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?
