I am looking for a Kripke frame condition corresponding to the McKinsey axiom M: $\Box\Diamond p \rightarrow \Diamond\Box p$. I read somewhere the following condition:

"For every partitioning of the set of worlds into two disjoint partitions, every world can see a world whose successors lie all in the same partition."

This follows from rewriting the formula as $\Diamond\Box \lnot p \lor \Diamond\Box p$. But it is difficult to use because it involves sets of worlds.

So I am looking for a frame condition for KM of the form $\forall w P(w)$, where $P$ is a predicate expressing visibility between $w$ and other worlds. In the case of M, this condition cannot be first-order, but it could still be second order. For example:

  • Formula $(p\land \Box(\Diamond p \rightarrow p)) \rightarrow \Box p$ corresponds to frames for which $\forall w$, if $wRw'$ then there is a finite sequence $w_0,...,w_n$ such that $w_0=w'$, $w_0Rw_1Rw_2...Rw_nRw$ and also $wRw_i$ for $1\le i \le n$. See article A Simple Incomplete Extension of T which is the Union of Two Complete Modal Logics with f.m.p. by Roy A. Benton.

  • Formula $\Box(\Box p \rightarrow p) \rightarrow \Box p$ (Loeb) corresponds to frames for which $\forall w$ we have $wRw' \land w'Rw'' \rightarrow wRw''$ (transitive) and also there is no infinite sequence of worlds $wRw_1Rw_2R...$ starting from $w$ (converse well-founded). See P. Blackburn Modal Logic pp 131; it is also shown there that both the Loeb and the McKinsey formulas do not correspond to a first order condition.

The above examples are not first-order conditions. But note that they describe their class of frames by stating what an arbitrary world can see, i.e. without using a partition or a valuation.

So my question is: is there a similar frame condition known for axiom M?

This should correspond to the frames of KM itself, i.e not in conjunction with other axioms. My hope is that in such a form it would be better suited for analyzing the extensions of KM − any extension, not just K4M.

  • $\begingroup$ How would you formalize the condition that "$P$ is a predicate expressing visibility between $w$ and other worlds" in a way that excludes the condition you gave, if you're allowing $P$ to be second-order? Do you mean that $P$ has access to an extra "reachability" relation, aside from the visibility one, but that no explicit second-order quantification is allowed? $\endgroup$ May 12, 2015 at 17:38
  • $\begingroup$ Let $P(w)$ say that $p$ holds at world $w$, and leq $R(w,v)$ say that $v$ is visibile from $w$. Then the McKinsey axiom is directly stated as $(\forall w)(\exists v)[R(w,v) \land P(v)] \to (\exists w)(\forall v)[R(w,v) \to P(v)]$. If that is not what you are looking for, can you clarify what you are looking for? $\endgroup$ May 12, 2015 at 17:51
  • $\begingroup$ @Gregory J. Puleo: I am not sure how to formalize this, but I would like something that does not quantify over partitions of worlds. Something like in the example I gave (there are formulas that correspond to such descriptions.) As I understand, $w$ sees $w'$ in one step or in two steps is first-order, but in a finite number of steps is not. The purpose is to get a condition which can be combined more easily with other frame conditions like transitivity, convergence etc. $\endgroup$
    – JuneA
    May 12, 2015 at 17:54
  • $\begingroup$ @Carl Mummert: I would like something that does not involve the valuations $P(v)$. Only who sees who if possible. $\endgroup$
    – JuneA
    May 12, 2015 at 17:58
  • $\begingroup$ Edited to clarify, added examples. $\endgroup$
    – JuneA
    May 13, 2015 at 4:55

2 Answers 2


I believe the question is answered by the following paper;

  • "A Note on Modal Formulae and Relational Properties", J. F. A. K. van Benthem, The Journal of Symbolic Logic, Vol. 40, No. 1 (Mar., 1975), pp. 55-58. DOI: 10.2307/2272270 URL: http://www.jstor.org/stable/2272270

which states:

Theorem 1. There is no first-order formula $\phi$ such that $F \vDash \phi \Leftrightarrow F \vdash \Box\Diamond p \to \Diamond \Box p$ for all $F$.

I found the result cited in

  • "The McKinsey Axiom is not Canonical", Robert Goldblatt, The Journal of Symbolic Logic, Vol. 56, No. 2 (Jun., 1991), pp. 554-562, DOI: 10.2307/2274699 URL: http://www.jstor.org/stable/2274699

Goldblatt attributes the result independently to van Benthem's paper above and to his own paper

  • "First-Order Definability in Modal Logic", R. I. Goldblatt, The Journal of Symbolic Logic, Vol. 40, No. 1 (Mar., 1975), pp. 35-40. DOI: 10.2307/2272267 URL: http://www.jstor.org/stable/2272267
  • $\begingroup$ I understand that the class of frames corresponding to M is not first-order definable. But does this mean there is no frame condition of the form I am asking? $\endgroup$
    – JuneA
    May 13, 2015 at 4:48

See :

  • Alexander Chagrov & Michael Zakharyaschev, Modal Logic (1997), page 82 :

A transitive frame $\mathfrak F$ validates the McKinsey formula iff satisfies the McKinsey condition

where the McKinsey condition is :

$\forall x \exists y(xRy \land \forall z(yRz \to y=z))$.

  • $\begingroup$ Yes, this is also called "every world sees a final world" in Hughes&Cresswell pp. 131. But it only works in conjunction with axiom 4, i.e. in K4M, or in S4.M if also reflexive. I am looking for a frame characterzation of KM. $\endgroup$
    – JuneA
    May 13, 2015 at 4:49

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