co.combinatorics – Does the non-extreme point operator of a closure space being idempotent, imply said space is a matroid (in flat lattice form)?

Given any set $X$ and some closure operator $text{cl}:2^Xto 2^X$ on $X$, we define $psi:2^Xto 2^X$ so that $psi(Q)={qin Q:qintext{cl}(Qsetminus{q})}$ for all $Qsubseteq X$, now if $forall Ssubseteq X(psi(psi(S))=psi(S))$ (i.e. if $psi$ is idempotent) must $text{cl}$ be the closure operator of some matroid?

Sorry if this is either trivially true or trivially false as I am still new to matroid theory.

I know that if $text{cl}$ is the closure operator of a matroid then $psi$ must be idempotent so the converse statement holds. Further I have been able to reduce the problem to proving $mathscr{C}={Csubseteq X:psi(C)neq emptysetlandbigcup_{Ssubset S}psi(C)=emptyset}$ is such that for all $C_1,C_2in mathscr{C}$ we have $ein C_1cap C_2implies exists C_3in mathscr{C}:C_3subseteq (C_1cup C_2)setminus {e}$ i.e. that the family of sets $mathscr{C}$ satisfies the matroid circuit axioms. This is because if $text{cl}$ is the closure operator of a matroid $M$ grounded on $X$, then $psi$ maps every $Ssubseteq X$ to the maximal cyclic set of $M$ in $S$ or in other words $psi(S)$ is the union of all circuits of $M$ contained in $S$, so in particular the closure operator $text{cl}^*:2^Xto 2^X$ of the dual matroid $M^*$ of $M$ satisfies $text{cl}^*(Q)=Xsetminuspsi(Xsetminus Q)$ for all $Qsubseteq X$.