Given a set $ X$ and a closure operator $ \text{cl}:2^X\to 2^X$ on $ X$ , if we define $ \psi:2^X\to 2^X$ s.t. $ \psi(Q)=\{q\in Q:q\in\text{cl}(Q\setminus\{q\})\}$ then is it true that $ \forall S\subseteq X(\psi(\psi(S))=\psi(S))$ (i.e. that $ \psi$ is idempotent) implies $ \text{cl}$ is the closure operator of some matroid?

I know if $ \text{cl}$ is the closure operator of a matroid $ M$ on $ X$ that $ \small\forall S\subseteq X(\psi(S)=\bigcup_{C\subseteq S:C\text{ is a circuit of }M}S)$ or equivalently the closure operator $ \text{cl}^*:2^X\to 2^X$ of the dual matroid $ M^*$ satisfies $ \forall S\subseteq X(\text{cl}^*(S)=X\setminus\psi(X\setminus S))$ …