I read that the schema $$\varphi\rightarrow\square\diamond\varphi$$ corresponds to the symmetric property (D. Palladino, C. Palladino, Logiche non classiche, 'non-classical logics', 2007) of the relation $R$ defined in a model of Kripke semantics.

I am not sure, but I suspect that it means that $\varphi\rightarrow\square\diamond\varphi$ 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 symmetric.

It is quite easy to verify that, if $R$ is symmetric, then $\varphi\rightarrow\square\diamond\varphi$ is valid.

Is the converse true? I suppose that the contrapositive could be used to prove that if $(W,R,I)\models\varphi\rightarrow\square\diamond\varphi$ then $R$ is symmetric, analogously to what has been done here by a very kind user, whom I thank again, but I have got some problems in building a model where $uRv$, $uRw$, $\lnot uRw$ and there is a world where $p\land\lnot\square\diamond p$ holds... Thank you very much for any answer!

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    $\begingroup$ You can see Alexander Chagrov & Michael Zakharyaschev,Modal Logic (1997), page 78 for this (an other) result. $\endgroup$ Feb 28, 2015 at 7:08

1 Answer 1


Reference :

  • Alexander Chagrov & Michael Zakharyaschev, Modal Logic (1997), page 78.

I'll copy-paste the proof from the above source:

Proposition 3.32 : $\mathscr F$ validates $p \to \square \diamond p$ iff $\mathscr F$ is symmetric.

Only $(\Rightarrow)$ requires a poof. If $\mathscr F= (W,R)$ is not symmetric then there are $x,y \in W$ such that $xRy$ and $\lnot yRx$. Define a valuation $V \in \mathscr F$ by taking $V(p) = \{ x \}$. Then we have $x \vDash p, y \nvDash \diamond p$, whence $x \nvDash \square \diamond p$ and $x \nvDash p \to \square \diamond p$.

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    $\begingroup$ ...where I'd say that $V(p)=\{x\}$ means that the only world where $p$ is true is $x$. Thank you very very much! $\endgroup$ Feb 28, 2015 at 11:41

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