\begin{align*} \operatorname {\mathrm{Hom}}_{\mathbf{Rel}\webleft (X,B\webright )}\webleft (R\mathbin {\diamond }S,T\webright ) & \cong \int _{x\in X}\int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (\webleft (R\mathbin {\diamond }S\webright )^{b}_{x},T^{b}_{x}\webright )\\ & \cong \int _{x\in X}\int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (\webleft (\int ^{a\in A}R^{b}_{a}\times S^{a}_{x}\webright ),T^{b}_{x}\webright )\\ & \cong \int _{x\in X}\int _{b\in B}\int _{a\in A}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{b}_{a}\times S^{a}_{x},T^{b}_{x}\webright )\\ & \cong \int _{x\in X}\int _{b\in B}\int _{a\in A}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (S^{a}_{x},\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{b}_{a},T^{b}_{x}\webright )\webright )\\ & \cong \int _{x\in X}\int _{a\in A}\int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (S^{a}_{x},\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{b}_{a},T^{b}_{x}\webright )\webright )\\ & \cong \int _{x\in X}\int _{a\in A}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (S^{a}_{x},\int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{b}_{a},T^{b}_{x}\webright )\webright )\\ & \cong \operatorname {\mathrm{Hom}}_{\mathbf{Rel}\webleft (X,A\webright )}\webleft (S,\int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{b}_{-_{1}},T^{b}_{-_{2}}\webright )\webright )\end{align*}
naturally in each $S\in \mathbf{Rel}\webleft (X,A\webright )$ and each $T\in \mathbf{Rel}\webleft (X,B\webright )$, showing that
\[ \int _{b\in B}\mathbf{Hom}_{\{ \mathsf{t},\mathsf{f}\} }\webleft (R^{b}_{-_{1}},S^{b}_{-_{2}}\webright ) \]
is right adjoint to the postcomposition functor $R\mathbin {\diamond }-$, being thus the right Kan lift along $R$. Here we have used the following results, respectively (i.e. for each $\cong $ sign):
This finishes the proof.
