## Proof that the EQI for $$c^n$$ is the sum of EQI for the individual $$c_k$$

Theorem: The EQI Decomposition Theorem:

When the Independent Evidence Conditions are satisfied,

$\EQI[c^n \pmid h_i /h_j \pmid b] = \sum^{n}_{k = 1} \EQI[c_k \pmid h_i /h_j \pmid b].$

Proof:

\begin{align} \EQI & [c^n \pmid h_i /h_j \pmid b]\\ & = \sum_{\{e^n\}} \QI[e^n \pmid h_i /h_j \pmid b\cdot c^n] \times P[e^n \pmid h_{i}\cdot b\cdot c^{n}] \\ & = \sum_{\{e^n\}} \log\left[ \frac{P[e^n \pmid h_{i}\cdot b\cdot c^n]}{P[e^n \pmid h_j\cdot b\cdot c^n]}\right] \times P[e^n \pmid h_i\cdot b\cdot c^{n}] \end{align} \begin{align} \phantom{\EQI} & = \sum_{\{e^{n-1}\} } \sum_{\{e_n\}}\left(\begin{aligned} \log\left[\frac{P[e_n \pmid h_{i}\cdot b\cdot c_n\cdot(c^{n-1}\cdot e^{n-1})]}{P[e_n \pmid h_j\cdot b\cdot c_n\cdot(c^{n-1}\cdot e^{n-1})}\right] \\ +\log\left[\frac{P[e^{n-1} \pmid h_{i}\cdot b\cdot c_n\cdot c^{n-1}]}{P[e^{n-1} \pmid h_j\cdot b\cdot c_n\cdot c^{n-1}]}\right] \end{aligned} \right) \\ &\qquad \times P[e_n \pmid h_{i}\cdot b\cdot c_n\cdot(c^{n-1}\cdot e^{n-1})] \\ &\qquad \times P[e^{n-1} \pmid h_{i}\cdot b\cdot c_n\cdot c^{n-1}] \\[1ex] & = \sum_{\{e^{n-1}\} } \sum_{\{e_n\}} \left(\begin{aligned} & \log\left[\frac{P[e_n \pmid h_{i}\cdot b\cdot c_n]}{P[e_n \pmid h_j\cdot b\cdot c_{n}]}\right]\\ & +\log\left[\frac{P[e^{n-1} \pmid h_{i}\cdot b\cdot c^{n-1}]}{P[e^{n-1} \pmid h_j\cdot b\cdot c^{n-1}]}\right] \end{aligned}\right) \\ &\qquad \times P[e_n \pmid h_{i}\cdot b\cdot c_{n}] \\ & \qquad \times P[e^{n-1} \pmid h_{i}\cdot b\cdot c^{n-1}] \\[1ex] & = \left(\begin{aligned} &\sum_{\{e_{ n}\}} \log\left[\frac{P[e_n \pmid h_{i}\cdot b\cdot c_n]}{P[e_n \pmid h_j\cdot b\cdot c_{n}]}\right] \\ &\times P[e_n \pmid h_{i}\cdot b\cdot c_{n}] \\ & \times \sum_{\{e^{n-1}\} } P[e^{n-1} \pmid h_{i}\cdot b\cdot c^{n-1}] \end{aligned} \right)\\ &\qquad+\left(\begin{aligned} &\sum_{\{e^{n-1}\}} \log\left[\frac{P[e^{n-1} \pmid h_{i}\cdot b\cdot c^{n-1}]}{P[e^{n-1} \pmid h_j\cdot b\cdot c^{n-1}]}\right]\\ & \times P[e^{n-1} \pmid h_{i}\cdot b\cdot c^{n-1}] \\ & \times \sum_{\{e_{ n}\} } P[e_n \pmid h_{i}\cdot b\cdot c_{n}] \end{aligned} \right) \\[3ex] & = \EQI[c_n \pmid h_i /h_j \pmid b] +\EQI[c^{n-1} \pmid h_i /h_j \pmid b] \\ & = \ldots \textrm{ (iterating this decomposition process)} \\ & = \sum^{n}_{k = 1} \EQI[c_k \pmid h_i /h_j \pmid b]. \end{align}