I'm working with von Neumann algebras and I stumbled with this statement in a work of Borchers (1999)

Since $\mathcal N \subseteq \mathcal M$, it follows by standard arguments that $\Delta_{\mathcal N} \ge \Delta_{\mathcal M}$

where we must think $\mathcal{N}$ and $\mathcal{M}$ as concrete von Neumann algebras over the same Hilbert space, and $\Delta_\mathcal{M}$ is the modular operator associated to $\mathcal{M}$ and similarly for $\Delta_\mathcal{N}$.

I suppose there is something I'm not catching because assuming $\mathcal{N}\subseteq\mathcal{M}$, I could only arrive to $S_{\mathcal{M}}(x)=S_\mathcal{N}(x)$ for $x\in \operatorname{Dom}(S_\mathcal{N})$ (to do that I assume there is a cyclic and separating vector for $\mathcal{N}$) where $S_R$ denotes the Tomita–Takesaki operator of the algebra $R$. This implies $\Delta_\mathcal{M}(x)=\Delta_\mathcal{N}(x)$, for $x\in \operatorname{Dom}(\Delta_\mathcal{N})$.

However, I couldn't find the proof of Borchers's assertion.