is the class NP closed under set difference?

Question

I know P is closed under all Boolean operations, but what about NP?

is NP closed under set difference and symmetric difference?

is this table accurate?

Edit: updated table:

Answer 1

It is unknown whether $\mathsf{NP}$ is closed under set-difference.

If $\mathsf{NP}$ were known to be closed under set-difference then we would know that $\mathsf{NP} = \textsf{co-NP}$. Indeed, for $L \in \textsf{co-NP}$, $L = \Sigma^* \setminus \overline{L}$ where $\overline{L} \in \textsf{NP}$ and hence $\textsf{co-NP} \subseteq \textsf{NP}$. Moreover, for $L \in \textsf{NP}$, $L = \Sigma^* \setminus \overline{L}$ where $\overline{L} \in \textsf{co-NP} \subseteq \textsf{NP}$ and hence $\textsf{NP} \subseteq \textsf{co-NP}$.

On the other hand, if $\mathsf{NP}$ was known not to be closed under set-difference then we would know that $\mathsf{P} \neq \mathsf{NP}$ (since $\mathsf{P}$ is closed under set difference).

Regarding the symmetric difference, notice that $\Sigma^* \, \Delta \, L = (\Sigma^* \setminus L) \cup (L \setminus \Sigma^*) = (\Sigma^* \setminus L) \cup \emptyset = \Sigma^* \setminus L = \overline{L}$.