Theorem 36.12: VC $\in$ NPC

Proof Sketch:

1.

VC $\in$ NP
Given V', check $\mid$V'$\mid$ = k, and for each edge (u,v) $\in$ E, check that either u $\in$ V' or v $\in$ V'.

2.

L' = CLIQUE

3.

CLIQUE VC
If graph G = (V, E) has clique V', then graph has vertex cover V - V'.
= (V, ) is the complement of G = (V,E), where
Reduction: G $\rightarrow$ (poly-time)

4.

x $\in$ CLIQUE(G) = V' $\longrightarrow$ f(x) $\in$VC( ) = V - V' ( | V' | = k)
Every edge (u,v) $\in$ implies (u,v) E, thus at least one of u and v V'. Thus, at least one of u,v belongs to V - V', which means edge (u,v) is covered by V - V'. Similar argument for other direction.