Definition of 'replication of $v$' is



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> Suppose $v \in V(G)$. Replication of $v$ is constructing $G'$ by adding a new vertex $v'$ such that $N\_{G'}(v')=N\_G(v) \cup \{v\}$.
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And the following statement is well-known fact(you can find a proof by googling).



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> Replication of a single vertex conserves perfectivity, i.e., $G'$ obtained from a perfect graph $G$ by replication of $v \in V(G)$ is still perfect.
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But I am curious whether the following statement is also true:



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> $G''$ obtained from $G$ by adding a new vertex $v''$ such that $N\_{G''}(v'')=N\_G(v)$ is perfect.
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Adding $v''$ looks similar to the replication, but it does not contain $v$ in neighborhood of $v''$.  

I could not find any counterexample to disprove it.  

And proving it seems to be different from the proof of the first statement, so I cannot follow its steps.  

Would you help me?



Pseudo-replication of a vertex in a perfect graph
If we pass to the complement, this is just a replication. But passing to a complement preserves the class of perfect graphs (a theorem of Lovasz, previously conjecture of Berge).


