# Does the modal logic S5 have a binary sole sufficient operator?

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?

Corrections:

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.
• 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. Commented Apr 5, 2023 at 22:03

• 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? Commented Apr 3, 2023 at 22:26