 Notice that $$\langle \mathrm{stab}_\circlearrowleft (J) \mid J \in \varphi \cdot \mathcal{J} \rangle= \varphi \left( \langle \mathrm{stab}_\circlearrowleft (J) \mid J \in \mathcal{J} \rangle \right) = \Phi \mathcal{H}.$$ 