Is it possible for the DNF and CNF to be the same Is it possible for a formula in propositional logic that its disjunctive normal form (DNF) and its conjunctive normal form (CNF) are the same?
For example, can $A \land C$ be both the conjunctive normal form and the disjunctive normal form of the formula below?

$(A \lor ((A \land B) \lor (\lnot A \land \lnot B \land  \lnot C))) \land C$

 A: The short answer is: yes! 
Indeed, $A \land C$ is both a CNF (because it is a conjunction of two disjunctive clauses, each one made of just one literal) and a DNF (with just one conjunctive clause). 
To prove that $A \land C$ is a CNF and a DNF of $(A \lor ((A \land B) \lor (\lnot A \land \lnot B \land  \lnot C))) \land C$, it remains to prove that $A \land C$ is logically equivalent to $(A \lor ((A \land B) \lor (\lnot A \land \lnot B \land  \lnot C))) \land C$. To prove it, I use logical equivalences listed here as rewriting rules (the use of the associativity is left implicit):
\begin{align} 
(A \lor (A \land B) \lor (\lnot A \land \lnot B \land  \lnot C)) \land C &\equiv (A \lor (\lnot A \land \lnot B \land  \lnot C)) \land C &\text{absortion law} \\
&\equiv (A \lor \lnot A) \land (A \lor \lnot B) \land (A \lor \lnot C) \land C &\text{distributivity law} \\
&\equiv (A \lor \lnot B) \land (A \lor \lnot C) \land C &\text{identity law} \\
&\equiv (A \lor (\lnot B \land \lnot C)) \land C &\text{distributivity law} \\
&\equiv (A\land C) \lor (\lnot B \land \lnot C \land C) &\text{distributivity law} \\
&\equiv A\land C &\text{identity law}
\end{align}
where the first identity law can be applied since $A \lor \lnot A$ is a tautology, and the second identity law can be applied because $\lnot B \land \lnot C \land C$ is a contradiction. 

Remark. Pay attention that a CNF of a given formula is not unique, and similarly for DNF. For instance, also $A \land C \land (B \lor \lnot B)$ is a CNF of $(A \lor ((A \land B) \lor (\lnot A \land \lnot B \land  \lnot C))) \land C$. So, properly speaking it is not clear what you refer to when you talk of the CNF (or the DNF) of a given formula.
More interesting remarks about the non-uniqueness of CNF (or DNF) of a given formula are in this question.  
A: A conjunctive normal form is a conjunction of a sequence of disjunctions of a sequence of literals or their negations.   Abreviated as: a conjunction of disjunctions of literals or their negations. 
A single literal is a conjunction of a sequence of one literal.   It is also a disjunction of a sequence of one literal. 
Thus $A\wedge C$ is a conjunction of two disjunctions of one literal; and also a disjunction of two conjunctions of one literal.   That is it is both a conjunctive normal form and a disjunctive normal form. 

Now, is it a CNF/DNF for $(A\vee((A\wedge B)\vee(\neg A\wedge\neg B\wedge\neg C)))\wedge C$?
Well, $C$ must be true for the statement to hold.   When we substitue $\top$ for $C$, that becomes $(A\vee((A\wedge B)\vee(\neg A\wedge\neg B\wedge \bot)))\wedge \top$ which simpliefies to $A$.   So $A$ must be true too.   The value of $B$ is irrelevant.
Therefore $A\wedge C$ is indeed a CNF/DNF equivalent to the statement.
