I never really studied modal logic, but to my better understanding this is similar to classical logic adding two modal operations:

  1. $\square P\ $ meaning necessarily $P$,
  2. $\lozenge P\ $ meaning possibly $P$.

Now if we consider Boolean-valued logic, in which we take a complete Boolean algebra and let the truth values be elements of the Boolean algebra where the evaluation uses the Boolean algebra, so:

  • $\|\psi\land\varphi\| = \|\psi\|\cdot\|\varphi\|$
  • $\|\psi\lor\varphi\|=\|\psi\|+\|\varphi\|$
  • $\|\lnot\psi\| = -\|\psi\|$
  • $\|\exists x\varphi(x)\| = \sum\|\varphi(y)\|$

If we take an ultrafilter on the Boolean algebra, we return to the usual two-valued logic by letting $\|\varphi\|\in\mathcal U$ being true, and false otherwise.

However if only take $\mathcal U$ to be a filter, can we think of it as a "necessary" predicate on $B$? that is $\|\square\varphi\|=1\iff\|\varphi\|\in\mathcal U$, and $\|\lozenge\varphi\|=1\iff\|\lnot\varphi\|\notin\mathcal U$ (both values are $0$ otherwise).

If the answer is indeed yes, is this a complete characterization of all modal logics, that is every quotient of a Boolean-valued logic is modal logic, and vice versa?

  • 2
    $\begingroup$ I find this a little dubious for the following reason: in this scheme, $\| \Box \varphi \|$ and $\| \Diamond \varphi \|$ can only ever be the maximum or minimum elements, so, if this scheme were complete for modal logic, $\Box \Box \varphi$ would be logically equivalent to $\Box \varphi$. But there are modal logics which aren't even monotonic! (Take the simplest possible modal logic which has no axioms other than the usual ones for classical logic.) Perhaps you would like to read up more about Kripke semantics. $\endgroup$
    – Zhen Lin
    Jan 5, 2012 at 2:27
  • $\begingroup$ As Zhen Lin has pointed out, there are many different modal logics, usually they are built on two main axioms: $\square (A \Rightarrow B) \Rightarrow (\square A \Rightarrow \square B)$ and $\square A \Rightarrow A$. I am not an expert on modal logic but I really liked this article in SEP: plato.stanford.edu/entries/logic-modal $\endgroup$
    – Daniil
    Jan 5, 2012 at 8:45

1 Answer 1


As Zhen Lin indicates, this cannot be a "complete characterization of all modal logics." It seems to be close to a characterization of one particular modal logic. The simplest non-trivial system is arguably the one called S5, in which iterations of modal operators are equivalent to single ones. You may be close to a characterization of this one system, though I haven't seen yours in print.

In what follows I apologize for not entering the appropriate symbols. (It's getting late here.)

In your question as I read it now, there appears to be a typo in the next-to-last paragraph. The condition for necessarily phi has "= 1" to the left of the double arrow. The condition for possibly phi should have "= 0". I assume that's what you intended.

One obtains S5 if one also adds the conditions that necessarily phi equals the zero element iff phi is absent from the filter, and possibly phi equals the unit element iff phi is absent from the filter. Since the filter need not be an ultrafilter, phi and its negation may both be absent, which corresponds to possibly P and possibly not-P both being true.

I believe this works out correctly for propositional S5. The quantifier of course introduces complications, so I'm not sure about that.

My email address is available on my profile page if you care to correspond about this.


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