Appendix: Proof of condition $$(*)$$ in Section 4 of Logic of Conditionals

For the sake of simplicity, we prove the version for acceptance of conditionals with respect to a premise set $$\Gamma$$, which uses the following condition: $\tag{PS$$_{acc}$$} \Gamma \Vdash_{acc} A>C \text{ iff for each } X \in \Gamma\bot |A|, \ \bigcap (X\cup |A|) \subseteq |C|.$ The version for truth at $$w$$ and with respect to $$\Gamma_w$$ (the premise set assigned to a world $$w$$) is analogous.

In the following, we assume that every proposition (subset of $$W$$) is expressible in our language.

Suppose that $$\Gamma$$ is a (possibly infinite) set of premises (propositions). We denote by $$\Gamma\bot\lvert A\rvert$$ the set of all subsets $$X$$ of $$\Gamma$$ that are maximally consistent with $$A$$, i.e., are maximal in having a non-empty intersection $$(\bigcap X) \cap \lvert A\rvert.$$

In the following, we suppose that $$\lvert A\rvert$$ and $$\lvert B\rvert$$ are non-empty.

Now suppose that $$(A\vee B) >C$$ is accepted w.r.t. $$\Gamma$$, i.e., for all elements $$X$$ of $$\Gamma\bot|A\vee B|$$, it holds that $$(\bigcap X)\cap |A\vee B| \subseteq \lvert C\rvert.$$ Equivalently,

$\tag{1} \left(\bigcup_{X\in \Gamma\bot|A\vee B|}\bigcap X\right)\cap |A\vee B| \subseteq \lvert C\rvert.$

We first show that

$\tag{2} \left(\bigcup_{Y\in \Gamma\bot|A|\\ Z\in \Gamma\bot|B|} \bigcap (Y \cup Z)\right) \cap |A|\cap|B| \ \subseteq \ \bigcup_{X\in \Gamma\bot|A\vee B|}\bigcap X.$

Take two arbitrary sets $$Y\in \Gamma\bot |A|$$ and $$Z\in \Gamma\bot |B|$$ and form the union $$Y\cup Z$$. Every subset of $$\Gamma$$ which is a superset of $$Y\cup Z$$ is inconsistent both with $$A$$ and with $$B$$, so it is inconsistent with $$A\vee B$$. Now if $$Y\cup Z$$ is consistent with $$A\vee B$$, then $$Y\cup Z$$ is itself an element of $$\Gamma\bot |A\vee B|$$, so it appears as an $$X$$ on the right-hand side of (2). If, on the other hand, $$Y\cup Z$$ is inconsistent with $$A\vee B$$ (which may well be the case), then the intersection of $$\bigcap (Y \cup Z)$$ with $$|A|\cap|B|$$ will leave only the empty set. This proves (2).

From (1) and (2), we can infer

$\tag{3} \left(\bigcup_{Y\in \Gamma\bot|A|\\ Z\in \Gamma\bot|B|} \bigcap (Y \cup Z)\right) \cap |A|\cap|B| \ \subseteq \ |C|.$

Now we make use of the following series of equations:

\begin{align}\tag{4} &\left(\bigcup_{Y\in \Gamma\bot|A|\\ Z\in \Gamma\bot|B|} \bigcap (Y \cup Z)\right) \cap |A|\cap|B| \\ &\qquad= \left(\bigcup_{Y\in \Gamma\bot |A|\\ Z\in \Gamma\bot |B|} \left(\bigcap Y \cap \bigcap Z\right)\right) \cap |A|\cap|B| \\ &\qquad= \left(\bigcup_{Y\in \Gamma\bot |A|} \bigcap Y\right) \cap \left(\bigcup_{Z\in \Gamma\bot |B|} \bigcap Z\right) \cap |A|\cap|B| \notag \\ &\qquad= \left(\bigcup_{Y\in \Gamma\bot |A|} \left(\bigcap Y \cap |A|\right)\right) \cap \left(\bigcup_{Z\in \Gamma\bot |B|} \left(\bigcap Z \cap |B|\right)\right)\notag \end{align}

From (3) and (4), it follows that

$\tag{5} \left(\bigcup_{Y\in \Gamma\bot |A|} \left(\bigcap Y \cap |A|\right)\right) \cap \left(\bigcup_{Z\in \Gamma\bot |B|} \left(\bigcap Z \cap |B|\right)\right) \subseteq |C|.$

Since we assume that every proposition is expressible in our language, we can choose two sentences $$D$$ and $$E$$ such that

$\lvert D\rvert = \bigcup_{Y\in \Gamma\bot\lvert A\rvert} \left(\bigcap Y \cap \lvert A\rvert\right)$

and

$\lvert E\rvert = \bigcup_{Z\in \Gamma\bot\lvert B\rvert} \left(\bigcap Z \cap \lvert B\rvert\right).$

(5) says that $$D$$ and $$E$$ together entail $$C$$. By the acceptance condition (PS$$_{acc}$$) for conditionals, we also know that $$A>D$$ and $$B>E$$ are accepted with respect to $$\Gamma$$. This proves $$(*)$$. $$\Box$$