Finding a kernel for d-Bounded degree deletion
Reduction Rule 1. Let $V$ be the set of vertices which are isolated. Convert the instance from $I = (G,k,d)$ to $I^{'} = (G -V, k,d)$. If $I^{'}$ is a yes instance, then so is $I$, because adding back the isolated vertices do not add on to the degree of other vertices. And isolated vertices have already have degree 0 ($\le d$ as $d \ge 0$). And if $I$ is a yes instance so is $I^{'}$, as $G-V$ is a sub-graph of $G$, and $k$ and $d$ remain same across the two instances.
Reduction Rule 2. If a vertex $v$ has degree $\ge k+d+1$ then it has to be kept in the deletion set. Otherwise $k+1$ of its neighbours have to be kept in the deletion set. Which cannot be done, as maximum size of deletion set is $k$. Thus after this reduction all vertices will have degree $\le k+d$.
Reduction Rule 3 If there are two vertices $v$ and $w$, such that degree of both of them is $\le d$. Then the edge $vw$ can be removed converting the instance from $I = (G, k,d)$ to $I^{'} = (G-\{vw\}, k,d)$. If $I$ is a yes instance so is $I^{'}$ as $G-\{vw\}$ is a sub-graph of $G$, with $k$ and $d$ remaining same across the instances. And if $I^{'}$ is a yes instance so is $I$, as adding back $vw$ can make the degree of $v$ and $w$ at most $d$ (as they initially had degree $\le d$).
Rule $1$ might be applied after applying Rule $3$ as well.
Now considering a graph where vertices with degree $\le d$ don't share an edge, no vertex is isolated and and all vertices have degree $\le k+d$ (ie. none of the above rules applied). Let $A$ be the set of vertices with degree $\le d$ and let $B$ be the set of vertices with degree $&gt; d$. We can argue that $|B| \le k(k+d) +k$, otherwise the output is no. Because if $|B| &gt; k(k+d) +k$, then as most $k$ elements will be chose from $B$ for the deletion set, and  $ &gt; k(k+d)$ elements will be left (in $B$). And as each of them has degree $&gt; d$, they each have at least one edge going into the deletion set. But the deletion set can only accommodate $k(k+d)$ edges. Also as each vertex in $A$ has at least one neighbor (and all of them in $B$), the maximum number of vertices in $A$, will be bounded as $|A| \le |B|(k+d)  \le (k(k+d)+k)(k+d) = k(k+d)(k+d+1)$.
So $|A| + |B| \le k(k+d) + k(k+d)(k+d+1) = k(k+d)(k+d+2)$.