(1,1)-bi-Engel implies second derived subring is in 2-torsion

Statement

Suppose $L$ is a (1,1)-bi-Engel Lie ring, i.e., $[[u,x],[u,y]]=0$ for all $u,x,y\in L$. Then, we have that $2[[L,L],[L,L]]=0$. More explicitly:

$2[[a,b],[c,d]]=0\ \forall \ a,b,c,d\in L$

Proof

Given: Lie ring $L$. $[[u,x],[u,y]]=0$ for all $u,x,y\in L$.

To prove: $2[[a,b],[c,d]]=0\ \forall \ a,b,c,d\in L$

Proof: Fix $a,b,c,d\in L$ for this proof.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$[[a,b],[c,d]]+[[c,b],[a,d]]=0$		Lie bracket is linear		Polarize the original identity, setting $x=b,y=d$. For $u$, take the values $a+c,a,c$ respectively, subtract, and simplify.
2	$[[a,b],[c,d]]=[[a,d],[c,b]]$		Lie bracket is skew-symmetric	Step (1)	In Step (1), use skew symmetry between $[a,d]$ and $[c,b]$.
3	$[[a,b],[c,d]]=[[b,a],[d,c]]$ and $[[a,d],[c,b]]=[[d,a],[b,c]]$.		Lie bracket is skew-symmetric		Use skew symmetry twice within each
4	$[[b,a],[d,c]]+[[d,a],[b,c]]=0$		Lie bracket is linear		Identical reasoning as Step (1), letter roles changed: now take $x=a,y=c$. For $u$, take the values $b+d,b,d$ respectively, subtract, and simplify.
5	$[[a,b],[c,d]]+[[a,d],[c,b]]=0$			Steps (3), (4)	Step-combination direct
6	$2[[a,b],[c,d]]=0$			Steps (2), (5)	Step-combination direct