**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 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.