Functionally Complete Logical Connectives

Theorem
These sets of logical connectives are functionally complete:


 * $$\left\{{\neg, \and}\right\}$$: Not and And;
 * $$\left\{{\neg, \or}\right\}$$: Not and Or;
 * $$\left\{{\neg, \implies}\right\}$$: Not and Implies;


 * $$\left\{{\uparrow}\right\}$$: NAND;
 * $$\left\{{\downarrow}\right\}$$: NOR.

There are others, but these are the main ones.

Proof
It suffices to consider binary operators:

Constant Functions

 * Two constant functions:
 * $$f_F \left({p, q}\right) = F$$;
 * $$f_T \left({p, q}\right) = T$$.

From Equivalence Properties and Exclusive Or Properties, we have that:


 * $$p \iff p \dashv \vdash \top \dashv \vdash f_T \left({p, q}\right)$$
 * $$p \oplus p \dashv \vdash \bot \dashv \vdash f_F \left({p, q}\right)$$

So both of the constant functions can be expressed in terms of $$\oplus$$ and $$\iff$$.

Equivalence and Non-Equivalence

 * The exclusive or: $$p \oplus q$$:

We note that by definition $$p \oplus q \dashv \vdash \neg \left({p \iff q}\right)$$.

Thus $$\oplus$$ can be expressed in terms of $$\neg$$ and $$\iff$$.


 * The equivalence: $$p \iff q$$:

We note that by definition $$p \iff q \dashv \vdash \left({p \implies q}\right) \and \left({q \implies p}\right)$$.

Thus $$\iff$$ can be expressed in terms of $$\implies$$ and $$\and$$.

Projections and Negated Projections

 * Two projections:
 * $$\operatorname{pr}_1 \left({p, q}\right) = p$$;
 * $$\operatorname{pr}_2 \left({p, q}\right) = q$$.

We note that: and similarly for $$\operatorname{pr}_2 \left({p, q}\right)$$.
 * $$p \and p \dashv \vdash p \dashv \vdash \operatorname{pr}_1 \left({p, q}\right)$$;
 * $$p \or p \dashv \vdash p \dashv \vdash \operatorname{pr}_1 \left({p, q}\right)$$

So the projections can be expressed in terms of either $$\and$$ or $$\or.$$.


 * Two negated projections:
 * $$\overline {\operatorname{pr}_1} \left({p, q}\right) = \neg p$$;
 * $$\overline {\operatorname{pr}_2} \left({p, q}\right) = \neg q$$.

It immediately follows from the above that these can be expressed in terms of either:
 * $$\neg$$ and $$\and$$, or:
 * $$\neg$$ and $$\or$$.

NAND and NOR

 * The NAND: $$p \uparrow q$$
 * The NOR: $$p \downarrow q$$

By definition:
 * $$p \uparrow q \dashv \vdash \neg \left({p \and q}\right)$$;
 * $$p \downarrow q \dashv \vdash \neg \left({p \or q}\right)$$.

So:
 * $$\uparrow$$ can be expressed in terms of $$\neg$$ and $$\and$$;
 * $$\downarrow$$ can be expressed in terms of $$\neg$$ and $$\or$$.

Conjunction, Disjunction, Conditional

 * The conjunction: $$p \and q$$
 * The disjunction: $$p \or q$$


 * Two conditionals:
 * $$p \implies q$$
 * $$q \implies p$$


 * Two negated conditionals:
 * $$\neg \left({p \implies q}\right)$$
 * $$\neg \left({q \implies p}\right)$$

All of the above are already expressed in terms of $$\neg, \and, \or, \implies$$.

Functionally Complete Sets
So we have shown that all sixteen of the binary operators can be expressed in terms of $$\neg, \and, \or, \implies$$.

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

And and Not
From Conjunction and Implication, we have that $$p \implies q \dashv \vdash \neg \left({p \and \neg q}\right)$$.

From De Morgan's laws, we have that $$p \or q \dashv \vdash \neg \left({\neg p \and \neg q}\right)$$.

So any instance of either $$\implies$$ or $$\or$$ can be replaced identically with one using just $$\neg$$ and $$\and$$.

It follows that $$\left\{{\neg, \and}\right\}$$ is functionally complete.

Or and Not
From De Morgan's laws, we have that $$p \and q \dashv \vdash \neg \left({\neg p \or \neg q}\right)$$.

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

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

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

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

So it follows that we can replace all occurrences of $$\and$$ 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 $$\and$$ 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 $$p \and q$$ 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 $$\or$$ 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 $$p \or q$$ 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.