Proof (cont.)

4. Is this a reduction?

Suppose f has a satisfying assignment. Then each clause C_r contains at least one literal l_i^r that is assigned 1, and each such literal corresponds to a vertex v_i^r.

Picking one such "true" literal from each clause yields a set V' of k vertices.

Is V' a clique? For any two vertices v_i^r, v_j^s, r $\neq$ 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 (v_i^r, v_j^s) 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 l_i^r such that v_i^r in V' without fear of assigning 1 to a literal and its complement. Each clause is satisfied, and f is satisfied.