[proofplan] We prove closure by taking two functions that are constant on interval types and grouping the convolution sum according to the two subinterval types determined by each cut point. The hereditary hypothesis says that the number of cut points of each ordered type pair depends only on the ambient interval type. This makes the convolution value a function only of the ambient type, and the same grouping gives the displayed product formula on type-functions. Finally, associativity follows from a finite reindexing of the ordinary interval convolution sums. [/proofplan] [step:Pass from reduced functions to functions on interval types] Let $f,g\in R(\mathcal F,\sim;k)$. Since $f$ and $g$ are constant on interval types, there are unique functions \begin{align*} a:\mathcal T\to k \end{align*} and \begin{align*} b:\mathcal T\to k \end{align*} such that, for every interval $J\in\mathcal F$, \begin{align*} f(J)=a(\operatorname{type}(J)) \end{align*} and \begin{align*} g(J)=b(\operatorname{type}(J)). \end{align*} The functions $a$ and $b$ are well-defined because $f$ and $g$ have the same value on equivalent intervals. [/step] [step:Group the convolution sum by the two subinterval types] Fix an interval $[x,y]\in\mathcal F$, and define its type by \begin{align*} \tau=\operatorname{type}([x,y]). \end{align*} For each $z\in[x,y]$, the closure of $\mathcal F$ under subintervals gives $[x,z]\in\mathcal F$ and $[z,y]\in\mathcal F$. Therefore the two types \begin{align*} \alpha_z=\operatorname{type}([x,z]) \end{align*} and \begin{align*} \beta_z=\operatorname{type}([z,y]) \end{align*} are defined elements of $\mathcal T$. Because $[x,y]$ is finite, the convolution sum is finite. Using the functions $a$ and $b$ from the previous step, we obtain \begin{align*} (f*g)([x,y]) = \sum_{z\in[x,y]} a(\alpha_z)b(\beta_z). \end{align*} Group this finite sum by the ordered pair $(\alpha_z,\beta_z)$. This gives \begin{align*} (f*g)([x,y]) = \sum_{\alpha,\beta} \left|\left\{z\in[x,y] : \alpha_z=\alpha \text{ and } \beta_z=\beta\right\}\right| a(\alpha)b(\beta). \end{align*} By the hereditary hypothesis, the displayed cardinality is exactly $N_{\alpha,\beta}^{\tau}$ and depends only on $\tau$, not on the representative interval $[x,y]$. Hence \begin{align*} (f*g)([x,y]) = \sum_{\alpha,\beta} N_{\alpha,\beta}^{\tau}a(\alpha)b(\beta). \end{align*} [guided] The goal is to show that $f*g$ is still reduced, meaning that its value on $[x,y]$ depends only on the type of $[x,y]$. Let \begin{align*} \tau=\operatorname{type}([x,y]). \end{align*} For a cut point $z\in[x,y]$, the convolution product uses the two subintervals $[x,z]$ and $[z,y]$. These subintervals are in $\mathcal F$ because $\mathcal F$ is assumed to be closed under subintervals, so their types are legitimate elements of $\mathcal T$. Define \begin{align*} \alpha_z=\operatorname{type}([x,z]) \end{align*} and \begin{align*} \beta_z=\operatorname{type}([z,y]). \end{align*} Since $f$ and $g$ are constant on interval types, the values of $f$ and $g$ on these subintervals are determined by the type-functions $a:\mathcal T\to k$ and $b:\mathcal T\to k$: \begin{align*} f([x,z])=a(\alpha_z) \end{align*} and \begin{align*} g([z,y])=b(\beta_z). \end{align*} Substituting these identities into the convolution formula gives \begin{align*} (f*g)([x,y]) = \sum_{z\in[x,y]} a(\alpha_z)b(\beta_z). \end{align*} Now we reorganize this finite sum. The interval $[x,y]$ is finite, so every reindexing below is a finite algebraic rearrangement in the commutative ring $k$. For each ordered pair $(\alpha,\beta)\in\mathcal T\times\mathcal T$, collect exactly those cut points $z$ for which the left subinterval has type $\alpha$ and the right subinterval has type $\beta$. Thus \begin{align*} (f*g)([x,y]) = \sum_{\alpha,\beta} \left|\left\{z\in[x,y] : \operatorname{type}([x,z])=\alpha \text{ and } \operatorname{type}([z,y])=\beta\right\}\right| a(\alpha)b(\beta). \end{align*} The hereditary hypothesis says precisely that this cardinality is $N_{\alpha,\beta}^{\tau}$ and that it depends only on the ambient type $\tau$, not on the particular representative interval $[x,y]$. Therefore \begin{align*} (f*g)([x,y]) = \sum_{\alpha,\beta} N_{\alpha,\beta}^{\tau}a(\alpha)b(\beta). \end{align*} The right-hand side is written entirely in terms of $\tau$, $a$, $b$, and the structure constants $N_{\alpha,\beta}^{\tau}$. This is the key point: no feature of the representative interval $[x,y]$ remains. [/guided] [/step] [step:Conclude that the convolution is reduced] Let $[x,y],[u,v]\in\mathcal F$ have the same interval type. Define \begin{align*} \tau=\operatorname{type}([x,y])=\operatorname{type}([u,v]). \end{align*} The formula from the previous step gives \begin{align*} (f*g)([x,y]) = \sum_{\alpha,\beta} N_{\alpha,\beta}^{\tau}a(\alpha)b(\beta) \end{align*} and \begin{align*} (f*g)([u,v]) = \sum_{\alpha,\beta} N_{\alpha,\beta}^{\tau}a(\alpha)b(\beta). \end{align*} Therefore \begin{align*} (f*g)([x,y])=(f*g)([u,v]). \end{align*} Thus $f*g$ is constant on interval types, so \begin{align*} f*g\in R(\mathcal F,\sim;k). \end{align*} This proves that $R(\mathcal F,\sim;k)$ is closed under convolution. [/step] [step:Identify the induced product on interval types] Let $a,b:\mathcal T\to k$ be type-functions, and let $f,g\in R(\mathcal F,\sim;k)$ be the corresponding reduced interval functions defined by \begin{align*} f(J)=a(\operatorname{type}(J)) \end{align*} and \begin{align*} g(J)=b(\operatorname{type}(J)) \end{align*} for every $J\in\mathcal F$. For $\tau\in\mathcal T$, choose any representative interval $[x,y]\in\tau$. Since $f*g$ is reduced, the value of $(f*g)([x,y])$ depends only on $\tau$. Define \begin{align*} (a*b)(\tau)=(f*g)([x,y]). \end{align*} The computation above gives \begin{align*} (a*b)(\tau) = \sum_{\alpha,\beta}N_{\alpha,\beta}^{\tau}a(\alpha)b(\beta), \end{align*} where only ordered pairs that occur inside a representative of $\tau$ contribute. Because every representative interval is finite, there are only finitely many such ordered pairs, so the sum is finite and defines an element of $k$. [/step] [step:Verify the associative algebra structure] It remains to check that the convolution product is associative on $R(\mathcal F,\sim;k)$. Let $f,g,h\in R(\mathcal F,\sim;k)$, and fix $[x,y]\in\mathcal F$. Since $[x,y]$ is finite, all sums below are finite. First, \begin{align*} ((f*g)*h)([x,y]) = \sum_{z\in[x,y]}\left(\sum_{u\in[x,z]} f([x,u])g([u,z])\right)h([z,y]). \end{align*} Reindexing this finite sum over pairs $(u,z)$ with $x\le u\le z\le y$ gives \begin{align*} ((f*g)*h)([x,y]) = \sum_{x\le u\le z\le y} f([x,u])g([u,z])h([z,y]). \end{align*} Similarly, \begin{align*} (f*(g*h))([x,y]) = \sum_{u\in[x,y]}f([x,u])\left(\sum_{z\in[u,y]} g([u,z])h([z,y])\right). \end{align*} Reindexing this finite sum over the same set of pairs $(u,z)$ with $x\le u\le z\le y$ gives \begin{align*} (f*(g*h))([x,y]) = \sum_{x\le u\le z\le y} f([x,u])g([u,z])h([z,y]). \end{align*} Hence \begin{align*} ((f*g)*h)([x,y])=(f*(g*h))([x,y]) \end{align*} for every $[x,y]\in\mathcal F$. Thus convolution is associative. Together with the pointwise $k$-module structure and the closure under convolution proved above, this shows that $R(\mathcal F,\sim;k)$ is an associative $k$-algebra under convolution. [/step]