The equivalent axiom of closure operator on a partial order set

Question

$\newcommand{\cl}{\operatorname{cl}}$ A function $\cl: P → P$ from a partial order $P$ to itself is called a closure operator if it satisfies the following axioms for all elements $x, y$ in $P$,

(a) $x ≤ \cl(x)$ ($\cl$ is extensive)

(b)$x ≤ y$ implies $\cl(x) ≤ \cl(y)$ ($\cl $ is increasing)

(c)$\cl(\cl(x)) = \cl(x)$($\cl$ is idempotent)

More succinct alternatives are available: the definition above is equivalent to the single axiom

$x ≤ \cl(y)$ if and only if $\cl(x) ≤ \cl(y)$ for all $x, y$ in $P$.

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The above is a copy of Wikipedia, I want to prove the equivalence.

Suppose we have the single axiom.

(a) Since $\cl(x)\leq \cl(x)$, $ x\leq\cl(x)$;

(b) If $x\leq y$, then $x\leq\cl(y)$ by (a), thus $\cl(x)\leq\cl(y)$;

(c) From (a) and (b), we know that $\cl(x)\leq\cl(\cl(x))$.

My Question:

How to verify $\cl(\cl(x))\leq \cl(x)$?

Answer 1

$\newcommand{\cl}{\operatorname{cl}}\cl(\cl(z))\le \cl(z)$ iff $\cl(z)\le \cl(z)$ (why?)

Take $x=\cl z$ and $y=z$.