I have seen this statement and was wondering how one might show that if a graph (with $|V(G)| = n$ vertices) is hamiltonian, that then it's $n$-closure is the $K_n$ (note that this is refering to simple graphs with $n \geq 3$)
The $n$-closure of $G$ is refering to the longest possible sequence (or rather the last element of the sequence) of graphs $G_0, G_1, \ldots, G_h$ $(G_0 := G)$ such that from $G_i$ to $G_{i+1}$ for $x,y \in V(G_i)$ non-adjacent with $d_{G_i}(x) + d_{G_{i}} (y) \geq n$ the edge $xy$ was added (it can be shown that this closure is well defined and unique).
The other implication follows with Ore's lemma as far as I know.
One could use Ore's lemma for the other direction aswell however I don't see why $d(x)+ d(y) \geq n$ should hold for every $x,y \in V(G)$. Especially since it is sufficient for $d(x) + d(y) \geq n-4$ and $G$ to be $1$-tough for $G$ to be hamiltonian.
As given by myself as an counterexample and identified by kabenyuk as such. The Graph $C_n$ (Cirlce with $n$ nodes) is a counter example to the claim: $G$ is hamiltonian $\implies$ $cl_n (G) \cong K_n$, since $cl_n (C_n) = C_n \not \cong K_n$.
As such the statement is wrong