Answer :
Final answer:
Double SAT, which involves finding at least two satisfying assignments for a satisfiability problem, is in NP because each solution can be verified in polynomial time, aligning with the properties of NP problems.
Explanation:
To show that double SAT is in NP, we need to consider that a problem belongs to NP (nondeterministic polynomial time) if a solution can be verified in polynomial time by a nondeterministic algorithm. Given two satisfying assignments, we can verify that both fulfill the conditions of the SAT (Boolean satisfiability) problem individually, in polynomial time.
Considering SAT to double SAT, the transition implies that instead of seeking just one satisfying assignment for a set of Boolean variables that make a Boolean expression true, we are seeking at least two distinct satisfying assignments. To verify this for double SAT, one could use a nondeterministic algorithm to guess two satisfying assignments and then verify both independently in polynomial time, which is within the bounds of NP complexity.
For example, given a Boolean formula represented by (S→ A) & {P↔ [(S✓ D) & (D → P)]}, we would first find two sets of assignments for the variables S, A, P, and D that satisfy this formula. Each assignment validating the formula can be checked quickly (in polynomial time), which aligns with the property of an NP problem.
