DNF and CNF look the same?

Question

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: $(s → \lnot q) ∨ \lnot s$

Answer 1

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.

The formula is indeed in both CNF and DNF, either as one clause with three literals, or as a disjunction of three one-literal clauses.

Answer 2

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.