I am looking for references about the complexity of *Boolean formula balancing* problem. In particular,

 1. Was it known that Boolean formulas can be balanced in $\mathsf{AC^0}$?
 2. Is there a simple proof of Boolean formula balancing being in $\mathsf{AC^0}$? 

By "simple" I mean a proof simpler than the one I mention below, in particular I am looking for a proof which doesn't depend on Boolean formula evaluation being in $\mathsf{NC^1}$.

----
### Background 
Here all mentioned complexity classes are the uniform ones.

> **BFB** (Boolean formula balancing):  
> Given a Boolean formula $\varphi$,  
> Find an equivalent balanced Boolean formula.

I am interested in the complexity of this problem, in particular simple proofs showing the problem is in $\mathsf{AC^0}$ (or even $\mathsf{TC^0}$ or $\mathsf{NC^1}$). The common balancing arguments like those based on Spira's lemma apply repeated structural modifications to the formula tree which seem to only give $BFB \in \mathsf{NC^2}$.

I have a proof for $BFB \in \mathsf{AC^0}$, however the proof is not simple and depends on the proof of $BFE \in \mathsf{NC^1}$.

> **BFE** (Boolean formula evaluation)   
> Given a Boolean formula $\varphi$ and a truth assignment $\tau$ for variables in $\varphi$,  
> Does $\tau$ satisfy $\varphi$ ($\tau \vDash \varphi$)? 

It is known from Sam Buss's celebrated result that *Boolean formula evaluation* ($BFE$) can be computed in $\mathsf{NC^1} = \mathsf{ALogTime}$ (see[ \[Buss87\]][1] and [ \[BCGR92\]][2]).

It follows (quite surprisingly, at least to me) that *Boolean formulas balancing* ($BFB$) is also in $\mathsf{NC^1}$:

The idea is that we can hardcode $\varphi$ in the input gates of $BFE$ to obtain a formula equivalent to $\varphi$ and this is a completely syntactic operation computable in $\mathsf{AC^0}$. Since $BFE$ has balanced formulas we obtain a equivalent balanced formula for $\varphi$. In other words, the algorithm is:

$$\ulcorner \varphi \urcorner \mapsto \ulcorner \lambda \vec{p}. Eval(\ulcorner \varphi \urcorner, \vec{p} )\urcorner$$

### Motivation
A simpler argument for $BFB$ being in $\mathsf{AC^0}$ (or $\mathsf{TC^0}$ or even $\mathsf{NC^1}$) would give a new simpler proof of $BFE \in \mathsf{NC^1}$ since it is easy to see that the *balanced* version of BFE can be solved in $\mathsf{NC^1}$ and we can compose it with $BFB$ and the result will be in $\mathsf{NC^1}$.

----
### Questions
1. Was it known that Boolean formulas can be balanced in $\mathsf{AC^0}$ ($BFB\in \mathsf{AC^1}$)?
2. Is there a simpler argument (e.g. not relying on $BFE\in \mathsf{NC^1}$) for $BFB\in\mathsf{AC^0}$?

  [1]: http://www.math.ucsd.edu/~sbuss/ResearchWeb/Boolean/
  [2]: http://euclid.ucsd.edu/~sbuss/ResearchWeb/Boolean2/