$p^1,p^2$ is straightforward, and the remaining process is inductive. Suppose that $p^1,\dots,p^{d-1}$ have been chosen satisfying the respective equations. Take the sum of all the terms on left hand side of which have $r \geq 2$. These give rise to a cocycle of bidegree $(d,2-d)$ in the following periodic complex of graded vector spaces: $$\begin{gathered}
 \label{eq:trivial-hh}
\cdots \rightarrow \mathit{Hom}_{H(A)}(Y,Y) \xrightarrow{[a] \mapsto [(-1)^{|a|}\mu^2_A(p^1,a) - \mu^2_A(a,p^1)] } \mathit{Hom}_{H(A)}(Y,Y) \longrightarrow \\ \xrightarrow{[a] \mapsto [(-1)^{|a|+1}\mu^2_A(p^1,a) - \mu^2_A(a,p^1) + a]} \mathit{Hom}_{H(A)}(Y,Y) \rightarrow \cdots \qquad\end{gathered}$$ Since that complex is acyclic, we can modify $p^{d-1}$ by adding a $\mu^1_A$-cocycle, so that the same sum of terms represents the zero class in $\mathit{Hom}_{H(A)}^{2-d}(Y,Y)$, and then choose $p^d$ so that holds.

The homotopy idempotent can then be used to define an $A_\infty$-module $M = (Y,p)^{\mathit{yon}}$, which in the category $H^0(A^{\mathit{mod}})$ is the direct summand of $Y^{\mathit{yon}}$ associated to the Yoneda image of $[p^1]$ (hence, independent of the choice of the homotopy idempotent up to quasi-isomorphism). It consists of the spaces $$\label{eq:q-splitting}
M(X) = \mathit{hom}_A(X,Y)[q]$$ where $q$ is a formal variable of degree $-1$, and has differential $$\mu^1_M(aq^j) = \sum_{r \geq 0} \sum_{k_1 + \cdots + k_r \leq j} \mu^{r+1}_A(p^{k_r},\dots,p^{k_1},a)\, q^{j-k_1-\cdots-k_r}
+ \begin{cases} 0 & \text{$j$ even,} \\ aq^{j-1} & 