Table 2.

Shows the decomposition of the quantities in the first row induced by the measures in the first column. For consistency, the decomposition of $I(S_{1,2};T_{1,3})$ should equal both the sum of the decomposition of $I(S_{1,2};T_1)$ and $I(S_{1,2};T_3|T_1)$, and the sum of the decomposition of $I(S_{1,2};T_3)$ and $I(S_{1,2};T_1|3)$. Note that the decomposition induced by $\widetilde{UI}$, $I_{\mathrm{red}}$ and ${\mathcal{S}}_{\mathrm{VK}}$ are not consistent. In contrast, $R_{\min }$ is consistent due to Theorem 5.

	$\mathit{I}({\mathit{S}}_{1,2};{\mathit{T}}_{1,3})$	$\mathit{I}({\mathit{S}}_{1,2};{\mathit{T}}_1)$	$\mathit{I}({\mathit{S}}_{1,2};{\mathit{T}}_3|{\mathit{T}}_1)$	$\mathit{I}({\mathit{S}}_{1,2};{\mathit{T}}_3)$	$\mathit{I}({\mathit{S}}_{1,2};{\mathit{T}}_1|{\mathit{T}}_3)$
$\begin{array}{c}\widetilde{UI},I_{\mathrm{red}},\\ {\mathcal{S}}_{\mathrm{VK}}\end{array}$	$\begin{array}{cc}\hfill U(S_1\backslash S_2\to T_{1,3})& =1\hfill \\ \hfill U(S_2\backslash S_1\to T_{1,3})& =1\hfill \end{array}$	$U(S_1\backslash S_2\to T_1)=1$	$U(S_2\backslash S_1\to T_3|T_1)=1$	$C(S_1,S_2\to T_3)=1$	$R(S_1,S_2\to T_1|T_3)=1$
$R_{\min }$	$\begin{array}{cc}\hfill R(S_1,S_2\to T_{1,3})& =1\hfill \\ \hfill C(S_1,S_2\to T_{1,3})& =1\hfill \end{array}$	$\begin{array}{cc}\hfill U(S_2\backslash S_1\to T_1)& =-1\hfill \\ \hfill R(S_1,S_2\to T_1)& =1\hfill \\ \hfill C(S_1,S_2\to T_1)& =1\hfill \end{array}$	$U(S_2\backslash S_1\to T_3|T_1)=1$	$C(S_1,S_2\to T_3)=1$	$R(S_1,S_2\to T_1|T_3)=1$