The claim is as follows: Let’s say we have a $ k$ -regular simple undirected graph $ G$ on $ n$ vertices. Then, does $ G$ then always have a $ d$ -factor for all $ d$ satisfying $ 1 \le d \lt k$ and $ dn$ being even.

I think its true, since we can construct a

- $ k+1$ -regular graph from $ G$ by adding $ \frac{n}{2}$ edges for even $ n$ or
- $ k+2$ -regular graph from $ G$ by adding $ n$ edges for odd $ n$

And the converse for the above. We can construct a $ k$ -regular graph on $ n$ vertices by

- removing $ \frac{n}{2}$ edges from a $ k+1$ -regular graph for even $ n$ .
- removing $ n$ edges from a $ k+2$ -regular graph for odd $ n$ .

Hence, using the converse argument, we can say that the original claim is true, since from a $ k$ -regular graph $ G$ we can construct

- $ 1$ -factor, $ 2$ -factor, … $ k-1$ -factor for even $ n$ or
- $ 2$ -factor, $ 4$ -factor, … $ k-2$ -factor for odd $ n$

Is my argument correct or is there any flaw in this argument?