Prove that nand is functionally complete

Question

Can anyone explain to me how to

Prove that nand is functionally complete.

(To wit: if we let $p ∗ q$ mean $¬(p ∧ q)$, show that the other connectives, $∧$, $∨$, $¬$ and $→$ are expressible in terms of $∗$.)

I understand that logical function on a fixed set of inputs has a finite number of cases, but unsure how to put that into context.

Answer 1

$¬p\equiv¬(p ∧ p)$

$p ∧ q\equiv¬(¬(p ∧ q))$

$p∨q\equiv¬(¬p ∧ ¬q)$

$p→q\equiv¬p∨q$

Therefore nand is functionally complete.

Answer 2

Two well-known basic facts:

Lemma 1: Every boolean function can be represented as a DNF.

Lemma 2: If $A = \{f_1, ... f_n\}$ if complete and every its function can be expressed in terms of functions from $B = \{g_1, ... g_m\}$, then $B$ is complete.

Lemma 1 $\implies \{\lnot, \lor, \land\}$ is complete. The following formulas

$¬p\equiv¬(p ∧ p)$

$p ∧ q\equiv¬(¬(p ∧ q))$

$p∨q\equiv¬(¬p ∧ ¬q)$

together with Lemma 2 imply that $\{NAND\}$ is complete.