finding the conjunctive normal form of a statement I am currently trying to find the conjunctive normal form of this statement
$(A \land B) \lor ((\lnot C \lor A) \land \lnot B) \lor B$. 
I think I have found the answer but I am unsure if it is correct. 
Am I correct in saying the answer is:
$A \lor B \lor \lnot C$ 
 A: Yes, $A \lor B \lor \lnot C$ is a conjunctive normal form (CNF) of $(A \land B) \lor ((\lnot C \lor A) \land \lnot B) \lor B$.
A way to prove it is just to observe that $A \lor B \lor \lnot C$ is a CNF and it is logically equivalent (you can check it by means of truth tables) to $(A \land B) \lor ((\lnot C \lor A) \land \lnot B) \lor B$.
An alternative proof is to use logical equivalences as rewriting rules leading from $(A \land B) \lor ((\lnot C \lor A) \land \lnot B) \lor B$ to $A \lor B \lor \lnot C$. More precisely, observe first that:
\begin{align}
 ((\lnot C \lor A) \land \lnot B) \lor B &\equiv (\lnot C \lor A \lor B) \land (\lnot B \lor B) &&\text{distributivity law} \\
&\equiv \lnot C \lor A \lor B &&\text{identity law (as $\lnot B \lor B$ is a tautology).}
\end{align}
Then (the use of the associativity law below is always implicit),
\begin{align}
 (A \land B) \lor ((\lnot C \lor A) \land \lnot B) \lor B &\equiv (A \land B) \lor (\lnot C \lor A \lor B) &&\text{substitution} \\ 
& \equiv (A \land B) \lor (A \lor B) \lor \lnot C  &&\text{Associative law} \\
&\equiv ((A \lor A \lor B) \land (B \lor A \lor B)) \lor \lnot C && \text{distributivity law} \\
&\equiv ((A \lor B) \land (A \lor B)) \lor \lnot C && \text{idempotent law} \\
&\equiv A \lor B \lor \lnot C &&\text{idempotent law.}
\end{align}
