4. Is this a reduction?
Suppose f has a satisfying assignment. Then each clause Cr contains
at least one literal lir that is assigned 1, and each such literal
corresponds to a vertex vir.
Picking one such "true" literal from each clause yields a set V' of
Is V' a clique? For any two vertices vir, vjs, r s, the
corresponding literals are mapped to 1 by the satisfying assignment and
thus the literals cannot be complements. By the construction of G, the
edge (vir, vjs) belongs in E.
Proving the other direction, if G has a clique V' of size k, no edges in G
connect vertices in the same triple, so V' contains exactly one vertex per
triple. Assign a 1 to each literal lir such that vir in V' without
fear of assigning 1 to a literal and its complement. Each clause is
satisfied, and f is satisfied.