q)_{L}(\nu^{2} q^{g}; q)_{L}}
        \\
        {_{4}}\phi_{3}
        \left[\begin{array}{cccc} q^{-\ell} & \mu^{-1} \nu^{-1} q &  \nu &  q^{-L}
        \\  & \nu^{-1}q^{1-g-L} &  \nu \mu^{-1} & q^{g-\ell+1} \end{array}; q, q
        \right]
        \frac{(\mu; q)_{\infty}(\nu^{2} q^{g}; q)_{\infty}}{(\nu;
        q)_{\infty}(\nu \mu q^{g}; q)_{\infty}}.
    \end{gathered}$$ The equality between and is termwise (when using the definition for $_4\phi_3$).

We will use Watson’s transformation formula [@GasperRahman (III.19)], $$\begin{gathered}
        \label{Watsontransformation}
        {_{4}}\phi_{3} \left[
            \begin{array}{cccc}
                q^{-n} & a & b & c  \\  & d & e & f
            \end{array}
            ; q, q \right]
        \\=
        \frac{(d/b, d/c; q)_{n}}{(d, d/bc; q)_{n}}\,
        {_{8}}\phi_{7} \left[
            \begin{array}{cccccccc} q^{-n} & \sigma & q \sigma^{\frac{1}{2}} & -q \sigma^{\frac{1}{2}} & f/a & e/a & b & c
            \\
            & \sigma^{\frac{1}{2}} &  - \sigma^{\frac{1}{2}} & e & f & ef/ab & ef/ac & e f q^{n}/a
        \end{array}; q, \frac{e f q^{n}}{bc} \right],
    \end{gathered}$$ where $d e f= a b c q^{1-n}$ and $\sigma = ef/aq$. Applying this formula to , we obtain $$\begin{gathered}
        \label{P_ell_g_rewrite_first_case_87}
                \mathbf{P}_{\ell,g}(L)
        =
        \frac
        {(\mu; q)_{\infty}(\nu^{2} q^{g}; q)_{\infty}}
        {(\nu; q)_{\infty}(\nu \mu q^{g}; q)_{\infty}}
        \,
        \mu^{L}
        \frac{(q^{g}; q^{-1})_{\ell} (\nu \mu^{-1}; q)_{L}(\nu q^{g}; q)_{L}(\nu^{-1}q^{1-g}, q)_{\ell}(\nu^{-2} q^{1-g-L}; q)_{\ell}}
        {(\mu \nu q^{g-1}; q^{-1})_{\ell} (q; q)_{L}(\nu^{2} q^{g}; q)_{L}(\nu^{-1} q^{1-g-L}; q)_{\ell}(\nu^{-2}q^{1-g}; q)_{\ell}}
        \\ \times
        {_{8}}\phi_{7} \left[\begin{array}{cccccccc}
                q^{-\ell} & \nu^{2} q^{g-\ell-1} & \nu q^{\frac{g-\ell+1}2} & -\nu 