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Does the modal logic K have a binary operator which, when composed only with itself, can produce $\wedge$, $\neg$ and $\square$?

I suspect that it doesn't: it probably makes a significant difference that K has infinitely many modalities (and hence infinitely many possible operators). I have seen that the modal logics S4 and S5 have sole sufficient operators (see here and here), but S4 and S5 have finitely many modalities.

I ask in part because a proof that K has no sole sufficient operator might help to answer one of my other questions (about a sole sufficient operator for the relation algebra), but I am also curious about the question for its own sake.

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