# Could a non simple graph have at least two vertices of the same degree?

I tried to prove that a simple graph $$G$$ with at least two vertices has two vertices of the same degree. It's easy to prove that this is true by the pigeonhole principle. If we remove the hypothesis that $$G$$ is simple can we affirm that $$G$$ necessarily has two vertices of the same degree? For example, if a $$G$$ has 3 vertices and between 1,2 there are two edges and between 2,3 only one we have one vertex of degree 2, one of degree 3 and the last of degree 1. Is it correct?

Not necessarily. Consider the graph on 3 vertices $$x,y,z$$ where there are 2 edges between $$x$$ and $$y$$, 1 edge between $$x$$ and $$z$$ (and no edges between $$y$$ and $$z$$). So $$d(x) = 2+1=3$$, $$d(y) = 2$$ and $$d(z)=1$$.