It is well-known that nand and nor are sole sufficient operators for classical logic.

I found a ternary operator earlier today that can express classical modal logic, given below:

$$ J(a,b,c) \;\;\text{is defined as}\;\; \square(a \barwedge b) \land (b \barwedge c) $$

With the addition of a truth value constant $\bot$, $\{\bot, J\}$ does generate all the classical connectives and the modal connectives $\square$ and $\lozenge$. However, without an explicit $\bot$, it is more difficult to see whether it does or not.

Here is my attempt to define a few connectives using $J$ and build my way back to $\bot$.

NOTE: an earlier version of this question erroneously said that $J(a,a,a)$ is equivalent to $\bot$.

  • $J(a,a,a)$ is $\square(\lnot a)$.
  • $J(\square(\lnot(a)), \square(\lnot(a)), \square(\lnot(a)))$ is $\square\lozenge(a)$ or $\lozenge a$ in S5.
  • $J(\lozenge(a), \square(\lnot(a)), \lozenge(a))$ is $\top$ in S5.
  • $\square\lnot(\top)$ is $\bot$.
  • $J(\bot, b, c)$ is $b \barwedge c$.
  • $b \barwedge b$ is $\lnot b$.
  • $J(\lnot a, \lnot a, \bot)$ is $\square a$.

From there, any classical connective as well as $\square$ and $\lozenge$ can be made, so $J$ is a sole sufficient operator.

Let's focus on the logic S5, where the accessibility relation is an equivalence relation, for concreteness.

My gut says that a binary sole sufficient operator $B$ for S5 is probably impossible since we need two positions to behave non-modally, at least under some circumstances, to make a nand or a nor. I'm struggling to prove it though.

Is there a binary sole sufficient operator for S5?


  1. $J(a,a,a)$ is NOT equivalent to bottom, it is equivalent to $\square(a \barwedge a) \land (a \barwedge a)$, which is equivalent to $\square \lnot a$ in S5.
  2. Added explicit language saying that $\{J, \bot\}$ generates S5. This is hardly special though; $\{\square, \barwedge\}$ does too.
  • $\begingroup$ The clause you give for Boolean negation says that $\neg b$ is defined as $b\overline{\wedge } b$. But this definition does not make use of the ternary operator, but uses just one conjunct of its definition. So you haven't provided a definition for Boolean negation via your ternary operator. $\endgroup$
    – sequitur
    Apr 5, 2023 at 22:03

1 Answer 1


There is a paper that proves that it is impossible to get a three-valued basis for S5. Dugundji, James. Note on a Property of Matrices for Lewis and Langford's Calculi of Propositions. Journal of Symbolic Logic 5 (1940), no. 4, 150--151. jstor, doi: 10.2307/2268175

You may get some form of modal logic out of it, but it won't be S5, or any other of the Lewis systems.

  • $\begingroup$ Thanks. I noticed a pretty glaring error in my original question because of this ($J(a,a,a)$ is not $\bot$). I suspect the three-place connective $J$ and a nullary connective $\bot$ will allow you to recover S5. Do you know whether the construction in the journal article you cite allows truth value constants or not? $\endgroup$ Apr 3, 2023 at 22:26
  • $\begingroup$ @Greg Nisbet I have not read the paper myself, but it has been cited as an objection to my own attempts at constructing a 3-valued modal logic. I suspect that it applies to S5, or perhaps the Lewis systems in general, and because Lewis's definition of the strict conditional doesn't work for the third logical value, but I don't know this for certain. $\endgroup$
    – Confutus
    Apr 3, 2023 at 22:47
  • 1
    $\begingroup$ Wait ... I'm talking about a sole sufficient operator that takes three arguments, not a three-valued semantics for modal logic. The underlying modal logic in my case is still two-valued at each world. Is your reference about a third truth value or three-place connectives? $\endgroup$ Apr 3, 2023 at 22:54
  • $\begingroup$ I was referring to a third truth value and three valued semantics, so I probably misunderstood the question, in which case I don't know. I can see how a three place operator might give you some or all 3^ 9 possibilities for the truth table for a binary operator. I do not see how it would yield a system which isn't truth functional at all. $\endgroup$
    – Confutus
    Apr 3, 2023 at 23:17

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