# How to interpret $\Box\Box A$ in the possible worlds

The possible world semantics (Kripke semantics) defines $$\Box A$$ as follows: $$v(\Box A, \omega)=T\iff \forall \omega'\in W\:(\omega R\omega'\land v(A,\omega')=T)$$ And so $$v(\Box\Box A, \omega)=T\iff \forall \omega'\in W\:\forall \omega''\in W\:(\omega R\omega'\land\omega' R\omega''\land v(A,\omega'')=T)$$ This is clear for temporal logic, i.e. $$\Box$$ means always. But if $$\Box$$ means necessary, it is unclear to me. So how is it interpreted for necessary? Any thought is welcome.

• $\Box$ isn't interpreted as possible, but rather necessary. – Noah Schweber Mar 30 at 0:08

The first line says "$$A$$ necessarily holds in wolrd $$w$$ iff $$A$$ holds in all worlds reachable from $$w$$". You can think of it as, $$A$$ necessarily holds in wolrd $$w$$ iff $$A$$ holds all around as far as one can see from $$w$$. If you've traveled everywhere in your reach and in all that map you see $$A$$, you believe that $$A$$ holds necessarily, it is a law for you.

The second line says "$$A$$ is necessarily necessary in $$w$$ iff $$A$$ holds in all worlds that are reachable from all worlds that are reachable from $$w$$".

Assume that $$R$$ is reflexive. If $$\Box\Box A$$, then by setting $$w'=w$$ in the interpretation, you get a result equivalent to the interpretation of $$\Box A$$. This justifies $$\Box\Box A \to \Box A$$.

Assume that $$R$$ is transitive. Quoting Alex Kruckman from the comments,

If $$A$$ is true at every world reachable from $$w$$, then $$A$$ is true at every world reachable from every world reachable from $$w$$, since all such worlds are reachable from $$w$$.

This justifies $$\Box A\to\Box\Box A$$. After all, if something is necessary, it is necessarily necessary.

• I think you have the implications corresponding to reflexivity and transitivity the wrong way round. – Alex Kruckman Mar 30 at 14:06
• @AlexKruckman I don't think so... Where exactly is the error? – frabala Mar 30 at 14:34
• Suppose $R$ is transitive. If $A$ is true at every world reachable from $w$, then $A$ is true at every world reachable from every world reachable from $w$, since all such worlds are reachable from $w$. So $\square A \rightarrow \square\square A$. – Alex Kruckman Mar 30 at 14:38
• On the other hand reflexivity is not sufficient to justify this implication. Suppose we have worlds $a$, $b$, $c$, each reachable from themselves, with $b$ reachable from $a$ and $c$ reachable from $b$, and no other relations. If $p$ is true at $a$ and $b$, but not at $c$, then $\square p$ is true at $a$, but $\square \square p$ is not true at $a$. – Alex Kruckman Mar 30 at 14:44
• That is, in your argument you reasoned about a particular world $w'$ reachable in two steps from $w$, which naturally gives you a conclusion about $\lozenge$. To get a conclusion about $\square$, you need to reason about all worlds reachable in two steps from $w$. – Alex Kruckman Mar 30 at 15:12