# Is it possible to get the CNF out of the DNF of this expression

Can i get the CNF of the following expression if i know the DNF?

I've the following expression:

$$\Bigl(\bigl(A\rightarrow (\overline A \land B) \bigr)\land \bigl((\overline A \land B)\rightarrow A\bigr)\Bigr)\rightarrow\bigl(B\land \overline B\bigr)$$

The DNF will be:

$$A\rightarrow B \Rightarrow \overline A \lor B:$$

$$\Rightarrow\Bigl(\bigl(\overline A \lor(\overline A \land B)\bigr)\land \bigl((A\lor\overline B)\lor A\bigr)\Bigr)\rightarrow \bigl(B\land \overline B\bigr)$$

$$\overline A \lor(\overline A \land B) \Rightarrow \overline A$$

$$(A\lor\overline B)\lor A \Rightarrow A\lor\overline B$$

$$\Rightarrow \bigl(\overline A\land (A\lor \overline B)\bigr)\rightarrow (B\land \overline B)$$

$$A\rightarrow B \Rightarrow \overline A \lor B:$$

$$\Rightarrow\overline{\bigl(\overline A \land(A\lor\overline B)\bigr)}$$

$$\Rightarrow \bigl( A\lor(\overline A\land B)\bigr)$$

$$\Rightarrow (A\lor B)$$

So, As we can see, The DNF of this expression is $$A\lor B$$, My question: Is it correct to say that the CNF of this expression will be $$\overline A\land \overline B$$?

I looked at Graham Kemp's Answer for this question, And didn't successfully understood how an expression that has only the operator $$\land$$ can be both conjunction of two disjunctions of one literal and disjunction of two conjunctions of one literal at the same time.

Because as i know (Please correct me if i wrong), A CNF has the form of $$A\land B\land C\land D$$ Where $$A,B,C,D$$ are expressions of the form $$x\lor y\lor z$$, And DNF has the form of $$A\lor B\lor C\lor D$$ Where $$A,B,C,D$$ are expressions of the form $$x\land y\land z$$.

Thanks!!!

• $\overline A\land \overline B$ is a negation of $A\lor B.$ In the given example are DNF and CNF equal. – user376343 Jan 29 '19 at 22:44
• CNF is a product if sums while DNF is a sum of products. In the above case one can consider $A\lor B$ as any of them. – user376343 Jan 29 '19 at 22:47

So, if the DNF is $$A \lor B$$, then the CNF cannot be $$\neg A \land \neg B$$, since that is the negation of it, i.e. not something that is equivalent.
Well, the CNF is just $$A \lor B$$ as well. Think of it as a conjunctin that consists of a single conjunct. And, since that conjunct is a disjunctin of literals, it fits the definition of CNF just fine.