Functionally Complete Logical Connectives

Theorem
These sets of logical connectives are functionally complete:

Conjunction, Negation and Disjunction

 * $\left\{{\neg, \land, \lor}\right\}$: Not, And and Or

Negation and Conditional

 * $\left\{{\neg, \implies}\right\}$: Not and Implies

NAND

 * $\left\{{\uparrow}\right\}$: NAND

NOR

 * $\left\{{\downarrow}\right\}$: NOR

There are others, but these are the main ones.

Thus, the only connectives that form singleton sets which are functionally complete are NAND and NOR.

Functionally Complete Sets
From Functionally Complete Logical Connectives: Negation, Conjunction, Disjunction and Implication, all sixteen of the binary operators can be expressed in terms of $\neg, \land, \lor, \implies$.

Next we show the sufficiency of the sets $\left\{{\neg, \land}\right\}, \left\{{\neg, \lor}\right\}, \left\{{\neg, \implies}\right\}$.

From these stronger results we can see that the set $\left\{{\neg, \land, \lor}\right\}$ is also functionally complete.

Implies and Not
From Conjunction and Implication, we have that:
 * $p \land q \dashv \vdash \neg \left({p \implies \neg q}\right)$

From the above, we can represent any boolean expression in terms of $\land$ and $\neg$.

So it follows that we can replace all occurrences of $\land$ by $\implies$ and $\neg$, and so $\left\{{\neg, \implies}\right\}$ is functionally complete.

NAND
From the above, we can represent any boolean expression in terms of $\land$ and $\neg$.

From Properties of NAND, we can express $\neg p$ in terms of $\uparrow$ as follows:
 * $\neg p \dashv \vdash p \uparrow p$

Having established that, we can then express $p \land q$ solely in terms of $\uparrow$ as follows:

So any boolean expression can be represented solely in terms of $\uparrow$.

Hence $\left\{{\uparrow}\right\}$ is functionally complete.

NOR
From the above, we can represent any boolean expression in terms of $\lor$ and $\neg$.

From Properties of NOR, we can express $\neg p$ in terms of $\downarrow$ as follows:
 * $\neg p \dashv \vdash p \downarrow p$.

Having established that, we can then express $p \lor q$ solely in terms of $\downarrow$ as follows:

So any boolean expression can be represented solely in terms of $\downarrow$.

Hence $\left\{{\downarrow}\right\}$ is functionally complete.

Only NAND and NOR
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 we look at $T \circ F$ 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.