negation of a nid-assignment to φ
Let φ be a 3cnf formula. A non-identical assignment, or nid-assignment, to the variables of φ is an assignment where each clause of φ contains two literals with different truth values, i.e., there cannot be a clause with the 3 true literals. Thus, a nid-assignment satisfies φ without assigning true to the 3 literals of any of the clauses.
(a) Show that the negation of a nid-assignment to φ is also a nid-assignment to φ.
(b) Let SATnid be the collection of 3cnf formulas that have a nid-assignment. Show that it is possible to obtain a polynomial-time reduction from 3SAT to SATnid by replacing each ci clause of the form (y1 ∨ y2 ∨ y3) by two clauses (y1 ∨ y2 ∨ zi) and (zi ∨ y3 ∨ b) where zi is a new variable for each ci clause, and b is a single additional new variable.
(c) Conclude that SATnid is NP-complete.