have been going the following thm ...help me with the proof
Theorem: Let $D = (d_{1},d_{2}, ...d_{n})$ a sequence of nonnegative integers , n > 1, $d_{1}\leq d_{2}\leq ...d_{n}$. Let $D' = (d'_{1},d'_{2},...d'_{n-1})$ be the sequence defined by $$d'_{i} = d_{i} \text{ for }i < n-d_{n},$$
$$d'_{i} = d_{i} - 1\text{ for }i \geq n-d_{n} $$
(For example, for D = (1; 1; 2; 2; 2; 3; 3) we have D' = (1; 1; 2; 1; 1; 2).) Then D is a graph degree sequence if and only if D′ is such.
Suppose you have a graph G such that the list of his degrees is D, such that $v_i$ has degree $d_i$.
We will call $S=\{v_1, \ldots, v_{n-d_n -1}\}$ the "small" vertices (because they have small degree) and $B$ the others, the "big" vertices. We denote by V the biggest vertex $v_{n}$ and by N its neighborhood. Note that $N=B$ iff $N \cap S = \emptyset $ by a matter of cardinality.
Simple case : $N=B$. Then the graph obtained by removing the vertex $V$ has list $D'$.
General case: $N\cap S \neq \emptyset$. Take an edge $e$ from V to a small vertex $w$, and a big vertex $u$ not linked to $V$ (why does It exist?). Then there exist a vertex $z$ that is linked with $u$ by an edge $f$ but not with $w$; otherwise we would have $deg(u) < deg(w)$ (why?), which is impossible because of the assumption on small/big vertices. Now move $e$ so that it links $V$ and $u$, and $f$ such that it links $z$ and $w$. This process does not modify the list of degrees D.
In this way tha cardinality of $N\cap S$ gets down; thus iterating this process we will reduce to the case $N=B$, which is the "simple case".
Conversely, given a graph G' with list of degrees D', add a vertex $v_n$ and link it with the $d_n$ bigger vertices in G'.