Can the CNF be the negation of the DNF?

I've got the following question: Asking for the CNF of the following formula: $$(\lnot X \lor Y)\to \bigl( Z\land\lnot(\lnot Y\land X) \bigr)$$ So the process should be like that: $$\lnot (\lnot X \lor Y)\lor \bigl( Z\land (Y\lor \lnot X)\bigr)$$ $$(X\land \lnot Y)\lor \bigl((Z\land Y)\lor (Z\land \lnot X)\bigr)$$ $$(X\land \lnot Y)\lor (Z\land Y)\lor (Z\land \lnot X)$$ As we can see, It's nothing other than the DNF of the above formula, Can i say that the CNF is just the negation of this DNF?

The negation of this DNF looks like the following: $$(\lnot X\lor Y)\land (\lnot Z\lor \lnot Y)\land (\lnot Z\lor X)$$ Is it correct to say that this is the CNF of the above formula?

Thanks!!!

• What you've found is a CNF, but not that of the above formula. It is a CNF of the negation. Commented Mar 27, 2019 at 20:25
• if you can use Karnaugh maps, do as follows: from K.map of the given formula you get easily the map of the negation, and also the DNF of the negation. Then you can use your trick to obtain the CNF it the given formula. Commented Mar 27, 2019 at 20:35

$$(X\land\lnot Y)\lor (Z\land Y)\lor (Z\land\lnot X)$$ is a DNF for the statement, as is the simpler $$(X\land\lnot Y)\lor Z$$.   (Use distribution, commutation, de Morgan's, and absorption.)
Thus a CNF for that statement would be $$(X\lor Z)\land(\lnot Y\lor Z)$$