\left(f, Q\right) \mapsto \left(\begin{array}{c} \begin{array}{l} \left[\forall_{b_{1}, b_{2}, \ldots, b_{j}~|~Q\left(b_{1}, b_{2}, \ldots, b_{j}\right)}~\left(f\left(b_{1}, b_{2}, \ldots, b_{j}\right) \in \mathbb{C}\right)\right] \Rightarrow \\ \left[\begin{array}{l}\forall_{a_{1}, a_{2}, \ldots, a_{i}, c_{1}, c_{2}, \ldots, c_{k} \in \mathbb{C}}~\\
\left(\begin{array}{c} \begin{array}{l} \left(a_{1} \cdot a_{2} \cdot \ldots \cdot a_{i} \cdot \left[\sum_{b_{1}, b_{2}, \ldots, b_{j}~|~Q\left(b_{1}, b_{2}, \ldots, b_{j}\right)}~f\left(b_{1}, b_{2}, \ldots, b_{j}\right)\right]\cdot c_{1} \cdot c_{2} \cdot \ldots \cdot c_{k}\right) = \\ \left[\sum_{b_{1}, b_{2}, \ldots, b_{j}~|~Q\left(b_{1}, b_{2}, \ldots, b_{j}\right)}~\left(a_{1} \cdot a_{2} \cdot \ldots \cdot a_{i} \cdot f\left(b_{1}, b_{2}, \ldots, b_{j}\right)\cdot c_{1} \cdot c_{2} \cdot \ldots \cdot c_{k}\right)\right] \end{array} \end{array}\right)\end{array}\right] \end{array} \end{array}\right)