$(p \wedge q) \wedge p$ convert to CNF I was doing some exercises to convert formulas to CNF by means of the axioms and I imagined the following exercise $(p \wedge q) \wedge p$, but for this exercise, I do not find an axiom that can take me to CNF. However, I know that if I use the truth table I could solve it. I would like to know if there is an axiom that I may not know, by which you can work the exercise.

$\begin{align}(p\wedge q)\wedge p &= p\wedge(p\wedge q) &&\text{commutivity}\\ &= (p\wedge p)\wedge q &&\text{associativity}\\ &=p\wedge q && \text{idempotence}\end{align}$
