If every diagonal value of an incidence function is a unit, recursive interval formulas construct both a left and a right convolution inverse

On a diagonal interval the equations $(f\ast g)(x,x)=1_R$ and $(h\ast f)(x,x)=1_R$ force $g(x,x)=h(x,x)=f(x,x{)}^{-1}$ by [L3].

Fix a natural $n$ and assume that $g$ and $h$ have been uniquely defined on every interval of cardinality less than $n$, with the required convolution equations there.

Let $x<y$ with $\mathrm{\mid }[x,y]\mathrm{\mid }=n$. Every $g(z,y)$ occurring in ${\sum }_{x<z\le y}f(x,z)g(z,y)$ belongs to the proper subinterval $[z,y]$, and every $h(x,z)$ in ${\sum }_{x\le z<y}h(x,z)f(z,y)$ belongs to the proper subinterval $[x,z]$; their cardinalities are less than $n$ by [F1] and [L2].

The displayed formulas in the Statement therefore assign unique values to $g(x,y)$ and $h(x,y)$, since the sums are finite and both diagonal inverses are unique.

Isolating the term $z=x$ in convolution gives $(f\ast g)(x,y)=f(x,x)g(x,y)+{\sum }_{x<z\le y}f(x,z)g(z,y)=0_R$ by the defining formula for $g(x,y)$.

Isolating the term $z=y$ gives $(h\ast f)(x,y)={\sum }_{x\le z<y}h(x,z)f(z,y)+h(x,y)f(y,y)=0_R$ by the defining formula for $h(x,y)$.

Steps 1.1 through 4.2, with strong induction on $\mathrm{\mid }[x,y]\mathrm{\mid }$, define $g$ and $h$ on every comparable pair and give $f\ast g=\delta$ and $h\ast f=\delta$.

Associativity and the identity law in [L4] now give $h=h\ast \delta =h\ast (f\ast g)=(h\ast f)\ast g=\delta \ast g=g$.

Hence the two recursive one-sided inverses coincide and their common value is a two-sided convolution inverse of $f$.