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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.

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

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  • $\begingroup$ I think you have the implications corresponding to reflexivity and transitivity the wrong way round. $\endgroup$ – Alex Kruckman Mar 30 at 14:06
  • $\begingroup$ @AlexKruckman I don't think so... Where exactly is the error? $\endgroup$ – frabala Mar 30 at 14:34
  • $\begingroup$ 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 $\square A \rightarrow \square\square A$. $\endgroup$ – Alex Kruckman Mar 30 at 14:38
  • $\begingroup$ 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 $a$. $\endgroup$ – Alex Kruckman Mar 30 at 14:44
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    $\begingroup$ 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 $w$. $\endgroup$ – Alex Kruckman Mar 30 at 15:12

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