I want to use Vertex Cover as a known NP Complete Problem for the reduction. The claim is that If a have a vertex cover in graph G with size $\le$ k, I will have a clique cover in $G\prime$  with total number of cliques $\le$ kI want to reduce vertex cover problem to clique cover problem in the following way:For each edge in graph G, I will create a corresponding vertex in $G\prime$. In the picture, each e$_i$ in G corresponds to v$_i$ in $G\prime$ for 1$\le$i$\le$5Two vertices in $G\prime$ will be connected by an edge if their corresponding two edges in G have one vertex in common. For example, there is an edge between v$_1$ to v$_2$ as their corresponding edges e$_1$ and e$_2$ have one vertex in common which is a.Each clique in $G\prime$ corresponds to a vertex in G and that vertex belongs to the vertex cover. For example, for the graph in picture I have vertex cover {a,c}. With vertex a, I get  a clique  {v$_1$, v$_2$, v$_5$} and with vertex c, I get clique {v$_3$, v$_4$} in $G\prime$.Now I don't know how to prove No instance in opposite direction: If $G\prime$ doesn't have a clique cover with $\le$ k cliques (i.e it needs at least k+1 cliques), G doesn't have a vertex cover size which is $\le$ k.Can anybody shed some light on this issue?
Prove that clique cover is NP Complete
np complete;reductions