Proof of Theorem MMA

A matrix equality, so we will go entry-by-entry, no surprise there. For $1\leq i\leq m$, $1\leq j\leq s$, \begin{align*} \matrixentry{A(BD)}{ij} &=\sum_{k=1}^{n}\matrixentry{A}{ik}\matrixentry{BD}{kj} &&\knowl{./knowls/theorem.EMP.knowl}{\text{Theorem EMP}}\\ &=\sum_{k=1}^{n}\matrixentry{A}{ik}\left(\sum_{\ell=1}^{p}\matrixentry{B}{k\ell}\matrixentry{D}{\ell j}\right) &&\knowl{./knowls/theorem.EMP.knowl}{\text{Theorem EMP}}\\ &=\sum_{k=1}^{n}\sum_{\ell=1}^{p}\matrixentry{A}{ik}\matrixentry{B}{k\ell}\matrixentry{D}{\ell j} &&\knowl{./knowls/property.DCN.knowl}{\text{Property DCN}}\\ \end{align*} We can switch the order of the summation since these are finite sums, \begin{align*} &=\sum_{\ell=1}^{p}\sum_{k=1}^{n}\matrixentry{A}{ik}\matrixentry{B}{k\ell}\matrixentry{D}{\ell j} &&\knowl{./knowls/property.CACN.knowl}{\text{Property CACN}}\\ \end{align*} As $\matrixentry{D}{\ell j}$ does not depend on the index $k$, we can use distributivity to move it outside of the inner sum, \begin{align*} &=\sum_{\ell=1}^{p}\matrixentry{D}{\ell j}\left(\sum_{k=1}^{n}\matrixentry{A}{ik}\matrixentry{B}{k\ell}\right) &&\knowl{./knowls/property.DCN.knowl}{\text{Property DCN}}\\ &=\sum_{\ell=1}^{p}\matrixentry{D}{\ell j}\matrixentry{AB}{i\ell} &&\knowl{./knowls/theorem.EMP.knowl}{\text{Theorem EMP}}\\ &=\sum_{\ell=1}^{p}\matrixentry{AB}{i\ell}\matrixentry{D}{\ell j} &&\knowl{./knowls/property.CMCN.knowl}{\text{Property CMCN}}\\ &=\matrixentry{(AB)D}{ij}&&\knowl{./knowls/theorem.EMP.knowl}{\text{Theorem EMP}} \end{align*}

So the matrices $(AB)D$ and $A(BD)$ are equal, entry-by-entry, and by the definition of matrix equality (Definition ME) we can say they are equal matrices.