplus } {\text {Acc}} \neg \psi $ is incoherent. By Classicality (complementarity) $m\,{\oplus } {\text {Acc}} \neg \psi \supset \neg \varphi \,{\oplus } {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ . So $m\vdash {\text {Acc}} (\neg \psi \supset \neg \varphi ) \supset (\varphi \rightarrow \psi )$ .

CN: Assume that for every coherent m: if $m\vdash {\text {Acc}}\ \varphi $ , then $m\vdash {\text {Acc}}\ \psi $ . By Strong Success $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \varphi $ , so either $m\,\circledast {\text {Acc}}\ \varphi $ is incoherent (in which case $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ ) or $m\,\circledast {\text {Acc}}\ \varphi $ is coherent and so by assumption $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ . So $m\vdash {\text {Acc}}(\varphi \rightarrow \psi )$ .

K: Assume that for every m: $m\vdash {\text {Acc}}\chi $ whenever $m\vdash {\text {Acc}}\ \psi _i$ (for all i: $1\leq i\leq n$ ). Take any m such that $m\vdash {\text {Acc}} (\varphi \rightarrow \psi _i)$ (for all i: $1\leq i\leq n$ ). Then $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi _i$ (for all i: $1\leq i\leq n$ ). By assumption $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \chi $ . So $m\vdash {\text {Acc}} (\varphi \rightarrow \chi )$ .

LLE: Assume that $\varphi $ and $\psi $ are non modal and that $\models _{\text {ICL}} \varphi \equiv \psi $ , that is, $\varphi $ and $\psi $ are classically equivalent. From the above we know that that for every coherent m: $m\vdash {\text {Acc}}\ \varphi $ iff $m\vdash {\text {Acc}}\ \psi $ . We need to show that $m\vdash {\text {Acc}}( \varphi \rightarrow \chi )$ iff $m\vdash {\text {Acc}}( \psi \rightarrow \chi )$ . Assume that $m\vdash {\text {Acc}} (\varphi \rightarrow \chi )$ . So $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\chi $ . By Strong Success $m\,\circledast {\text {Acc}}\ \varphi \Vdash {\text {Acc}}\ \varphi $ , so $m\,\circledast {\text {Acc}}\ \varphi \Vdash {\text {Acc}}\ \psi $ . By Weak Redundancy $m\,\circledast {\text {Acc}}\ \varphi = m\,\circledast {\text {Acc}}\ \varphi \,\circledast {\text {Acc}}\ \psi $ . By R-Permutation $m\,\circledast {\text {Acc}}\ \varphi = m\,\circledast {\text {Acc}}\ \psi \,\circledast {\text {Acc}}\ \varphi $ . As $m\,\circledast {\text {Acc}}\ \psi \Vdash {\text {Acc}}\ \varphi $ , by Weak Redundancy $m\,\circledast {\text {Acc}}\ \varphi = m\,\circledast {\text {Acc}}\ \psi $ . So $m\,\circledast {\text {Acc}} (\psi \vdash {\text {Acc}}\chi )$ and so $m\vdash {\text {Acc}}( \psi \rightarrow \chi )$ . The converse direction follows analogously.

CEM: By Success $m\,{\oplus } {\text {Rej}} (\varphi \rightarrow \psi )\vdash {\text {Rej}}( \varphi \rightarrow \psi )$ . If $m\,{\oplus } {\text {Rej}} (\varphi \rightarrow \psi )$ is incoherent then $m\,{\oplus } {\text {Rej}}( \varphi \rightarrow \psi )\vdash ({\text {Acc}}\ \varphi \rightarrow \neg \psi )$ . So assume that $m\,{\oplus } {\text {Rej}}\ \varphi \rightarrow \psi $ is coherent. By the rejection condition for the conditional: $m\,{\oplus } {\text {Rej}}\ \varphi \rightarrow \psi \,\circledast {\text {Acc}}\ \varphi \vdash {\text {Rej}} \psi $ , and so $m\,{\oplus } {\text {Rej}}( \varphi \rightarrow \psi )\,\circledast {\text {Acc}}( \varphi \vdash {\text {Acc}} \neg \psi )$ and so $m\,{\oplus } {\text {Rej}} (\varphi \rightarrow \psi )\vdash {\text {Acc}} (\varphi \rightarrow \neg \psi )$ . An analogous argument yields $m\,{\oplus } {\text {Rej}}( \varphi \rightarrow \neg \psi )\vdash {\text {Acc}} (\varphi \rightarrow \psi )$ . So $m\vdash {\text {Acc}}( \varphi \rightarrow \psi \vee \varphi \rightarrow \neg \psi )$ .

Import-Export: Assume that $\varphi $ and $\psi $ are factual; so ${\text {Acc}}\ \varphi $ and ${\text {Acc}}\ \psi $ are factual. It is enough to show that $m\,\circledast {\text {Acc}}\ \varphi \,\circledast {\text {Acc}}\ \psi = m\,\circledast {\text {Acc}} (\varphi \wedge \psi )$ . By Strong Success, $m\,\circledast {\text {Acc}}\ \varphi \,\circledast {\text {Acc}}\ \psi \Vdash {\text {Acc}} (\varphi \wedge \psi )$ . Thus, by Weak Redundancy, $m\,\circledast {\text {Acc}}\ \varphi \,\circledast {\text {Acc}}\ \psi = m\,\circledast {\text {Acc}}\ \varphi \,\circledast {\text {Acc}}\ \psi \,\circledast {\text {Acc}} (\varphi \wedge \psi )$ . By R-Permutation, $m\,\circledast {\text {Acc}}\ \varphi \,\circledast {\text {Acc}}\ \psi \,\circledast {\text {Acc}}( \varphi \wedge \psi ) = m\,\circledast {\text {Acc}} (\varphi \wedge \psi )\,\circledast {\text {Acc}}\ \varphi \,\circledast {\text {Acc}}\ \psi $ . As $m\,\circledast {\text {Acc}} (\varphi \wedge \psi )\Vdash {\text {Acc}}\ \varphi $ and $m\,\circledast {\text {Acc}} (\varphi \wedge \psi )\Vdash {\text {Acc}}\ \psi $ , it follows that $m\,\circledast {\text {Acc}} (\varphi \wedge \psi )\,\circledast {\text {Acc}}\ \varphi \,\circledast {\text {Acc}}\ \psi = m\,\circledast {\text {Acc}}( \varphi \wedge \psi )$ , so $m\,\circledast {\text {Acc}}\ \varphi \,\circledast {\text {Acc}}\ \psi = m\,\circledast {\text {Acc}} (\varphi \wedge \psi )$ .

[rMP] Assume that $\varphi $ and $\psi $ are syntactically factual. Take any m such that $m\vdash {\text {Acc}}\ \varphi \rightarrow \psi $ and $m\vdash {\text {Acc}}\ \varphi $ . We have $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ . By R-Inclusion $m\,{\oplus } {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ . By Redundancy $m\,{\oplus } {\text {Acc}}\ \varphi = m$ so $m\vdash {\text {Acc}}\ \psi $ .

[rCS] Assume that $\varphi $ and $\psi $ are syntactically factual. Take any m such that $m\vdash {\text {Acc}}\ \varphi $ and $m\vdash {\text {Acc}}\ \psi $ . If $m\,{\oplus }{\text {Acc}}\ \varphi $ is incoherent then by Explosion $m\,{\oplus }{\text {Acc}}\ \varphi \vdash {\text {Acc}}(\varphi \rightarrow \psi )$ . So assume that $m\,{\oplus }{\text {Acc}}\ \varphi $ is coherent. By R-Vacuity $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ and so $m\vdash {\text {Acc}} (\varphi \rightarrow \psi )$ .□

Proof of Theorem 3.

Begin by noting that any sentence $\varphi $ is ICL-equivalent to a sentence $\varphi '$ where all negations have been showed inward, that is where every occurrence of $\neg $ either applies to an atom or has the form $\neg (\psi \rightarrow \perp )$ (this is a standard property of conditional logics that satisfy CEM). For instance, $\neg (p\rightarrow q)$ is logically equivalent to $\neg (p\rightarrow \perp ) \vee p\rightarrow \neg q$ . So we will only consider sentences where $\neg $ has been showed inward. Let $\bar {\varphi }$ denote a sentence that is logically equivalent to $\neg \varphi $ but where all negations have been showed inward. Define a function $^*$ on sentences that returns a set of sets of sentences of either the form $\varphi \rightarrow a$ (where a is a possibly negated atomic sentence) or the form $\neg (\varphi \rightarrow \perp )$ . As follows:

1. 1. $(a)^* = \{\{\top \rightarrow a\}\}$ , here a is an atomic sentence (possibly negated) and $\top $ is an arbitrary tautology.

2. 2. $(\neg (\varphi \rightarrow \perp ))^* = \{\{\neg (\varphi \rightarrow \perp )\}\}$ .

3. 3. $(\varphi \wedge \psi )^* =\{X \cup Y\,|\, X\in \varphi ^* \text { and } Y\in \psi ^*\}$ .

4. 4. $(\varphi \vee \psi )^* = (\varphi \wedge \psi )^* \cup (\varphi \wedge \bar {\psi })^* \cup (\bar {\varphi }\wedge \psi )^*$ .

5. 5. $X\in (\varphi \rightarrow \psi )^*\; \mathrm{iff}\; X\; = \{\varphi \rightarrow \perp\}$ or there is a $Y\in \psi ^*$ such that:

   $ Y=\{\sigma _1\rightarrow a_1, \ldots , \sigma _n\rightarrow a_n, \neg (\theta _1\rightarrow \perp ), \ldots , \neg (\theta _m\rightarrow \perp )\},$

   and:

   $X=\{(\varphi \wedge \sigma _1)\rightarrow a_1, \ldots , (\varphi \wedge \sigma _n)\rightarrow a_n, \neg ((\varphi \wedge \theta _1)\rightarrow \perp ), \ldots , \neg ((\varphi \wedge \theta _m)\rightarrow \perp )\}. $

Let $D(\varphi ) = \bigvee \{\bigwedge X\,|\, X\in \varphi ^*\}$ . Each set in $\varphi ^*$ thus corresponds to a conjunction of its elements and multiple sets indicate that these in $D(\varphi )$ are connected through disjunction. Note that $D(\varphi )$ has the form $\delta _1 \vee \cdots \vee \delta _n$ and that each $\delta _j$ has the form:

$$ \begin{align*} \bigwedge_{1\leq i\leq k_j}(\psi^j_i\rightarrow a^j_i) \wedge \bigwedge_{1\leq i \leq m_j} \neg(\chi^j_i\rightarrow \perp ), \end{align*} $$

Moreover, if $\varphi $ is the formula $\sigma \rightarrow \theta $ , each $\delta _j$ has the form:

$$ \begin{align*} \bigwedge_{1\leq i\leq k_j}((\sigma\wedge\psi^j_i)\rightarrow a^j_i) \wedge \bigwedge_{1\leq i \leq m_j} \neg((\sigma\wedge\chi^j_i)\rightarrow \perp ). \end{align*} $$

That is, the antecedent $\sigma $ in $\varphi $ occurs everywhere as an antecedent in the subformulae of $D(\varphi )$ .

From (4) and (5) the only clauses that introduce new disjunctive sets – it is evident that the disjuncts will be mutually logically exclusive. We need to prove by induction over the length of $\varphi $ that $D(\varphi )$ is logically equivalent to $\varphi $ . But all the cases except possibly the last are straightforward. For the last case note that $\varphi \rightarrow (\delta _1 \vee \cdots \vee \delta _n)$ is logically equivalent to $\varphi \rightarrow \delta _1 \vee \cdots \vee \varphi \rightarrow \delta _n$ , and that

$$ \begin{align*} \varphi\rightarrow \left( \bigwedge_{1\leq i\leq k_j}(\psi^j_i\rightarrow a^j_i) \wedge \bigwedge_{1\leq i \leq m_j} \neg(\chi^j_i\rightarrow \perp )\right ) \end{align*} $$

is equivalent to

$$ \begin{align*} \bigwedge_{1\leq i\leq k_j}(\varphi\rightarrow( \psi^j_i\rightarrow a^j_i)) \wedge \bigwedge_{1\leq i \leq m_j} (\varphi\rightarrow\neg(\chi^j_i\rightarrow \perp )) \end{align*} $$

which, by Import-Export and CEM, is equivalent to:

$$ \begin{align*} (\varphi\rightarrow \perp)\vee \left(\bigwedge_{1\leq i\leq k_j}((\varphi\wedge \psi^j_i)\rightarrow a^j_i) \wedge \bigwedge_{1\leq i \leq m_j}\neg((\varphi\wedge \chi^j_i)\rightarrow \perp ). \right)\end{align*} $$

Proof of Theorem 4.

Take any expressivist model $\mathcal {E}$ on $L_R$ . Let the ICL-model $M_c = (W_c, \mathcal {B}_c, V_c)$ be the canonical model of the proof of Theorem 1.

Define:

$$ \begin{align*} {||\!{\text{Acc}}\ \varphi||} & = {|\varphi|}_{M_c}\\ {||\!{\text{Rej}}\ \varphi||} & = {|\neg\varphi|}_{M_c}\\ {||m||} & = \{f: \text{ for every sentence } \varphi, \text{ if } m\vdash {\text{Acc}}\ \varphi, \text{ then } f\models \varphi\}. \end{align*} $$

One needs to show that this is a semantic interpretation of $\mathcal {E}$ .

Begin by showing that ${||m||}$ is modally reflexive, i.e. (1) that it has a fixed modal background and (2) that it prefers its own factual content.

(1) Take any selection functions f and $f'$ in ${||m||}$ . If one can show that $f(F({|\varphi |}))$ is defined iff $f'(F({|\varphi |}))$ is defined, for all factual $\varphi $ , then f and $f'$ have the same modal background. So assume that $f(F({|\varphi |}))$ is defined and for reductio that $f'(F({|\varphi |}))$ is not defined. As $f(F({|\varphi |}))$ is defined, ${\scriptstyle \mathrm {MB}}_f\cap F({|\varphi |})$ is nonempty. As a result $f\not \models \varphi \rightarrow {\perp }$ . So $m\not \vdash {\text {Acc}}\ \varphi \rightarrow {\perp }$ . So $m\,\circledast {\text {Acc}}\ \varphi $ is coherent. By Theorem 2, $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\neg {\perp }$ and so $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Rej}} {\perp }$ . So $m\vdash {\text {Rej}} (\varphi \rightarrow {\perp })$ and so $m\vdash {\text {Acc}} \neg (\varphi \rightarrow {\perp })$ . By the construction $f'\models \neg (\varphi \rightarrow {\perp })$ . But as ${\scriptstyle \mathrm {MB}}_{f'}\cap F({|\varphi |})$ is empty, $f'\models \varphi \rightarrow {\perp }$ and we have a contradiction.

(2) Let $X = F({||m||})$ . Assume that $f\in {||m||}$ . Assume that $A\cap F({||m||})\neq \emptyset $ . There is thus some $f'\in {||m||}$ such that $f'(W)\in A$ . So $m\not \vdash {\text {Acc}}\neg S(A)$ ( $m\not \vdash {\text {Rej}} S(A)$ ). Assume for reductio that $f(A)\not \in F({||m||})$ . There is then some B such that $f(A)\in B$ and $B\cap F({||m||}) = \emptyset $ . So $m\vdash {\text {Acc}} \neg S(B)$ but $m\not \vdash {\text {Acc}} (S(A)\rightarrow S(B))$ . But given Redundancy this violates R-Vacuity. For by Redundancy $m\,{\oplus } {\text {Acc}}\ S(A)\vdash {\text {Acc}}\neg S(B)$ and so by R-Vacuity $m\,\circledast {\text {Acc}} S(A)\vdash {\text {Acc}} \neg S(B)$ . So we have a contradiction: $f(A)\in F({||m||})$ .

Note that as the set of accepted sentences in m is closed under $\models _{\text {ICL}}$ (follows from the above Theorem 2) we have (this can be shown by familiar techniques from modal logic), where $a = {\text {Acc}}\ \varphi $ or $a = {\text {Rej}}\ \varphi $ :

$$ \begin{align*} m\vdash a \text{ iff } {||m||}\subseteq{||a||}. \end{align*} $$

Proof of Corollary 1.

Assume that mental states are represented as closed propositions (so ${||m||}= \bigcap \{{|\varphi |}\,:\, {||m||}\subseteq {|\varphi |}\}$ ). It follows that if ${||m||}\subseteq {|\varphi |}$ iff ${||m'||}\subseteq {|\varphi |}$ for all $\varphi $ , then ${||m||}= {||m'||}$ .

1. ${||m\,{\oplus } {\text {Acc}}\ \varphi ||}\subseteq {|\psi |}$ iff ${||m\,{\oplus } {\text {Acc}}\ \varphi ||}\subseteq {||\!{\text {Acc}}\ \psi ||}$ iff $m\,{\oplus } {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ iff $m\vdash {\text {Acc}}( \varphi \supset \psi )$ iff ${||m||}\subseteq {||\!{\text {Acc}}\ \varphi \supset \psi ||}$ iff ${||m||}\subseteq {|\varphi \supset \psi |}$ iff ${||m||}\cap {|\varphi |}\subseteq {|\psi |}$ iff ${||m||}\cap {||\!{\text {Acc}}\ \varphi ||}\subseteq {|\psi |}$ . So ${||m\,{\oplus } {\text {Acc}}\ \varphi ||} = {||m||}\cap {||\!{\text {Acc}}\ \varphi ||}$ .

2. ${||m\,\circledast {\text {Acc}}\ \varphi ||}\subseteq {|\psi |}$ iff ${||m\,\circledast {\text {Acc}}\ \varphi ||}\subseteq {||\!{\text {Acc}}\ \psi ||}$ iff $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ iff $m\vdash {\text {Acc}} (\varphi \rightarrow \psi )$ iff ${||m||}\subseteq {||\!{\text {Acc}}\ \varphi \rightarrow \psi ||}$ iff ${||m||}\subseteq {|\varphi \rightarrow \psi |}$ . By showing that ${||m||}\subseteq {|\varphi \rightarrow \psi |}$ iff ${||m||}/F({|\varphi |})\subseteq {|\psi |}$ , we thus show that ${||m\,\circledast {\text {Acc}}\ \varphi ||} = {||m||}/F({|\varphi |})$ . So assume that ${||m||}\subseteq {|\varphi \rightarrow \psi |}$ . If $m/F({|\varphi |})$ is empty, then trivially it is a subset of ${|\psi |}$ . So assume that $m/F({|\varphi |})$ is nonempty. Take any $f\in {||m||}/F({|\varphi |})$ . There is some $f'\in {||m||}$ such that $f = f'/F({|\varphi |})$ . As $f'\in {|\varphi \rightarrow \psi |}$ , $f'/F({|\varphi |}) = f \in {|\psi |}$ . So ${||m||}/F({|\varphi |})\subseteq {|\psi |}$ . For the other direction assume that ${||m||}/F({|\varphi |})\subseteq {|\psi |}$ . If ${||m||}/F({|\varphi |})$ is empty then ${||m||}\subseteq {|\varphi \rightarrow \psi |}$ . So assume that ${||m||}/F({|\varphi |})$ is nonempty. Take any $f\in {||m||}$ . As $f/F({|\varphi |})\in {|\psi |}$ , $f\in {|\varphi \rightarrow \psi |}$ . So ${||m||}\subseteq {|\varphi \rightarrow \psi |}$ .

3. A direct consequence of (2) and the construction.□

Proof of Theorem 5.

The claim can be proven by showing how to construct a signature expressivist model for any ICL-model on either L or $L_R$ . The construction provided equates mental states and attitudes with their propositional content (so mental states and attitudes become modal propositions). As a result, for any mental state m, ${||m||} = m$ , and for any attitude a, ${||a||} = a$ (this is ‘cheating’ somewhat as logically equivalent attitudes such as ${\text {Acc}} \neg \varphi $ and ${\text {Rej}}\ \varphi $ will turn out to be identical rather than merely equivalent but this won’t affect the main result).

Take any ICL-model M. Let the set of mental states $\mathcal {M}$ be the set of propositions that are modally reflexive. A mental state m is thus a modal proposition with a fixed modal background; we can denote this modal background by ${\scriptstyle \mathrm {MB}}_m$ .

The set of incoherent mental states is defined: $\mathcal {C} = \{\emptyset \}$ . Define:

$$ \begin{align*} {\text{Acc}}\ \varphi & =_{df} {|\varphi|}_M.\\ {\text{Rej}}\ \varphi & =_{df} {|\neg\varphi|}_M. \end{align*} $$

The set of attitudes $\mathcal {A}$ is thus the set of modal propositions that can be expressed with a sentence. Define (for $a\in \mathcal {A}$ ):

$$ \begin{align*} m\vdash a & \text{ iff}_{df} \, m\subseteq a.\\ m\,{\oplus} a & =_{df} m\cap a.\\ m\,\circledast a & =_{df} m\!/\!\!/\! a. \end{align*} $$

Finally, for mental states m and $m'$ :

$$ \begin{align*} mRm' \text{ iff}_{\text{df}} \text{ there is some } A\in \mathcal{B} \text{ such that } m'\subseteq m/A. \end{align*} $$

This ends the construction. One now needs to show that it satisfies all the requirements on expressivist models for the language L (and so for the language $L_R$ ). It is easy to see that R is reflexive and transitive and that $mR(m\,\circledast a)$ and $mR(m\,{\oplus } a)$ .

Consider the requirements governing acceptance and rejection. The proof that the boolean connectives satisfies the stated requirements on acceptance and rejection is left as an exercise.

$\rightarrow $ Acc: $m\vdash {\text {Acc}}\ \varphi \rightarrow \psi $ iff for all $f\in m$ : $f\models \varphi \rightarrow \psi $ for all $f\in m$ iff ${\scriptstyle \mathrm {MB}}_m \cap {|\varphi |} = \emptyset $ , or for all $f\in m$ : $f\!/\!\!/\! {|\varphi |}\models \psi $ iff (as $m\,\circledast {\text {Acc}}\ \varphi = m\!/\!\!/\! {|\varphi |}) m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ .

$\rightarrow $ Rej: $m\vdash {\text {Rej}}\ \varphi \rightarrow \psi $ iff $f\not \models \varphi \rightarrow \psi $ for all $f\in m$ iff $m\!/\!\!/\! {|\varphi |} \neq \emptyset $ and, for all $f\in m$ : $f\!/\!\!/\! {|\varphi |}\not \models \psi $ iff $m\,\circledast {\text {Acc}}\ \varphi $ is coherent and $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Rej}}\ \psi $ .

Now the structural requirements. The proof that $\,{\oplus }$ as defined satisfies the requirements is left as an exercise, as are the requirements [Classicality 1], [Classicality 2] and [Explosion].

[Strong Success] Assume that $(m\,\circledast {\text {Acc}}\ \varphi )Rm'$ . By construction $m'\subseteq (m\,\circledast {\text {Acc}}\ \varphi )/A$ , for some A, i.e. by construction, $m'\subseteq (m\!/\!\!/\!{|\varphi |})/A$ . Take any $f\in m'$ . There is some $f'\in m$ such that $f = (f'\!/\!\!/\!{|\varphi |})/A$ . By definition $f'\!/\!\!/\!{|\varphi |}$ forces ${|\varphi |}$ . So $f\in {|\varphi |}$ , i.e. $f\models \varphi $ . So $m'\models \varphi $ . So $m\,\circledast {\text {Acc}}\ \varphi \Vdash {\text {Acc}}\ \psi $ .

[Weak Redundancy] Assume that $m\Vdash {\text {Acc}}\ \varphi $ . So $m'\models \varphi $ for every $m'$ such that $mRm'$ . Take any $f\in m$ . Assume for reductio that f does not force ${|\varphi |}$ . So there is some A such that $f/A\not \in {|\varphi |}$ . But then $m/A\not \subseteq {|\varphi |}$ , i.e. $m/A\not \models \varphi $ . But $mR(m/A)$ and so we have a contradiction. So f forces ${|\varphi |}$ ; so $f\!/\!\!/\!{|\varphi |} = f$ . This holds for all $f\in m$ so $m\!/\!\!/\!{|\varphi |} = m\,\circledast {\text {Acc}}\ \varphi = m$ .

[R-Permutation] Follows trivially from the construction (as $f/A/B = f/B/A$ for any A and B).

[R-Inclusion] Assume that $m\,\circledast {\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ (where $\varphi $ and $\psi $ are factual). Two cases. (1) $m\cap {|\varphi |} = \emptyset $ . Trivially: $m\cap {|\varphi |}\subseteq {|\psi |}$ so, $m\,{\oplus }{\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ . (2) $m\cap {|\varphi |} \neq \emptyset $ . Take any $f\in m\cap {|\varphi |}$ , $f(W)= f(F({|\varphi |})) = f/F({|\varphi |})(W)$ and by assumption $f/F({|\varphi |})\in {|\psi |}$ . So $f\in {|\psi |}$ . So $m\cap {|\varphi |}\subseteq {|\psi |}$ and so $m\,{\oplus }{\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ .

[R-Vacuity] Assume that $m\,{\oplus }{\text {Acc}}\ \varphi $ is coherent and $m\,{\oplus }{\text {Acc}}\ \varphi \vdash {\text {Acc}}\ \psi $ , where $\varphi $ and $\psi $ are factual. So $m\cap {|\varphi |}\neq \emptyset $ and $m\cap {|\varphi |}\subseteq {|\psi |}$ . Take any $f\in m$ . As m prefers its own factual content $f(F({|\varphi |})) \in F(m)$ . So there is some $f'\in m$ such that $f(F({|\varphi |})) = f'(W)$ . So $f'\in m\cap {|\varphi |}$ . As $m\cap {|\varphi |}\subseteq {|\psi |}$ , we have $f'\in {|\psi |}$ . So, as $\psi $ is factual, $f/F({|\varphi |})\in {|\psi |}$ . So $m/F({|\varphi |}) = m\,\circledast {\text {Acc}}\ \varphi \subseteq {|\psi |}$ .□

Proof of Observation 3.

(i) is trivial. (ii) By definition $p({|\neg \varphi |}_\psi \cap {|\psi |}) = p({|\psi \wedge (\psi \rightarrow \neg \varphi )|}) $ . Note that ${|\psi \wedge (\psi \rightarrow \neg \varphi )|} = {|\psi \wedge \neg (\psi \rightarrow \varphi )|}$ . So: $p({|\psi \wedge \psi \rightarrow \neg \varphi )|}) = p({|\psi |}) -p({|\psi \wedge (\psi \rightarrow \varphi )|})$ . But then

$$ \begin{align*} \text{Pr}(\neg\varphi\,|\,\psi) & = \frac{p({|\varphi|}_\psi\cap{|\psi|})}{p({|\psi|})}= \frac{p({|\psi|}) -p({|\psi\wedge (\psi\rightarrow \varphi)|})}{p({|\psi|})}\\ & = 1- \frac{p(|\varphi)|_{\psi}\cap{|\psi|)}}{p({|\psi|})} = 1- \text{Pr}(\varphi\,|\,\psi). \end{align*} $$

(iii) Assume that $\varphi $ and $\chi $ are logically incompatible. Note that ${|\psi \wedge (\psi \rightarrow (\varphi \vee \chi ))|} = {|\psi \wedge (\psi \rightarrow \varphi )|} \cup {|\psi \wedge (\psi \rightarrow \chi )|}$ . Moreover, ${|\psi \wedge (\psi \rightarrow \varphi )|} \cap {|\psi \wedge (\psi \rightarrow \chi )|} = \emptyset $ . From this it follows in just a few steps that $\text {Pr}(\varphi \vee \chi \,|\,\psi ) = \text {Pr}(\varphi \,|\,\psi ) + \text {Pr}(\chi \,|\,\psi )$ .□

Proof of Observation 4.

First prove the claim for base conditionals. It is assumed that if ${p}(|\varphi |) = 0$ , then ${p}(|\varphi \rightarrow \psi |) = 1$ . Note that ${|\varphi \rightarrow \psi |}_\chi \cap {|\chi |}= {|\chi \rightarrow (\varphi \rightarrow \psi )|} \cap {|\chi |} = $ (by Import-Export) ${|((\chi \wedge \varphi )\rightarrow \psi )\wedge \chi |}$ . We find

$$ \begin{align*} {p}(|((\chi\wedge\varphi)&\rightarrow \psi)\wedge\chi|) = {p}({|(\chi\wedge\varphi)\rightarrow \psi|}) - {p}({|((\chi\wedge\varphi)\rightarrow \psi)\wedge\neg\chi}|)\\ & = {p}({|(\chi\wedge\varphi)\rightarrow \psi|}) - {p}({|(\chi\wedge\varphi)\rightarrow \psi|})\times {p}({|\neg\chi|}) \quad (\text{by SI)}\\ & = {p}({|(\chi\wedge\varphi)\rightarrow \psi|}) - {p}({|(\chi\wedge\varphi)\rightarrow \psi|})\times (1- {p}(|\chi|))\\ & = {p}({|(\chi\wedge\varphi)\rightarrow \psi|}) - {p}({|(\chi\wedge\varphi)\rightarrow \psi|}) + {p}(|(\chi\wedge\varphi)\rightarrow \psi|)\times p(|\chi|)\\ & = {p}(|(\chi\wedge\varphi)\rightarrow \psi|)\times p(|\chi|). \end{align*} $$

So (where $\text {Pr}(\varphi \wedge \chi )> 0$ ):

$$ \begin{align*} \text{Pr}(\varphi\rightarrow\psi\,|\,\chi) & = \frac{{p}({|\varphi\rightarrow\psi|}_\chi \cap {|\chi|})}{{p}({|\chi|})} = \frac{{p}(|(\chi\rightarrow(\varphi\rightarrow\psi))\wedge \chi|)}{ {p}(|\chi|)}\\&=\frac{{p}(|((\chi\wedge\varphi)\rightarrow\psi)\wedge \chi|)}{ {p}(|\chi|)} = \frac{{p}(|(\chi\wedge\varphi)\rightarrow \psi|)\times p(|\chi|)}{ {p}(|{\chi}|)} \\&= p(|(\chi\wedge\varphi)\rightarrow\psi|) = \frac{{p}(|(\chi\wedge\varphi)\rightarrow\psi|)\times p(|\chi\wedge\varphi|)}{ {p}(|{\chi\wedge\varphi}|)}\\& = \frac{{p}(|((\chi\wedge\varphi)\rightarrow\psi)|\cap |\chi\wedge\varphi|)}{ {p}(|{\chi\wedge\varphi}|)} = \frac{{p}(|\psi|_{\chi\wedge\varphi} \cap|\chi\wedge\varphi|)}{ {p}(|{\chi\wedge\varphi}|)} \\ &= \text{Pr}(\psi\,|\,\varphi\wedge\chi). \end{align*} $$

Next, note that by Import-Export an arbitrarily right-hand nested conditional $\varphi _1\rightarrow (\varphi _2\rightarrow \cdots (\varphi _n\rightarrow \psi )\cdots )$ is logically equivalent to a base conditional $(\varphi _1\wedge \cdots \wedge \varphi _n)\rightarrow \psi $ . Thus ${|\varphi _1\rightarrow (\varphi _2\rightarrow \cdots (\varphi _n\rightarrow \psi )\cdots )|}_\chi = {|(\varphi _1\wedge \cdots \wedge \varphi _n)\rightarrow \psi |}_\chi $ and so they will get the same conditional probability $\text {Pr}(\cdot \,|\,\chi )$ .□

Proof of Observation 5.

It is assumed that if ${p}(|\varphi |) = 0$ , then ${p}(|\varphi \rightarrow \psi |) = 1$ . So consider the following form of (GI):

(A) $$ \begin{align} {p}\left( \left|\neg\varphi\wedge \bigwedge_{1\leq i\leq n}((\varphi\wedge\psi_i)\rightarrow a_i) \right |\right)= {p}(|\neg\varphi|)\times {p}\left(\left|\bigwedge_{1\leq i\leq n}((\varphi\wedge\psi_i)\rightarrow a_i)\right|\right). \end{align} $$

Here each $a_i$ is an atomic sentence or a negated atomic sentence (in light of the original assumption we do not need the restriction that the antecedents have nonzero probability). From (A) we can establish:

(B) $$ \begin{align} {p}\left (\left|\neg\varphi\wedge \bigwedge_{1\leq i\leq n}((\varphi\wedge\varphi_i)\rightarrow a_i) \wedge \bigwedge_{1\leq i \leq m} \neg((\varphi\wedge\chi_i)\rightarrow \perp ) \right|\right )= \nonumber\\ \quad\quad {p}(|\neg\varphi|)\times {p}\left(\left|\bigwedge_{1\leq i\leq n}((\varphi\wedge\varphi_i)\rightarrow a_i) \wedge \bigwedge_{1\leq i \leq m} \neg((\varphi\wedge\chi_i)\rightarrow \perp )\right|\right). \end{align} $$

For the probability of any conjunction containing negated terms can always be written out as the sum or difference of probabilities of conjuncts containing only unnegated terms. E.g. ${p}(|\neg \varphi \wedge ((\varphi \wedge \psi )\rightarrow a) \wedge \neg ((\varphi \wedge \chi )\rightarrow {\perp })|) = {p}(|\neg \varphi \wedge ((\varphi \wedge \psi )\rightarrow a)|) - {p}(|\neg \varphi \wedge ((\varphi \wedge \psi )\rightarrow a) \wedge ((\varphi \wedge \chi )\rightarrow {\perp })|)$ . So (B) can be written as a series of terms where there are no negated conditionals, in which case (GI) applies to each term.

From (B) it follows by basic probabilistic principles (see previous observation) that:

(C) $$ \begin{align} {p}\left (\left|\varphi\wedge \bigwedge_{1\leq i\leq n}((\varphi\wedge\varphi_i)\rightarrow a_i) \wedge \bigwedge_{1\leq i \leq m} \neg((\varphi\wedge\chi_i)\rightarrow \perp ) \right|\right )= \nonumber\\ \quad\quad {p}(|\varphi|)\times {p}\left(\left|\bigwedge_{1\leq i\leq n}((\varphi\wedge\varphi_i)\rightarrow a_i) \wedge \bigwedge_{1\leq i \leq m} \neg((\varphi\wedge\chi_i)\rightarrow \perp )\right|\right). \end{align} $$

Recall (Theorem 3) that the disjunctive normal form of a conditional $\varphi \rightarrow \psi $ is a disjunction $D(\varphi \rightarrow \psi )$ of mutually inconsistent disjuncts $\delta _1\vee \cdots \vee \delta _n$ where each $\delta _j$ has the form

$$ \begin{align*} \bigwedge_{1\leq i\leq k_j}((\varphi\wedge\psi^j_i)\rightarrow a^j_i) \wedge \bigwedge_{1\leq i \leq m_j} \neg((\varphi\wedge\chi^j_i)\rightarrow \perp ). \end{align*} $$

Together with claim (C) this ensures that for any sentence $\varphi \rightarrow \psi $ :

(D) $$ \begin{align} {p}(|\varphi&\wedge (\varphi\rightarrow\psi)|) \nonumber\\[3pt] & = p({|\varphi\wedge D(\varphi\rightarrow \psi)|} = p({|(\varphi \wedge \delta_1)\vee\cdots\vee (\varphi\wedge\delta_n)|}) = \sum_{1\leq j\leq n}p({|\varphi \wedge \delta_j|})\nonumber\\[3pt] & = \sum_{1\leq j\leq n} {p}\left(\left | \varphi\wedge\bigwedge_{1\leq i\leq k_j}((\varphi\wedge\psi^j_i)\rightarrow a^j_i) \wedge \bigwedge_{1\leq i \leq m_j} \neg((\varphi\wedge\chi^j_i)\rightarrow \perp )\right|\right).\nonumber\\[3pt] & = {p}(|\varphi|)\times\sum_{1\leq j\leq n} {p}\left(\left| \bigwedge_{1\leq i\leq k_j}((\varphi\wedge\psi^j_i)\rightarrow a^j_i) \wedge \bigwedge_{1\leq i \leq m_j} \neg((\varphi\wedge\chi^j_i)\rightarrow \perp )\right|\right).\nonumber\\[3pt] & = {p}(|\varphi|)\times{p}(|\varphi\rightarrow \psi|). \end{align} $$

Now assume that $\psi $ has the form $\chi \rightarrow \sigma $ , where $\chi $ is factual and $\sigma $ can be an arbitrary sentence of $L_R$ . By the definition of $\text {Pr}(\cdot \,|\, \cdot )$ it follows that (where $\text {Pr}(\varphi \wedge \chi )> 0$ ):

$$ \begin{align*} \text{Pr}(\chi\rightarrow\sigma\,|\,\varphi) & = \frac{{p}(|\chi\rightarrow\sigma|_\varphi \cap |\varphi|)}{{p}(|\varphi|)} = \frac{{p}(|(\varphi\rightarrow (\chi\rightarrow\sigma))\wedge\varphi|)}{{p}(|\varphi|)}\\[6pt] & = \frac{{p}(|\varphi\rightarrow (\chi\rightarrow\sigma)|)\times{p}(|\varphi|)}{{p}(|\varphi|)} \quad \text{(by (D))} \\[6pt] & = {p}(|\varphi\rightarrow(\chi\rightarrow \sigma)|) = {p}(|(\varphi\wedge\chi)\rightarrow\sigma|)\\[6pt] & = \frac{{p}(|(\varphi\wedge\chi)\rightarrow \sigma|)\times{p}(|\varphi\wedge\chi|)}{{p}(|\varphi\wedge\chi|)}\\[6pt] & = \frac{{p}(|((\varphi\wedge\chi)\rightarrow \sigma)\wedge\varphi\wedge\chi|)}{{p}(|\varphi\wedge\chi|)} \quad\text{(by (D))}\\ & = \frac{{p}(|\sigma|_{\varphi\wedge\chi}\cap |\varphi\wedge\chi|)}{{p}(|\varphi\wedge\chi|)} \\ & = \text{Pr}(\sigma\,|\,\varphi\wedge\chi). \end{align*} $$

□

Proof of Observation 6.

Assume that $X,\,f\models \text {Might}\, (\varphi \rightarrow \psi )$ . Then there is an $f'\in X$ such that $X,\, f'\models \varphi \rightarrow \psi $ . The case when $f'/\varphi $ is empty is trivial (recall that X has a fixed modal background). So consider the case when $X/\varphi , f'/\varphi \models \psi $ . As $f/\varphi \in X/\varphi $ : $X/\varphi , f/\varphi \models \text {Might }\psi $ , so $X,\, f\models \varphi \rightarrow \text {Might }\psi $ .

Assume instead that $X,\,f\models \varphi \rightarrow \text {Might}\, \psi $ . The case when $f/\varphi $ is empty is trivial. So consider the case when $X/\varphi , f/\varphi \models \text {Might}\,\psi $ . So there is some $f'\in X/\varphi $ such that $X/\varphi , f'\models \psi $ . But there is some $f''\in X$ such that $f''/\varphi = f'$ . So $X/\varphi , f''/\varphi \models \psi $ . So $X,\, f''\models \varphi \rightarrow \psi $ . But then $X,\,f\models \text {Might}\, (\varphi \rightarrow \psi )$ .□

Proof of Observation 7.

Let ${\perp } A = \{f\,|\, f/A = \emptyset \}$ . Note first:

(A) $$ \begin{align} \{f'\,|&\, f'/{\scriptstyle \mathrm{MB}}_f\in{|\varphi\rightarrow \psi|}\} \cup {\perp} {\scriptstyle \mathrm{MB}}_f\nonumber\\ & = \{f'\,|\, [f'/{\scriptstyle \mathrm{MB}}_f\cap F({|\varphi|})]\in{|\psi|}\} \cup{\perp}({\scriptstyle \mathrm{MB}}_f\cap F({|\varphi|})) \cup {\perp}{\scriptstyle \mathrm{MB}}_f\nonumber\\ & = \{f'\,|\,[ f'/{\scriptstyle \mathrm{MB}}_f\cap F({|\varphi|})]\in{|\psi|}\}\cup {\perp}({\scriptstyle \mathrm{MB}}_f\cap F({|\varphi|})). \end{align} $$

The last step holds in virtue of ${\perp } {\scriptstyle \mathrm {MB}}_f\subseteq {\perp }({\scriptstyle \mathrm {MB}}_f\cap F({|\varphi |}))$ .

(1) Take any f such that $f\models \neg (\varphi \rightarrow {\perp })$ . $ f\models P_n(\varphi \rightarrow \psi )$ iff $ p*{\scriptstyle \mathrm {MB}}_f ({|\varphi \rightarrow \psi |}) = V(n)$ iff $p(\{f'\,|\, f'/{\scriptstyle \mathrm { MB}}_f\in {|\varphi \rightarrow \psi |}\} \cup {\perp } {\scriptstyle \mathrm {MB}}_f) = V(n)$ iff (by (A)) $p*({\scriptstyle \mathrm {MB}}_f\cap F({|\varphi |}))({|\psi |}) = V(n)$ iff (as ${\scriptstyle \mathrm {MB}}_{f/F({|\varphi |})} = {\scriptstyle \mathrm {MB}}_f\cap F({|\varphi |})$ ) $f/F({|\varphi |})\models P_n(\psi )$ iff (as $f/F({|\varphi |})\neq \emptyset $ ) $f\models \varphi \rightarrow P_n(\psi )$ . So: $\models \neg (\varphi \rightarrow {\perp })\supset (P_n(\varphi \rightarrow \psi )\equiv (\varphi \rightarrow P_n(\psi )))$ .

(2) $f\models P_n(\psi \,|\,\varphi )$ iff $p*({\scriptstyle \mathrm { MB}}_f\cap F({|\varphi |}))({|\psi |}) = V(n)$ or $f/F({|\varphi |}) = \emptyset $ iff $f/F({|\varphi |})\models P_n(\psi )$ or $f/F({|\varphi |}) = \emptyset $ iff $f\models \varphi \rightarrow P_n(\psi )$ .