2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ We can use (4) and (2) to prove u==w $\endgroup$ – JonMark Perry Feb 27 '15 at 13:23
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you very much! I didn't take using the contrapositive into account because I've never seen its use in the introductory book I'm following in similar cases. Very interesting technique! $\endgroup$ – Self-teaching worker Feb 27 '15 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.