I an trying to learn about $DisNP$ Complexity class. I couldn't find this class in any book. But, regarding it, I found this definition in a paper:Definition: A disjoint NP-pair (NP-pair for short) is a pair of nonempty sets $A$ and $B$ such that $A, B \in NP$ and $A \cap B = Empty$. $DisNP$ denote the class of all disjoint NP-pairs. Now I am aware that when a problem like $SAT$ etc. is said to be in $NP$, it means that assuming we have a NDTM we can get the solution in polynomial time. Or in other words, if we can guess the solution, then the verification takes polynomial time.But, what does it mean when we say a set $A \in NP$ or a set $B \in P$. Does the elements of the sets represent the solution for which the corresponding $SAT$ instance for each set is true i.e. set $A$ and $B$ contain the solutions of some $SAT$ instances $SAT_A$ and $SAT_B$ such that $SAT_A$ is satisfiable iff its solution is in the set $A$?I am not clear how we use sets (which I assume have binary strings) for defining $P, NP, NPC$?
Clarification of definition of class $DisNP$
complexity theory
A disjoint NP pair consists of two NP languages $A,B$ which are disjoint. For example, $A$ could consist of all graphs which are $k(n)$-colorable (where $n$ is the number of vertices and $k$ is an arbitrary function), and $B$ could consist of all graphs having a clique of size $k(n)+1$.Disjoint pairs of NP languages are used in proof complexity. For example, the disjoint pair discussed above (for $k(n) = n^c$) is used to prove exponential lower bounds on the size of proofs in the cutting planes proof system. See for example these notes. (For the state-of-the-art regarding cutting planes, see the recent paper Random CNFs are Hard for Cutting Planes.)