Baileyのterminating 10φ9変換公式

まず, Jacksonの${}_8\phi_7$和公式 より, $w=a^2q/bcd$としたとき,
\begin{align} \Q87{w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,aq^n,q^{-n}}{\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,wq^{1-n}/a,wq^{n+1}}{q}&=\frac{(b,c,d,wq;q)_n}{(aq/b,aq/c,aq/d,a/w;q)_n} \end{align}
より,
\begin{align} \frac{(a,b,c,d;q)_n}{(aq/b,aq/c,aq/d,q;q)_n}&=\frac{(a,a/w;q)_n}{(wq,q;q)_n}\sum_{k=0}^n\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,aq^n,q^{-n})_k}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,wq^{1-n}/a,wq^{n+1},q;q)_k}q^k\\ &=\sum_{k=0}^n\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a)_k(a;q)_{n+k}(a/w;q)_{n-k}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,q;q)_k(wq;q)_{n+k}(q;q)_{n-k}}\left(\frac aw\right)^k \end{align}
であるから,
\begin{align} &\Q{10}9{a,\sqrt aq,-\sqrt aq,b,c,d,e,f,g,q^{-n}}{\sqrt a,-\sqrt a,aq/b,aq/c,aq/d,aq/e,aq/f,aq/g,aq^{n+1}}{q}\\ &=\sum_{0\leq k}\frac{(\sqrt aq,-\sqrt aq,e,f,g,q^{-n})_k}{(\sqrt a,-\sqrt a,aq/e,aq/f,aq/g,aq^{n+1};q)_k}q^k\sum_{j=0}^k\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a;q)_j(a;q)_{k+j}(a/w;q)_{k-j}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,q;q)_j(wq;q)_{k+j}(q;q)_{k-j}}\left(\frac aw\right)^j\\ &=\sum_{j=0}^n\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a;q)_j}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,q;q)_j}\left(\frac aw\right)^j\sum_{k=j}^n\frac{(\sqrt aq,-\sqrt aq,e,f,g,q^{-n})_k(a;q)_{k+j}(a/w;q)_{k-j}}{(\sqrt a,-\sqrt a,aq/e,aq/f,aq/g,aq^{n+1};q)_k(wq;q)_{k+j}(q;q)_{k-j}}q^k\\ &=\sum_{j=0}^n\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,\sqrt aq,-\sqrt aq,e,f,g,q^{-n};q)_j(a;q)_{2j}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,\sqrt a,-\sqrt a,aq/e,aq/f,aq/g,aq^{n+1},q;q)_j(wq;q)_{2j}}\left(\frac {aq}w\right)^j\\ &\qquad\qquad\cdot \sum_{k=0}^{n-j}\frac{(aq^{2j},\sqrt aq^{j+1},-\sqrt aq^{j+1},eq^j,fq^j,gq^j,a/w,q^{j-n})_k}{(\sqrt aq^j,-\sqrt aq^j,aq^{j+1}/e,aq^{j+1}/f,aq^{j+1}/g,aq^{n+j+1},wq^{2j+1},q;q)_k}q^k\\ \end{align}
ここで, Jacksonの${}_8\phi_7$和公式より, $a^3q^{n+2}=bcdefg$のとき,
\begin{align} &\sum_{k=0}^{n-j}\frac{(aq^{2j},\sqrt aq^{j+1},-\sqrt aq^{j+1},eq^j,fq^j,gq^j,a/w,q^{j-n})_k}{(\sqrt aq^j,-\sqrt aq^j,aq^{j+1}/e,aq^{j+1}/f,aq^{j+1}/g,aq^{n+j+1},wq^{2j+1},q;q)_k}q^k\\ &=\frac{(aq^{2j+1},aq/ef,aq/eg,aq/fg;q)_{n-j}}{(aq^{j+1}/e,aq^{j+1}/f,aq^{j+1}/g,aq^{1-j}/efg;q)_{n-j}}\\ &=\frac{(aq;q)_{n+j}}{(aq;q)_{2j}}\frac{(aq/e,aq/f,aq/g;q)_j}{(aq/e,aq/f,aq/g;q)_n}\frac{(eq^{-n}/w,fq^{-n}/w,gq^{-n}/w;q)_{n-j}}{(q^{-n-j}/w;q)_{n-j}}\\ &=\frac{(aq;q)_{n+j}}{(aq;q)_{2j}}\frac{(aq/e,aq/f,aq/g;q)_j}{(aq/e,aq/f,aq/g;q)_n}\frac{(wq^{j+1}/e,wq^{j+1}/f,wq^{j+1}/g;q)_{n-j}}{(wq^{2j+1};q)_{n-j}}\left(\frac{efgq^{j-2n}}{w^2}\right)^{n-j}q^{2\binom{n-j}2}\\ &=\frac{(aq;q)_{n+j}}{(aq;q)_{2j}}\frac{(aq/e,aq/f,aq/g;q)_j}{(aq/e,aq/f,aq/g;q)_n}\frac{(wq^{j+1}/e,wq^{j+1}/f,wq^{j+1}/g;q)_{n-j}}{(wq^{2j+1};q)_{n-j}}\left(\frac{aq^{1+j-n}}{w}\right)^{n-j}q^{2\binom{n-j}2}\\ &=\frac{(aq;q)_{n+j}}{(aq;q)_{2j}}\frac{(aq/e,aq/f,aq/g;q)_j}{(aq/e,aq/f,aq/g;q)_n}\frac{(wq/e,wq/f,wq/g;q)_{n}}{(wq/e,wq/f,wq/g;q)_j}\frac{(wq;q)_{2j}}{(wq;q)_{n+j}}\left(\frac{a}{w}\right)^{n-j} \end{align}
より,
\begin{align} &\Q{10}9{a,\sqrt aq,-\sqrt aq,b,c,d,e,f,g,q^{-n}}{\sqrt a,-\sqrt a,aq/b,aq/c,aq/d,aq/e,aq/f,aq/g,aq^{n+1}}{q}\\ &=\frac{(aq,wq/e,wq/f,wq/g;q)_n}{(wq,aq/e,aq/f,aq/g;q)_n}\sum_{j=0}^n\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,\sqrt aq,-\sqrt aq,e,f,g,q^{-n};q)_j(a;q)_{2j}}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,\sqrt a,-\sqrt a,aq/e,aq/f,aq/g,aq^{n+1},q;q)_j(wq;q)_{2j}}\\ &\qquad\qquad\cdot \frac{(aq/e,aq/f,aq/g,aq^{n+1};q)_j}{(wq/e,wq/f,wq/g,wq^{n+1};q)_j}\frac{(wq;q)_{2j}}{(aq;q)_{2j}}\left(\frac{a}w\right)^{n-j}\\ &=\frac{(aq,wq/e,wq/f,wq/g;q)_n}{(wq,aq/e,aq/f,aq/g;q)_n}\left(\frac aw\right)^n\sum_{j=0}^n\frac{(w,\sqrt wq,-\sqrt wq,wb/a,wc/a,wd/a,e,f,g,q^{-n};q)_j}{(\sqrt w,-\sqrt w,aq/b,aq/c,aq/d,wq/e,wq/f,wq/g,wq^{n+1},q;q)_j}q^j \end{align}
最後に, $g=awq^{n+1}/ef$であることから,
\begin{align} \frac{(wq/g;q)_n}{(aq/g;q)_n}&=\frac{(efq^{-n}/a;q)_n}{(efq^{-n}/w;q)_n}=\frac{(aq/ef;q)_n}{(wq/ef;q)_n}\left(\frac wa\right)^n \end{align}
であることから定理を得る.