NP-complete: To prove that $B$ is NP-complete, choose a known NP-complete problem and show $A \leq_P B$.
Reduction: To prove that $A \leq_P B$, i.e. $A$ can be reduced to $B$, show that
Some transformations:
Note: A maximal clique is a subgraph $S$ such that
Solution: We will show that INDEPENDENT-SET is reducible to MAX-CLIQUE.
Lemma: INDEPENDENT-SET $\leq_P$ MAX-CLIQUE.
Proof: For input graph for INDEPENDENT-SET, $G(V,E)$ and integer $k$, we construct $G’(V, E’)$ where $E’ = K(|V|) \setminus E$. That is, removing of the edges in $E$ from the complete graph of $|V|$ vertices. This will be our input for MAX-CLIQUE.
$(\Rightarrow)$ If we have a YES-instance of INDEPENDENT-SET of size $k$, then there exists some $S \subset G$ where $S$ is completely disconnected from each other. Thus in $G’$, the induced subgraph of the $k$ vertices in $S$ must be completely connected, i.e. a maximal clique of size $k$. Hence we have a YES-instance of MAX-CLIQUE.
$(\Leftarrow)$ If we have a YES-instance of MAX-CLIQUE of size $k$, then there exists some $S \subset G$ where $S$ is completely connected. Then in the
If MAX-CLIQUE is can be solved in polynomial time, then INDEPENDENT-SET can be solved in polynomial time too, and thus we have a contradiction. Hence, MAX-CLIQUE is NP-complete.