Showing another result on distributive, complemented lattice

Question

Currently reading Bell & Slomson's Models & Ultraproducts. Looking for clues, tips hints for showing the following:

Let $L$ be a distributive, complemented lattice. Then for any $x$ $\in$ $L$, (($x$*) *) = $x$. (Where * indicates complement.)

Proof: (Here's what I have so far)

Take $x$ $\in$ $L$. By lemma 1.14 (i.e. In every lattice $L$ that's complemented and distributive, each element $x$ $\in$ $L$ has unique complement $y$ $\in$ $L$) that $x$ has a unique complement $x$*. Thus $x$ * has a unique complement as well, say (($x$*) *).

Where do I go from here? Not sure if what I have here is the correct strategy.

NOTE: Courtesy of Berci, whose answer is below, I learned the proof requires only another two lines. By axiom L4 (complemented lattice), $L$ satisfies $x$ v $x$* = 1 and $x$ ^ $x$* = $\emptyset$. Now we can substitute (($x$*) *) for the left part of the join / meet and get the same thing. This together with the uniqueness of complements in the lattice yields the desired result.

Answer 1

There are defining equations that the complement must satisfy. If I remember correctly, these are:

$x^*\land x=0\ $ and $\ x^*\lor x=1$.

Now as both $x$ and $x^{**}$ satisfy these for $x^*$, they must be equal by the cited lemma.