There is a Lemma: If densely defined operator $A$ and $B$ in a Banach space $\mathscr{X}$ are the infinitesimal generator of a strongly continuous contract semi-group and $B \subset A$, then $B=A$.
From the Hille-Yosida Theorem, we have $\forall \lambda >0$, $$R(\lambda-B)=\mathscr{X}=R(\lambda-A).$$ But how to illustrate the $B=A$?
If $B \subset A$, choose $x\in\mathcal{D}(A)\setminus\mathcal{D}(B)$. Then $(B-\lambda I)y = (A-\lambda I)x$ for some $y\in\mathcal{D}(B)$. Because $B\subset A$, then $Ay=By$, which gives $(A-\lambda I)(x-y)=0$. So $x=y$ because $A-\lambda I$ is invertible, which contradicts $x\in\mathcal{D}(A)\setminus\mathcal{D}(B)$.