circuits complexity

Question

I ran into this claim and I dont really understand why it is true ( and I need to use it for proving other things)

Every function from $(0, 1)^n$ to $(0, 1)$ is computable by a $2^n10n$-sized circuit.

Answer 1

If you write the function in disjunctive normal form, it has at most $2^n$ clauses with at most $n$ atoms in each.

To express this as a circuit you need $n$ NOT-gates to make the negated atoms, at most $2^n(n-1)$ AND gates to make the clauses, and $2^n-1$ OR gates to combine them into the final result. Each of these counts is less than $2^nn$, so their sum is at most $2^n3n$, with plenty of room to spare.