View source for 1Phi0(a;;z)1Phi0(b;;az)=1Phi0(ab;;z)

==Theorem==
The following formula holds:
$${}_1\phi_0(a;;z){}_1\phi_0(b;;z)={}_1\phi_0(ab;;z),$$
where ${}_1\phi_0$ denotes [[basic hypergeometric phi]].

==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=1Phi0(a;;z) as infinite product
|next=(z/(1-q))2Phi1(q,q;q^2;z)=Sum z^k/(1-q^k)}}: $4.8 (5)$

[[Category:Theorem]]
[[Category:Unproven]]