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:
and
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:
| $v \left({p}\right)$ | $v \left({q}\right)$ | $v \left({p \circ q}\right)$ |
|---|---|---|
| $F$ | $F$ | $T$ |
| $F$ | $T$ | $T$ |
| $T$ | $F$ | $F$ |
| $T$ | $T$ | $F$ |
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:
| $v \left({p}\right)$ | $v \left({q}\right)$ | $v \left({p \circ q}\right)$ |
|---|---|---|
| $F$ | $F$ | $T$ |
| $F$ | $T$ | $F$ |
| $T$ | $F$ | $F$ |
| $T$ | $T$ | $F$ |
... 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:
| $v \left({p}\right)$ | $v \left({q}\right)$ | $v \left({p \circ q}\right)$ |
|---|---|---|
| $F$ | $F$ | $T$ |
| $F$ | $T$ | $F$ |
| $T$ | $F$ | $T$ |
| $T$ | $T$ | $F$ |
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:
| $v \left({p}\right)$ | $v \left({q}\right)$ | $v \left({p \circ q}\right)$ |
|---|---|---|
| $F$ | $F$ | $T$ |
| $F$ | $T$ | $T$ |
| $T$ | $F$ | $T$ |
| $T$ | $T$ | $F$ |
... 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.
$\blacksquare$
Sources
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.4.2$: Theorem $2.37$
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.10$: Exercise $2.10$