Show that any monotone Boolean function is computable by a circuit containing only AND and OR gates A Boolean function $f : \{0, 1\}^n → \{0, 1\}$ is called monotone if changing any of the $n$ input bits $x_1, \ldots , x_n$ from $0$ to $1$ can only ever change the output $f(x_1, \ldots ,x_n)$ from $0$ to $1$, never from $1$ to
$0$.

I know how to do a simple proof of exhaustion for $f : \{0, 1\}^1$ and $f : \{0, 1\}^2$, but I do not know how to prove the following statement for $f:\{0,1\}^n$: any monotone Boolean function is computable by a circuit containing only AND and OR gates.
Any Boolean function can be written as a DNF. Each clause in the DNF specifies one truth assignment for which the function holds. For example, the DNF form of XOR is $(x \land \lnot y) \lor (\lnot x \land y)$.

The main observation is that if the function is monotone, you can remove all the negated literals (why?). Once you do that, you get a formula for the function which uses only AND and OR. Some clauses in this formula subsume others (for example, $x$ subsumes $x \land y$). If you drop all the subsumed clauses, you arrive at the minterm representation of the function. (A minterm is a minimal satisfying assignment.)

It's a bit more confusing, but you can also do everything using CNFs. This way you arrive at the maxterm representation.