Prove or disprove: There exists a non trivial graph $D$ in which no two vertices of $D$ have the same out degree but every two vertices of $D$ have the same in degree.
I don't think this statement is true. At least I can't find such graph. However, I can't find a way to disprove it either. I wonder if anyone can give me a hint.
On
Take two vertices $A,B$. At both of these vertices, draw two oriented half-edges pointing towards them: two towards $A$, two towards $B$. Next, draw one oriented half-edge pointing away from $A$, and draw three pointing away from $B$. Altogether you have four oriented half-edges that point towards vertices, and four oriented half edges that point that point away from vertices; and now, put these together in pairs to form four oriented edges.
What about a graph o<---o<===>o? The leftmost node has no out degree, one in degree, the middle most has two out degree and one in degree, and the rightmost node has one out degree, one in degree.