We require the following well-known lemma (e.g. Dawid [15]).

**Lemma 1**. *We have* $X\phantom{\rule{thinmathspace}{0ex}}\mathrm{\perp}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{\perp}\phantom{\rule{thinmathspace}{0ex}}A\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}B$ *if and only if*
$p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}a,b)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b)$*for all* $x,a,b$ *such that* $p(a,b)>0$.

*Proof*. (of Proposition 1) To simplify notation we write ${u}^{c}:={u}^{c}(a,b)$. We have by Lemma 1
$p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}a,b,c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}a,c)$(7)for all $x,a,b,c$ with $p(a,b,c)>0$. As the main argument we show that
$p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}\tilde{b},c)$(8)for all $x,b,\tilde{b},c$ with $b,\tilde{b}\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{B}({U}_{i}^{c})$ for the same *i*.

**Step 1**, we prove eq. (8) for $b,\tilde{b}\in {Z}_{i}^{c}$. We first show that there is a path $\mathrm{\lambda}:t\mapsto (a(t),b(t))$, such that $p(a(t),b(t),c)>0$ for all $0\le t\le 1$, and $b(0)=b$ and $b(1)=\tilde{b}$. Since the interval $[0,1]$ is compact and $\mathrm{\lambda}$ is continuous, the path $\{(a(t),b(t)):0\le t\le 1\}$ is compact, too (for notational simplicity we identify the path $\mathrm{\lambda}$ with its image). Define for each point $(a(t),b(t))$ on the path an open ball with radius small enough such that all $(a,b)$ in the ball satisfy $p(a,b,c)>0$ (this is possible because $(a,b,c)\phantom{\rule{thinmathspace}{0ex}}\mapsto \phantom{\rule{thinmathspace}{0ex}}p(a,b,c)$ is assumed to be continuous). Because these balls are path-connected, they also lie in ${Z}_{i}^{c}$. They form an open cover of the path $\{(a(t),b(t)):0\le t\le 1\}$, and we can thus choose a finite subset of balls, of size *n* say, that still provides an open cover of the path. Without loss of generality let $(a(0),b(0))$ be the centre of ball 1 and $(a(1),b(1))$ be the centre of ball *n*. It suffices to show that eq. (8) holds for the centres of two neighbouring balls, say $({a}_{1},{b}_{1})$ and $({a}_{2},{b}_{2})$. Choose a point $({a}^{\ast},{b}^{\ast})$ from the non-empty intersection of those two balls. Since $d(({a}_{1},{b}_{1}),({a}^{\ast},{b}_{1}))<d(({a}_{1},{b}_{1}),({a}^{\ast},{b}^{\ast}))$ and $d(({a}_{2},{b}_{2}),({a}_{2},{b}^{\ast}))<d(({a}_{2},{b}_{2}),({a}^{\ast},{b}^{\ast}))$ for the Euclidean metric *d*, we have that $p({a}_{1},{b}_{1},c)$, $p({a}^{\ast},{b}_{1},c)$, $p({a}^{\ast},{b}^{\ast},c)$, $p({a}_{2},{b}^{\ast},c)$ and $p({a}_{2},{b}_{2},c)$ are all greater than zero. Therefore, using eq. (7) several times,
$\begin{array}{rl}p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}{b}_{1},c)& =p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}{a}_{1},c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}{a}^{\ast},c)\\ & =p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}{b}^{\ast},c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}{a}_{2},c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}{b}_{2},c)\end{array}$This shows eq. (8) for $b,\tilde{b}\in {Z}_{i}^{c}$.

**Step 2**, we prove eq. (8) for $b\in {Z}_{i}^{c}$ and $\tilde{b}\in {Z}_{i+1}^{c}$, where ${Z}_{i}^{c}$ and ${Z}_{i+1}^{c}$ are coordinate-wise connected (and thus equivalent). If ${b}^{\ast}\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{B}({Z}_{i}^{c})\cap \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{B}({Z}_{i+1}^{c})$, we know that
$p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}{b}^{\ast},c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}\tilde{b},c)$from the argument given in step 1. If ${a}^{\ast}\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({Z}_{i}^{c})\cap \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({Z}_{i+1}^{c})$, then there is a ${b}_{i},{b}_{i+1}$ such that $({a}^{\ast},{b}_{i})\in {Z}_{i}^{c}$ and $({a}^{\ast},{b}_{i+1})\in {Z}_{i+1}^{c}$. By eq. (7) and the argument from step 1 we have
$p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}{b}_{i},c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}{b}_{i+1},c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}\tilde{b},c).$

We can now combine these two steps in order to prove the original claim from eq. (

8). If

$b,\tilde{b}\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{B}({U}_{i}^{c})$ then

$b\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{B}({Z}_{1}^{c})$ and

$\tilde{b}\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{B}({Z}_{n}^{c})$, say. Further, there is a sequence

${Z}_{1}^{c},\dots ,{Z}_{n}^{c}$ with

${Z}_{k}^{c}$ and

${Z}_{k+1}^{c}$ being coordinate-wise connected for

$k=1,\dots ,n-1$. Combining steps 1 and 2 proves eq. (

8).

Consider now $x,b,c$ such that $p(b,c)>0$ (which implies $p(c)>0$) and consider ${u}^{c}=i$, say. Observe further that $p(a,c)>0$ for $a\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({U}_{i}^{c})$. We thus have
$\begin{array}{rl}p(x,{u}^{c}\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)& ={\int}_{a}p(x,a,{u}^{c}\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da={\int}_{a\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({U}_{i}^{c})}p(x,a\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da\\ & ={\int}_{a\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({U}_{i}^{c})}\frac{p(x,a,c)p(a,c)}{p(c)p(a,c)}\phantom{\rule{thickmathspace}{0ex}}da\\ & ={\int}_{a\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({U}_{i}^{c})}p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}a,c)p(a\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da\\ & ={\int}_{a\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({U}_{i}^{c}),p(a,b,c)>0}p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}a,c)p(a\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da\\ & \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+{\int}_{a\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({U}_{i}^{c}),p(a,b,c)=0}p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}a,c)p(a\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da\\ & \stackrel{(7)}{=}p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c){\int}_{a\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({U}_{i}^{c}),p(a,b,c)>0}p(a\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da+{\int}_{{\mathcal{A}}_{b}}p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}a,c)p(a\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da\\ & =:(\mathrm{\#})\end{array}$with ${\mathcal{A}}_{b}=\{a\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({U}_{i}^{c}):p(a,b,c)=0\}$. It is the case, however, that for all $a\in {\mathcal{A}}_{b}$ there is a $\tilde{b}(a)\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{B}({U}_{i}^{c})$ with $p(a,\tilde{b}(a),c)>0$. But since also $b\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{B}({U}_{i}^{c})$ we have $p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}\tilde{b},c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c)$ by eq. (8). Ergo,
$\begin{array}{rl}(\mathrm{\#})& =p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c){\int}_{a\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({U}_{i}^{c}),p(a,b,c)>0}p(a\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da+{\int}_{{\mathcal{A}}_{b}}p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}a,\tilde{b}(a),c)p(a\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da\\ & =p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c){\int}_{a\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({U}_{i}^{c}),p(a,b,c)>0}p(a\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da+p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c){\int}_{{\mathcal{A}}_{b}}p(a\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da\\ & =p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c){\int}_{a\in \mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{A}({U}_{i}^{c})}p(a\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\phantom{\rule{thickmathspace}{0ex}}da\\ & =p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c)\phantom{\rule{thickmathspace}{0ex}}p({u}^{c}\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c)\end{array}$This implies
$p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c,{u}^{c})=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}b,c)\phantom{\rule{thickmathspace}{0ex}}.$Together with eq. (7) this leads to
$p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}a,b,c,{u}^{c})=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}a,b,c)=p(x\phantom{\rule{1pt}{0ex}}|\phantom{\rule{1pt}{0ex}}c,{u}^{c})\phantom{\rule{1pt}{0ex}}.$□

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