# Proof of a graph problem on degree sequence

Prove that: order of natural number $d_1 , d_2 , \ldots , d_n$ as descending is a ordering graph if and only if the order was sorted as descending $d_2-1, d_3 -1 , \ldots ,d_d -1,\ldots ,d_n$ be ordering graph. $d_d$ in sequence is d $d_1$. (arrangement=order).

Hint: If you know that $d=d_1\ge d_2 \ge \dots \ge d_n$ are degrees of vertices of some graph, what can you say about degrees of the graph where the vertex of degree $d_1$ is omitted? Can you go in opposite direction, too? (I.e. can you obtain from a graph with degrees $d_1-1,\dots,d_d-1,d_{d+1},\dots,d_n$ a graph with degrees $d_1,d_2,\dots,d_n$?)