Inner derivation from Lazard ideal is exponentiable

Statement

Suppose $L$ is a Lie ring and $I$ is a Lazard ideal in $L$. Then, for any $u\in I$, the inner derivation of $L$ arising via the adjoint action of $u$ is an exponentiable derivation of $L$.

Proof

Given: Lie ring $L$, ideal $I$ of $L$. A natural number $c$ such that the 3-local nilpotency class of $I$ is at most $c$ and $I$ is powered over all primes less than or equal to $c$. An element $u\in I$.

To prove: The adjoint map $\operatorname {ad} u=x\mapsto [u,x]$ is an exponentiable derivation of $L$.

Proof: The current version of the proof is incomplete and inaccurate, some fixing needs to be done

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$\operatorname {ad} u$ is $(c+1)$-step-nilpotent, i.e., $(\operatorname {ad} u)^{c+1}=0$.		The 3-local nilpotency class of $I$ is at most $c$ and $I$ is an ideal of $L$.		[SHOW MORE] For $x\in L$, let $v=[u,x]$. Then, $v\in I$ because $I$ is an ideal. Now, $(\operatorname {ad} u)^{c+1}(x)=(\operatorname {ad} u)^{c}(v)=[u,[u,\dots ,u,[u,v]\dots ]]$. This Lie bracket is completely inside $I$, has length $c+1$, and involves two elements $u,v$ of $I$. The 3-local class condition tells us that the value must be zero.
2	$\operatorname {ad} u$ is $(c+1)$-step-binilpotent, i.e., $[(\operatorname {ad} u)^{i}(x),(\operatorname {ad} u)^{j}(y)]=0$ for all $x,y\in L$ and all positive integers $i,j$ such that $i+j\geq c+1$.		The 3-local nilpotency class of $I$ is at most $c$, and $I$ is an ideal of $L$.		[SHOW MORE] For $x,y\in L$, let $v=[u,x]$ and $w=[u,y]$. We get that $v,w\in I$ because $I</mth>isanideal.Wecanrewrite<math>[(\operatorname {ad} u)^{i}(x),(\operatorname {ad} u)^{j}(y)]=[(\operatorname {ad} u)^{i-1}(v),(\operatorname {ad} u)^{j-1}(w)]$