Show that, at each possible world $\Gamma$ of a modal model, $\Gamma \Vdash \square X \equiv \sim \diamondsuit \sim X$

Question

Show that, at each possible world $\Gamma$ of a modal model, $\Gamma \Vdash \square X \equiv \sim \diamondsuit \sim X$.

I'm not exactly sure how to proceed here.

I know that a modal model $\mathcal{M}=(\mathcal{g},\mathcal{R},\Vdash)$, is composed of the nonempty set of possible worlds, $\mathcal{g}$; a binary accessibility relation, $\mathcal{R}$; and $\Vdash$, a relation between possible worlds and propositional letters.

However, I'm not exactly sure how to start the proof, or what method would work best. It looks like aiming for a contradiction would be best, since $\square X = \sim \diamondsuit \sim X$ (?).

Answer 1

Given the definitions $$\begin{align} w \Vdash \diamondsuit X &:= \exists w'\,(w R w' \land w'\Vdash X), \\ w \Vdash \square X &:= \forall w'\,(w R w' \to w'\Vdash X) \\ \end{align}$$ for any world $w$. In the metatheory, we can use equivalences of classical logic:

• $\neg(p\to q) \equiv(p \land \neg q)$,
• $\forall = \neg\exists\neg$ and $\exists = \neg\forall\neg$.

Putting these together, $$\begin{align} w \Vdash \square X &:= \forall w'\,(w R w' \to w'\Vdash X) \\ &\equiv \neg\exists w'\neg\,(w R w' \to w'\Vdash X) \\ &\equiv \neg\exists w'\,(w R w' \land \neg (w'\Vdash X)) \\ &\equiv \neg\exists w'\,(w R w' \land w'\Vdash \neg X) \\ &\equiv \neg w\Vdash\diamondsuit \neg X\\ &\equiv w\Vdash\neg\diamondsuit \neg X.\\ \end{align}$$

So the duality law $\square = \neg\diamondsuit\neg$ holds in any world $w$. Similarly, or using $\neg\neg X \equiv X$, its dual $\diamondsuit = \neg\square\neg$ holds in any world.