A proof strategy for $L_XY=[X,Y]$ I'm trying to prove that $$L_XY=[X,Y]$$ I do realize that there are other proofs given of this assertion on stackexchange. However, I'm looking for ways to prove it using my strategy, as given below:

Let $\phi(t)$ be the flow of the vector $X$ at the point $p$. Then $$L_XY=\lim\limits_{t\to 0}\frac{\phi^*_{-t}Y(\phi(t))-Y(p)}{t}$$ Now $\phi^*_{-t}Y(\phi(t))$ can be written as $V(\phi(t))+\int_t^0\nabla_{-\phi'(s)}Y(\phi(s))ds$. Hence, $$\lim\limits_{t\to 0}\frac{\phi^*_{-t}Y(\phi(t))-Y(p)}{t}=\lim\limits_{t\to 0}\frac{V(\phi(t))+\int_t^0\nabla_{-\phi'(s)}Y(\phi(s))ds-V(p)}{t}$$ This . can be simplified as $$\lim\limits_{t\to 0}\frac{\int_t^0\nabla_{-\phi'(s)}Y(\phi(s))ds}{t}+\lim\limits_{t\to 0}\frac{V(\phi(t))-V(p)}{t}$$ 

Can we write $$\lim\limits_{t\to 0}\frac{\int_t^0\nabla_{-\phi'(s)}Y(\phi(s))ds}{t}=-\nabla_YX$$ Because then we'll be done, as $\lim\limits_{t\to 0}\frac{V(\phi(t))-V(p)}{t}=\nabla_XY$

Let's just act on a function:
$$ L_X(Y) (f) = X (Y(f)) -Y(X(f)),$$
by the Leibnitz rule and the property $L_X(f)=X(f)$.
