In geometric algebra, contraction by a vector $$ is an anti-derivation, meaning for all (possibly inhomogeneous) multivectors $A, B$, we have $$ \DeclareMathOperator{\lc}{\rfloor} \lc (AB) = ( \lc A)B + A^\star(\lc B) \label{1}\tag{1} ,$$ where $^\star$ denotes involution ($^\star = -$ and $(AB)^\star = A^\star B^\star$) and the left contraction $\lc$ is defined by $$ A \lc B = \sum_{p,q} ⟨⟨A⟩_p ⟨B⟩_q⟩_{q-p} $$ where $⟨\phantom{A}⟩_k$ is the grade $k$ projection.
I am struggling to prove eq. \eqref{1}.
I have however proven that $ \lc$ is an anti-derivation with respect to the wedge product; $$ \lc (A∧B) = ( \lc A)∧B + A^\star∧(\lc B) \label{2}\tag{2} .$$
Can anyone show how \eqref{2} implies \eqref{1}, or otherwise prove eq. \eqref{1}?
Making use of the identity $$ \DeclareMathOperator{\lc}{\rfloor} \lc A = \frac12(A - A^\star ) $$ makes for a very simple proof:
\begin{align} \lc (AB) &= \frac12(AB - (AB)^\star ) \\ &= \frac12(AB - A^\star B + A^\star B - A^\star B^\star ) \\ &= ( \lc A)B + A^\star ( \lc B) \end{align}