Showing there is a node in the graph with only one edge

Question

I saw this question recently, Showing there is a node in the graph with one and only one edge and I am just wondering how the approach would be different if we added the following restraint:
We have an undirected simple graph with n vertices where for every pair of vertices $v_1,v_2$, if $d(v_1)=d(v_2)$ then the set of neighbours of $v_1$ is disjoint from the set of neighbours of $v_2$ where there may be an edge between $v_1$ and $v_2$. Assuming the graph contains at least one edge, prove that there is a vertex of degree exactly 1 in the graph.

Answer 1

I think there'll be no different. You mean that $v_2\in N(v_1)$ and $v_1\in N(v_2)$ is allowable. While $v_1\notin N(v_1),$ $v_2\notin N(v_2),$ $N(v_1)\cap N(v_2)=\varnothing$ is still right.