# DNF and CNF look the same?

When constructing both a DNF and CNF of the below, my solutions look the same. I must be off somewhere.

This is what they look like: $$\lnot s ∨ \lnot q ∨ \lnot s$$

How would you construct a DNF and CNF of this: $$(s → \lnot q) ∨ \lnot s$$

• In both cases I'd simplify to $\neg s\lor\neg q$, but this formula is in both conjunctive and disjunctive normal form. For CNF: It's the conjunction of a single clause, which is the disjunction of two literals. ForDNF: It's the disjunction of two clauses, each of which is the conjunction of a single literal. Commented Jul 2, 2019 at 17:41
• incredibly helpful, thank you! Commented Jul 2, 2019 at 17:56

Since your formula has no $$\land$$ at all, it makes no difference whether the place they are absent from is above or below the $$\lor$$s in the syntax tree.
First, note that the expression $$\lnot s \lor \lnot q \lor \lnot s$$ can be further simplified to $$\lnot s \lor \lnot q$$ by the commutativity of $$\lor$$ operator.
Then, $$\lnot s \lor \lnot q$$ is both in CNF and DNF form.
• For CNF form, we can consider $$\lnot s \lor \lnot q$$ as a clause so the CNF form has $$1$$ clause.
• For DNF form, we can consider $$\lnot s$$ and $$\lnot q$$ as clauses so the DNF form has $$2$$ clauses.