$(p \wedge q) \wedge p$ convert to CNF

Question

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.

Answer 1

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

Answer 2

$ (p \land q) \land p$ is already in CNF ...

And if you don't like the parentheses there, then please know that CNF is define as a generalized conjunction of generalised disjunctions of literals.

As such, $(p \land q) \land p$ may be better put simply as $p \land q \land p$ ... which can of course be simplified to $p \land q$, but does not need to be to count as being in CNF

Also, if you are wondering where the disjunctions are ... $p$ is a disjunction consisting of exactly one disjunct, and that disjunct is a literal $p$. Thus, $p \land q \land p$ is a conjunction of three conjuncts, where each conjunct is a disjunction of exactly one disjunct, and where each disjunct is a literal, and so again, yes, it is in CNF