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?



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