Let $A$, $B$, $X$, and $Y$ be sets.

  1. Interaction With Inverses. Let
    \begin{align*} R & \colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}A,\\ S & \colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}X \end{align*}

    We have

    \[ \webleft (R\times S\webright )^{\dagger } = R^{\dagger }\times S^{\dagger }. \]
  2. Interaction With Composition. Let
    \begin{align*} R_{1} & \colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B,\\ S_{1} & \colon B\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}C,\\ R_{2} & \colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}Y,\\ S_{2} & \colon Y\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}Z \end{align*}

    be relations. We have

    \[ \webleft (S_{1}\mathbin {\diamond }R_{1}\webright )\times \webleft (S_{2}\mathbin {\diamond }R_{2}\webright )=\webleft (S_{1}\times S_{2}\webright )\mathbin {\diamond }\webleft (R_{1}\times R_{2}\webright ). \]

Item 1: Interaction With Inverses
Unwinding the definitions, we see that:
  1. We have $\webleft (a,x\webright )\sim _{\webleft (R\times S\webright )^{\dagger }}\webleft (b,y\webright )$ iff :
    • We have $\webleft (b,y\webright )\sim _{R\times S}\webleft (a,x\webright )$, i.e. iff :
      • We have $b\sim _{R}a$;
      • We have $y\sim _{S}x$;
  2. We have $\webleft (a,x\webright )\sim _{R^{\dagger }\times S^{\dagger }}\webleft (b,y\webright )$ iff :
    • We have $a\sim _{R^{\dagger }}b$ and $x\sim _{S^{\dagger }}y$, i.e. iff :
      • We have $b\sim _{R}a$;
      • We have $y\sim _{S}x$.

These two conditions agree, and thus the two resulting relations on $A\times X$ are equal.

Item 2: Interaction With Composition
Unwinding the definitions, we see that:
  1. We have $\webleft (a,x\webright )\sim _{\webleft (S_{1}\mathbin {\diamond }R_{1}\webright )\times \webleft (S_{2}\mathbin {\diamond }R_{2}\webright )}\webleft (c,z\webright )$ iff :
    1. We have $a\sim _{S_{1}\mathbin {\diamond }R_{1}}c$ and $x\sim _{S_{2}\mathbin {\diamond }R_{2}}z$, i.e. iff :
      1. There exists some $b\in B$ such that $a\sim _{R_{1}}b$ and $b\sim _{S_{1}}c$;
      2. There exists some $y\in Y$ such that $x\sim _{R_{2}}y$ and $y\sim _{S_{2}}z$;
  2. We have $\webleft (a,x\webright )\sim _{\webleft (S_{1}\times S_{2}\webright )\mathbin {\diamond }\webleft (R_{1}\times R_{2}\webright )}\webleft (c,z\webright )$ iff :
    1. There exists some $\webleft (b,y\webright )\in B\times Y$ such that $\webleft (a,x\webright )\sim _{R_{1}\times R_{2}}\webleft (b,y\webright )$ and $\webleft (b,y\webright )\sim _{S_{1}\times S_{2}}\webleft (c,z\webright )$, i.e. such that:
      1. We have $a\sim _{R_{1}}b$ and $x\sim _{R_{2}}y$;
      2. We have $b\sim _{S_{1}}c$ and $y\sim _{S_{2}}z$.

These two conditions agree, and thus the two resulting relations from $A\times X$ to $C\times Z$ are equal.


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