Incidence convolution is associative and distributes over pointwise addition

Expanding the left bracketing and distributing the factor $h(v,y)$ through the inner sum gives $((f\ast g)\ast h)(x,y)={\sum }_{x\le v\le y}{\sum }_{x\le u\le v}f(x,u)g(u,v)h(v,y)$.

Put $D:=\{(u,v)\in [x,y{]}^2:u\le v\}$. Expanding the right bracketing gives $(f\ast (g\ast h))(x,y)={\sum }_{x\le u\le y}{\sum }_{u\le v\le y}f(x,u)g(u,v)h(v,y)$.

For every $x\le y$, $((f+g)\ast h)(x,y)={\sum }_{x\le z\le y}(f(x,z)+g(x,z))h(z,y)={\sum }_{x\le z\le y}f(x,z)h(z,y)+{\sum }_{x\le z\le y}g(x,z)h(z,y)$ by distributivity in $R$ and additivity of a finite sum; hence $(f+g)\ast h=f\ast h+g\ast h$.

The same calculation with the sum in the right factor gives $f\ast (g+h)=f\ast g+f\ast h$.

Extend the displayed summand by $0_R$ from $D$ to $[x,y{]}^2$. Splitting each finite inner sum into the admissible indices and the zero terms identifies steps 1.1 and 1.2 with its two iterated sums over $[x,y{]}^2$. Finite Fubini makes those iterated sums equal.

Since steps 2.1 and 1.2 agree for every comparable $(x,y)$, $(f\ast g)\ast h=f\ast (g\ast h)$.

Steps 3.1, 1.3 and 1.4 prove associativity and both distributive laws.