WebSep 19, 2015 · Thus every world visible from w ′ satisfies ϕ. But, because of S5, visibility is an equivalence relation. Thus, we have that every world visible from w is visible from w ′, and vice versa. Thus every world visible from w satisfies ϕ. Thus w satisfies ϕ. We conclude every world w satisfies ϕ → ϕ. Now, use the completeness proof for S5 ...
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