Does there exist a graph on 9 vertices whose degrees are 1, 1, 1, 1, 1, 2, 4, 5, 6?

Question

Mathematics

Is my proof correct? What to do for large sequences? What is the algorithm for doc-va such tasks?
I prove it as follows.
Let's say that all vertices have degree 0
0 0 0 0 0 0 0 0 0
Then to get a vertex with degree 6 we need to add edges
0 0 1 1 1 1 1 1 6
Here I connected the last vertex with 3,4,5, 6,7,8 vertices, and their degree has increased by 1
Now we need to get a vertex with degree 5
1 1 1 1 1 2 2 5 6
Next we need to get a vertex with degree 4
1 1 1 1 2 3 4 5 6
Contradiction

Answer 1

Erdős-Gallay theorem
1. 8 ≥ 6 ≥ 5 ≥ 4 ≥ 2 ≥ 1 ≥ 1 ≥ 1 ≥ 1 ≥ 1
2. 6 + 5 + 4 + 2 + 1 + 1 + 1 + 1 + 1 = 22
This sequence is correct .
k = 1, 6 ≤ (0 + 8), satisfied,
k = 2, 11 ≤ (2 + 9), satisfied,
k = 3, 15 ≤ (6 + 7), not satisfied.
This means that this sequence is not graphic, and it is impossible to construct a simple graph from it.

Answer 2

Specifically, here you can think of it like this:
vertices 4, 5, 6, even if they are connected to each other in pairs (which will take 6 degrees), they still have 9 free degrees that are not covered by the rest of the vertices.
but this is provided that there are no multiple edges (when edges A and B are connected by more than one vertex), and loops (edges A-A). Do you have the exact same case?