# Proving a logic axiom involving counterfactuals

I just read David Lewis' (1973) "Counterfactuals and Comparative Possibility" where he states an axiom (p. 441) I'd like to prove.

$$((a \vee b) \Box\rightarrow a) \vee ((a \vee b) \Box\rightarrow b) \vee (((a \vee b) \Box\rightarrow c) \leftrightarrow (a \Box\rightarrow c) \wedge (b \Box\rightarrow c))$$

Suppose for reductio that the negated axiom is true. Then $$\neg((a \vee b) \Box\rightarrow a)$$ $$\neg((a \vee b) \Box\rightarrow b)$$ $$\neg(((a \vee b) \Box\rightarrow c) \leftrightarrow (a \Box\rightarrow c) \wedge (b \Box\rightarrow c))$$

Given the negated biconditional, it must also be true that either

$$((a \vee b) \Box\rightarrow c)$$ $$\neg((a \Box\rightarrow c) \wedge (b \Box\rightarrow c))$$ or $$\neg((a \vee b) \Box\rightarrow c)$$ $$(a \Box\rightarrow c) \wedge (b \Box\rightarrow c)$$

According to Lewis (p. 424), "$$A \Box\rightarrow C$$ is true at $$i$$ iff some (accessible) $$AC$$-world is closer to $$i$$ than any $$A\neg C$$-world, if there are any (accessible) $$A$$-worlds."

So, in order to complete the proof I gather that I need to assume a certain distribution of truths at $$i$$ and the closet worlds to $$i$$ and then show that a contradiction follows from any possible distribution of truths. Is this correct and if so, is there an easy way to do this?