# In modal logic, is $\lnot\square P\equiv\lozenge\lnot P$?

"Possibly" and "necessarily" seem very much like "exists" and "for-all", but does the following hold true: $\neg \square P \equiv \lozenge \neg P$ in the same way as $\neg\forall P \equiv \exists\neg P$ ?

From the definition of "necessarily" in my book, being that $P$ should be true in all possible worlds, it seems to be so.

A corner case will be if there are no possible worlds. In this case, the "necessarily" is vacuously true, so its negation is false; at the same time, there are no possible worlds whatsoever, so there's no world where $\lnot P$ is possible, so the right-hand statement is false as well.

Is this so, or is my reasoning flawed?

• Yes. The modal operators are really just a restricted form of quantification and your argument is fine. – Qiaochu Yuan Mar 6 '12 at 1:15
• @QiaochuYuan That is correct. If you express your comment as an answer, I will upvote it. – user22805 Mar 6 '12 at 1:31
• @Ilya: What are your axioms? Do you define $\Box P = \lnot \Diamond \lnot P$? Is the law of excluded middle valid? – Zhen Lin Mar 6 '12 at 7:38
• Yes, $\square P \equiv \neg \lozenge \neg P$ -- which also leads me towards trusting my conclusion. – Ilya Mar 6 '12 at 12:04

Yes. There's a strong convention that $\Box$ and $\Diamond$ are always each other's duals, even in special-purpose modal logics. When the propositional substratum is classical, this implies that $\neg\Box\equiv \Diamond\neg$ and $\Box\neg\equiv\neg\Diamond$.
For example, when $\Box P \leftrightarrow \neg\Diamond\neg P$ is an axiom (or the definition of $\Box$ as an abbreviation), you get these laws by either negating both sides or letting $P$ be $\neg Q$, and then applying double-negation elimination.