Converse of Axiom K Just for clarity, this is the specific formulation I am referring to.  
$K$: $\Box(p\to q)\to(\Box p\to\Box q)$
Naturally, we might consider the converse of this statement:  
$K'$: $(\Box p\to\Box q)\to\Box(p\to q)$
The question is simple. Given $K$, does $K'$ follow? An accompanying proof of the answer would be greatly appreciated.
 A: $K$ does not imply $K'$; indeed, I can't think of a reasonable system which does include $K'$, which doesn't trivialize the modality.
Think of a Kripke frame with three worlds $a, b, c$ where


*

*$a$ sees all three worlds,

*neither $b$ nor$c$ see any worlds, and

*$b\models p\wedge\neg q$, $c\models q\wedge\neg p$.
$K$ is true in this frame, but $K'$ is not: "$\Box p\rightarrow\Box q$" is true at $a$ (since $\Box p$ is false at $a$), but "$\Box(p\rightarrow q)$" is false at $a$ (since $a$ sees $b$ and $b\models\neg(p\rightarrow q)$).

EDIT: It's not hard to pin down exactly what Kripke frames validate $K'$: they are the frames in which each world sees at most one world (possibly itself).
Showing that such frames verify $K'$ is easy. If a world $a$ sees no worlds, then we trivially have $a\models\Box (p\implies q)$; conversely, if $aRb$ and $a$ sees no worlds besides $b$, then "$a\models \Box t$" is the same as "$b\models t$," so $K'$ holds at $a$ since $(p\implies q)\implies (p\implies q)$ holds at $b$.
Conversely, set $p=\neg q$ in $K'$, and note that $K'$ tells us in particular that if $\neg\Box p$ then $\Box (p\implies q)$ - which in our case translates to $(\Diamond q)\implies(\Box q)$ (since "$\neg\Box \neg$"="$\Diamond$"
 and "$\neg q\implies q$"="$q$"). So every frame validating $K'$ also validates the "trivialization" principle $$(*)\quad(\Diamond q)\implies (\Box q).$$ But this statement clearly implies that in such a frame, no world sees more than one world (if $aRb, aRc,$ and $b\not=c$, set $q$ true at $b$ so that $a\models\Diamond q$ and set $q$ false at $c$ so that $a\models\neg\Box q$).
Note that we've in fact shown that in Kripke frames, validating $K'$ is the same as validating $(*)$. I don't know off the top of my head what the weakest modal system $X$ is such that $X+(*)$ is equivalent to $X+K'$, but I suspect it's quite weak.
