View source for B(x,y)B(x+y,z)=B(z,x)B(x+z,y)

==Theorem==
The following formula holds:
$$B(x,y)B(x+y,z)=B(z,x)B(x+z,y),$$
where $B$ denotes the [[beta]] function.

==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=B(x,y)B(x+y,z)=B(y,z)B(y+z,x)|next=B(x,y)B(x+y,z)B(x+y+z,u)=Gamma(x)Gamma(y)Gamma(z)Gamma(u)/Gamma(x+y+z+u)}}: $\S 1.5 (7)$

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