## Appendix A Proof of Theorems 1 and 2
### B.1. Proof of Theorem 3
#### b.1.1. \(H(A|Z)\geq I(U;Y)-I(U;Z)\)  
In this case, we need to prove that any rate-equivocation triple \((R_{1},R_{2},R_{e})\) satisfying the following constraints are achievable.  
\[R_{1} \leq H(B)\] \[R_{1} +R_{2}\leq I(U;Y)\] \[R_{e} \leq R_{1}+R_{2}\] \[R_{e} \leq I(U;Y)-I(U;Z)\]  
It is sufficient to show that the rate triples \((R_{1},R_{2},R_{e}=I(U;Y)-I(U;Z))\) are achievable. The coding scheme includes codebook generation, encoding and decoding. Then, we give the equivocation analysis.  
_Codebook generation and encoding_: Let \(R_{1}=H(B)-\theta_{1}\) and \(R_{1}+R_{2}=I(U;Y)-\theta_{2}\), where \(\theta_{1}\), \(\theta_{2}\) are fixed positive numbers. Since \(R_{e}\leq R_{1}+R_{2}\), it is easy to get \(\theta_{2}\leq I(U;Z)\). For each \(m_{1}\in\{1,2,...,2^{NR_{1}}\}\), an independent and identically distributed (i.i.d) codeword, \(b^{N}(m_{1})\), is generated according to \(p(b^{N})=\prod\limits_{i=1}^{N}p(b_{i})\). Then, \(2^{NR_{2}}\) action sequences \(a^{N}(m_{1},m_{2})\) are i.i.d generated for each \(b^{N}(m_{1})\) according to \(p(a^{N}(m_{1},m_{2})|b^{N}(m_{1}))=\prod\limits_{i=1}^{N}p(a_{i}|b_{i})\), where \(m_{2}\in\{1,2,...,2^{NR_{2}}\}\). For each \(a^{N}(m_{1},m_{2})\), we generate \(\|\mathcal{T}_{u}\|=2^{N(I(U;Y)-R_{1}-R_{2}-\epsilon)}\) i.i.d codewords \(u^{N}(m_{1},m_{2},t_{u})\) according to \(p(u^{N}(m_{1},m_{2},t_{u})|a^{N}(m_{1},m_{2}))=\prod\limits_{i=1}^{N}p(u_{i}|a_ {i})\). Note that \(t_{u}\) is the index of codeword \(u^{N}\). To send the message \((m_{1},m_{2})\) with the action sequence, \(a^{N}(m_{1},m_{2})\), and corresponding state sequence \(s^{N}\), the encoder randomly chooses an index \(t_{u}^{*}\in\{1,2,...,\|\mathcal{T}_{u}\|\}\). Then, the input sequence of the channel is generated by \(p(x^{N}|u^{N}(m_{1},m_{2},t_{u}^{*}),s^{N})=\prod\limits_{i=1}^{N}p(x_{i}|u_{i },s_{i})\).  
_Decoding and error probability analysis_: Decoder 1 can decode the message, \(m_{1}\), correctly, since \(R_{1}\leq H(B)\). For the receiver, he tries to find a unique sequence \(u^{N}(\hat{\hat{m}}_{1},\hat{\hat{m}}_{2},\hat{\hat{t}}_{u})\), such that \((u^{N}(\hat{\hat{m}}_{1},\hat{\hat{m}}_{2},\hat{\hat{t}}_{u}),a^{N}(\hat{\hat{ m}}_{1},\hat{\hat{m}}_{2}),y^{N})\in T^{N}_{UAY}\). It is easy to show the decoding error probabilities \(P_{e1}\leq\epsilon\) and \(P_{e2}\leq\epsilon\), and therefore, we omit the proof here. We mainly focus on the analysis of equivocation.  
_Equivocation analysis_:  
\[H(M_{1},M_{2}|Z^{N}) =H(M_{1},M_{2},Z^{N})-H(Z^{N})\] \[=H(M_{1},M_{2},Z^{N},U^{N})-H(U^{N}|M_{1},M_{2},Z^{N})-H(Z^{N})\] \[=H(M_{1},M_{2},U^{N})+H(Z^{N}|M_{1},M_{2},U^{N})-H(U^{N}|M_{1},M_ {2},Z^{N})-H(Z^{N})\] \[\geq H(U^{N})+H(Z^{N}|U^{N})-H(U^{N}|M_{1},M_{2},Z^{N})-H(Z^{N}) \tag{73}\] \[=H(U^{N})-I(U^{N};Z^{N})-H(U^{N}|M_{1},M_{2},Z^{N})\] \[\geq I(U^{N};Y^{N})-I(U^{N};Z^{N})-H(U^{N}|M_{1},M_{2},Z^{N})\] \[\geq NI(U;Y)-NI(U;Z)-H(U^{N}|M_{1},M_{2},Z^{N}) \tag{74}\]  
where Equation (73) is from the Markov chain \((M_{1},M_{2})\to U^{N}\to Z^{N}\), and Equation (74) is from that the codewords, \(u^{N}\), are i.i.d and the channels are discrete memoryless. The conditional entropy, \(H(U^{N}|M_{1},M_{2},Z^{N})\), is calculated as follows. Given the message \((M_{1},M_{2})\), the number of \(U^{N}\) is \(\|\mathcal{T}_{u}\|=2^{N(I(U;Y)-R_{1}-R_{2}-\epsilon)}=2^{N(\theta_{2}- \epsilon)}\leq 2^{N(I(U;Z)-\epsilon)}.\) Therefore, \(H(U^{N}|M_{1},M_{2},Z^{N})\to 0\) as \(N\rightarrow\infty\). Substituting this result into Equation (74) and utilizing Equation (3), we finish the proof of \(\lim\limits_{N\rightarrow\infty}\triangle\geq R_{e}\) for the model in Figure 2 with causal channel states.