## Appendix A Proof of Theorems 1 and 2
### Proof of Theorem 1
#### a.1.1 \(H(A|Z)\geq I(U;Y)-\max\{I(U;Z),I(U;S|A)\}\)  
In this case, we need to prove that any rate-equivocation triple \((R_{1},R_{2},R_{e})\) satisfying the following constraints are achievable. \[R_{1} \leq H(B)\] \[R_{1}+R_{2}\leq I(U;Y)-I(U;S|A)\] \[R_{e} \leq R_{1}+R_{2}\] \[R_{e} \leq I(U;Y)-\max\{I(U;Z),I(U;S|A)\}\]  
It is sufficient to show that the rate triples \((R_{1},R_{2},R_{e}=I(U;Y)-\max\{I(U;Z),I(U;S|A)\})\) are achievable. The coding scheme includes codebook generation, encoding and decoding. Then we give the equivocation analysis. _Codebook generation and encoding_: Let \(R_{1}=H(B)-\tau_{1}\) and \(R_{1}+R_{2}=I(U;Y)-I(U;S|A)-\tau_{2}\), where \(\tau_{1}\), \(\tau_{2}\) are fixed positive numbers. Since \(R_{e}\leq R_{1}+R_{2}\), it is easy to get \(\max\{I(U;S|A),I(U;Z)\}-I(U;S|A)\). 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 \(\|\mathcal{T}\|=2^{N(I(U;Y)-R_{1}-R_{2}-\epsilon)}\) i.i.d codewords \(u^{N}(m_{1},m_{2},t_{b},t_{u})\) according to \(p(u^{N}(m_{1},m_{2},t_{b},t_{u})|a^{N}(m_{1},m_{2}))=\prod\limits_{i=1}^{N}p( u_{i}|a_{i})\). These codewords are put into \(\|\mathcal{T}_{b}\|=2^{N(\max\{I(U;S|A),I(U;Z)\}-I(U;Z)+\epsilon^{\prime})}\) bins, such that each bin contains \(\|\mathcal{T}\|/\|\mathcal{T}_{b}\|\) codewords. Note that \(t_{b}\), \(t_{u}\) are the indexes of the bin and codeword, respectively. The codebook structure is shown in Figure 11. 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 chooses a \(u^{N}(m_{1},m_{2},t_{b},t_{u})\) from the \(\|\mathcal{T}\|\) sequences, such that \((u^{N}(m_{1},m_{2},t_{b},t_{u}),a^{N}(m_{1},m_{2}),s^{N})\in T^{N}_{US|A}\). If no such sequence exists, it picks \((t_{b},t_{u})=(1,1)\). Then, the input sequence of the channel is generated by \(p(x^{N}|u^{N},s^{N})=\prod\limits_{i=1}^{N}p(x_{i}|u_{i},s_{i})\). _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 \(u^{N}(\hat{\hat{m}}_{1},\hat{\hat{m}}_{2},\hat{\hat{t}}_{b},\hat{\hat{t}}_{u})\), such that \((u^{N}(\hat{\hat{m}}_{1},\hat{\hat{m}}_{2},\hat{\hat{t}}_{b},\hat{\hat{t}}_{u} ),a^{N}(\hat{\hat{m}}_{1},\hat{\hat{m}}_{2}),y^{N})\in T^{N}_{UAY}\). It is easy to show the decoding error probabilities \(P_{e1}\leq\epsilon\) and \(P_{e2}\leq\epsilon\) by similar arguments in [11,12]. We mainly focus on the analysis of equivocation. _Equivocation analysis_:  
\[H(M_{1},M_{2}|Z^{N}) =H(M_{1},M_{2},Z^{N})-H(Z^{N})\] \[=H(M_{1},M_{2},Z^{N},U^{N})-H(U^{N}|M_{1},M_{2},Z^{N})-H(Z^{N})\] \[=H(M_{1},M_{2},U^{N})+H(Z^{N}|M_{1},M_{2},U^{N})-H(U^{N}|M_{1},M_{2 },Z^{N})-H(Z^{N})\] \[\geq H(U^{N})+H(Z^{N}|U^{N})-H(U^{N}|M_{1},M_{2},Z^{N})-H(Z^{N}) \tag{51}\] \[\geq H(U^{N})-I(U^{N};Z^{N})-H(T_{b},U^{N}|M_{1},M_{2},Z^{N})\] \[R_{1}+R_{2}\leq I(U;Y)-I(U;S|A)\]0 \[R_{1}+R_{2}\leq I(U;Y)-I(U;S|A)\]1 \[R_{1}+R_{2}\leq I(U;Y)-I(U;S|A)\]2  
where Equation (51) is from the Markov chain \((M_{1},M_{2})\to U^{N}\to Z^{N}\), and Equation (52) is from that the codewords, \(u^{N}\), are i.i.d and the channels are discrete memoryless. Next, we bound \(H(T_{b})\) and \(H(U^{N}|M_{1},M_{2},Z^{N},T_{b})\). Since \(\|\mathcal{T}_{b}\|=2^{N(\max\{I(U;S|A),I(U;Z)\}-I(U;Z)+\epsilon^{\prime})}\), we have \(H(T_{b})\leq log\|\mathcal{T}_{b}\|=N(\max\{I(U;S|A),I(U;Z)\}-I(U;Z)+\epsilon^ {\prime})\). The explanation for bounding \(\frac{1}{N}H(U^{N}|M_{1},M_{2},Z^{N},T_{b})\) is presented as follows.