How can I prove DP completeness?

Question

We defined the class $\text{DP}$ like this:

$$\text{DP} := \{ A \setminus B : A, B \in \text{NP} \}$$

We say a problem $P$ is $\text{DP}$ complete iff $P \in \text{DP}$ and $X \leq P \forall X \in \text{DP}$, meaning that every language $X \in \text{DP}$ can be reduced to $P$ in polynomial time.

For a particular problem (which I do not expect you to solve, so I won't explain all the details), I already proved that $P \in \text{DP}$, but I do not know how to continue. Also, I could not find more information about this class.

Could someone give me a hint how to start such a proof or point me to further information about this complexity class which might help me to achieve it?

Edit: The definition is from a lecture about complexity of algorithms. The task is to prove that

$$\text{SAT-UNSAT} = \{\, (F,G) : F \text{ is satisfiable}, G \text{ is not}\, \}$$

is DP-complete. I proved that $\text{SAT-UNSAT} \in \text{DP}$, but I failed to prove the completeness.

Answer 1

The DP-hardness of SAT-UNSAT is a direct consequence of the fact that SAT is NP-hard.

Suppose that $L \in \mathrm{DP}$. Then there exist languages $A \in \mathrm{NP}$ and $B \in \mathrm{coNP}$ such that $L = A \cap B$. Since SAT is NP-complete, there exists a polytime reduction $f$ such that $x \in A$ iff $f(x) \in \mathrm{SAT}$. Since UNSAT is coNP-complete, there exists a polytime reduction $g$ such that $x \in B$ iff $g(x) \in \mathrm{coSAT}$. You can check that the polytime reduction $h(x) = (f(x),g(x))$ satisfies $x \in L$ iff $h(x) \in \mathrm{SAT\text{-}UNSAT}$. This shows that SAT-UNSAT is DP-hard.

To complete the proof that SAT-UNSAT is DP-complete, you have to show that SAT-UNSAT is in DP. Presumably you already know how to do that.

Note that DP contains both SAT and UNSAT, and in particular differs from NP unless NP=coNP.