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For every Boolean function, f in l variables there exists an l-variable CNF formula Phi with f(u)=Phi(u) for all arguments u. And we can ensure that the size of this CNF formula Phi is at most l*2^l. Where the size of a CNF formula here is measured as the number of and and or symbols that are contained in the CNF formula. To prove this, let's look at the behavior of a single clause that contains all the l variables, either themselves or in negated form. Suppose for example l = 2, so we only have two variables, let's say x and y. Now, a clause that involves all those variables would be x or y. Or it could be x or not y, or it could be not x or y. Or it could be not x or not y. And what we observe is that a single such clause is false for exactly a single variable assignment, so there's only a single way of assigning truth values to all the variables in such a way that the clause is false. For example, if we have the clause x or not y, then the only way to make this clause false is to make x false and to make not y fals as well. And not y is false if y is true. The clause will be true for all other possible assignments. Now consider the function f. And look at all the inputs that lead to an output of 0 or false. How can we make this happen? Consider a specific input u for which f(u) is 0. Then what we can do to in our construction of our CNF formula Phi is that we simply include a clause that makes the formula flase for this input u. Due to our previous observation, we know what such a clause would have to look like. So we add that clause to our CNF formula. And we continue to do this for all inputs u for which f(u) is equal to 0. What we end up with is a formula Phi that is zero exactly when f is. So the only question is how big this formula is. Every time we add a clause to our formula Phi, we increased the size of the formula by l. This is because each of our clauses consists of l literals and then we need to connect that clause to all the remaining clauses with an and symbol, except for the very first clause we add. But in total we only add at most 2^l many clauses because there only two to the 2^l possible inputs u and therefore the total size of the formula we generate is not larger than l*2^l.