Functionally Complete Singleton Sets

Theorem
The only binary logical connectives that form singleton sets which are functionally complete are NAND and NOR.

Proof
Let $\uparrow$ and $\downarrow$ denote NAND and NOR respectively.

From:
 * NAND is Functionally Complete

and
 * NOR is Functionally Complete

the singleton sets $\left\{{\uparrow}\right\}$ and $\left\{{\downarrow}\right\}$ are functionally complete.

Suppose $\circ$ is a binary logical connective such that $\left\{{\circ}\right\}$ is a functionally complete set.

The unary logical connective $\neg$ has to be equivalent to some formula:
 * $\cdots \circ \left({p \circ p}\right) \circ \cdots$

Suppose some interpretation $v$ assigns $T$ to $p$.

Then $v \left({\neg p}\right) = F$

So:
 * $v \left({\cdots \circ \left({p \circ p}\right) \circ \cdots}\right) = F$

So it has to be true that $T \circ T = F$, otherwise:
 * $v \left({\cdots \circ \left({T \circ T}\right) \circ \cdots}\right) = v \left({\cdots \circ \left({T}\right) \circ \cdots}\right) = T$

Similarly:
 * $F \circ F = T$

Now consider $T \circ F$ and $F \circ T$.

Suppose $T \circ F = F$.

If $F \circ T = T$, then we have the function defined by this truth table:

which is $\neg p$, and hence only negation would be definable.

So if $T \circ F = F$ we need $F \circ T = F$.

This gives the truth table for the NOR function:

... which we have seen is functionally complete.

Similarly, Suppose $T \circ F = T$.

If $F \circ T = F$, then we have the function defined by this truth table:

which is $\neg q$, and hence only negation would be definable.

So if $T \circ F = T$ we need $F \circ T = T$.

This gives the truth table for the NAND function:

... which we have seen is functionally complete.

Thus it follows that there can be no functionally complete binary logical connectives apart from NAND and NOR.