Euclidean relations and $\diamond P\rightarrow\square\diamond P$

Question

I read* that the formula $$\diamond \varphi\rightarrow\square\diamond\varphi$$is valid in a structure $(W,R)$, intended as in Kripke semantics, -i.e. that it is true for any interpretation $I$ and in any world $u\in W$ of a model $(W,R,I)$- if and only if relation $R$ is Euclidean, i.e. if and only if relation $R$ is such that, if $uRw$ and $uRv$, then $vRw$ and $wRv$.

I would be interested in proving it to myself, but, although the converse is quite straightforward even for me, I cannot find a way to show that if $\diamond \varphi\rightarrow\square\diamond\varphi$ is valid then $R$ is Euclidean. I have tried to define a particular interpretation $I$ such that the Euclidean character of $R$ is shown, but I cannot find one. I have tried, for example: once a world $u\in W$ has been fixed, defining $I(P,w)=1$, i.e. $P$ true in world $w$, if and only if $w$ is accessible from $u$ ($ uRw$). But my trials have been fruitless until now. Could anybody explain how to prove that $\diamond \varphi\rightarrow\square\diamond\varphi$ is valid only if $R$ is Euclidean? Thank you very much!

*D. Palladino, C. Palladino, Logiche non classiche.

Answer 1

We want to show that validity of $\diamond p\implies \Box\diamond p$ implies Euclideanness. I'll prove the contrapositive.

Suppose I have a Kripke frame $K$ with three worlds $u, v, w\in K$ such that $uRv, uRw,$ but $\neg vRw$ - that is, $u, v, w$ witness a failure of Euclideanness.

Then consider some valuation which assigns $p$ "True" at $w$, and "false" at every world that $v$ sees. (Note that such a valuation exists since $\neg vRw$; in particular we can just assign $p$ "True" at all worlds but $w$.) We have:

• $u\models\diamond p,$ but

• $v\models\neg\diamond p$; so

• $u\models\neg\Box\diamond p$.

So we are done.