$f(x,y) - f(x,z) = g(y,z)$ implies $f(x,y) = a(x) + b(y)$

Question

A result I need is:

If $f(x,y) - f(x,z) = g(y,z)$ for all $(x,y,z)$, then $f(x,y) = a(x) + b(y)$ for some functions $(a,b)$.

This seems almost obvious, and I've constructed a proof, but that proof seems unnecessarily complicated and is remarkably tedious (and so isn't included here). I'd like pointers or ideas leading to something more simple and elegant. Surely this is a known result, or a special case of a well-known result?

Answer 1

Plugging $z=0$ into equation gives that $$ f(x,y)-f(x,0)=g(y,0), $$ so $$ f(x,y)=f(x,0)+g(y,0). $$ Now, just denote $a(x):=f(x,0)$ and $b(y):=g(y,0)$ and we get the desired result.

Answer 2

take $a(x)=f(x,z)$, $b(x)=g(x,z)$ for some $z$. then $f(x,y)=f(x,z)+g(y,z)=a(x)+b(y)$.