## Security Analysis
### Formal Security Proof
: The games \(Game_{i}(0\leq i\leq 4)\) are defined to simulate the attacks initiated by \(\mathcal{A}\). The variable \(Win_{i}(0\leq i\leq 4)\) represents \(\mathcal{A}\)'s success in guessing the random bit \(r\) in the \(Game_{i}\). The games are defined as follows:  
* \(Game_{0}\): In this game, we simulate the actual attacks initiated by \(\mathcal{A}\). Based on the given definition, we obtain the following results: \[\text{Adv}^{A}_{P}=\left|2\text{Pr}[\text{Win}_{0}]-1\right|\] (1)
* \(Game_{1}\): In this game, we simulate eavesdropping attacks where \(\mathcal{A}\) has access to all the messages transmitted publicly. Then, \(\mathcal{A}\) obtains all the messages transmitted between \(IN^{i}_{VUI}\), \(IN^{i}_{RSU}\), and \(IN^{i}_{TA}\) by executing \(Execute\left(IN^{i}_{VUI},IN^{i}_{RSU},IN^{i}_{TA}\right)\). Then, \(\mathcal{A}\) executes \(Test\left(IN^{i}_{VUI},IN^{i}_{RSU},IN^{i}_{TA},r\right)\) and attempts to guess whether the returned value corresponds to the session key. However, due to the presence of random numbers and ECDLP, the attacker is unable to determine the correlation between the captured messages and the session keys. Thus, we obtain the following: \[\text{Pr}[\text{Win}_{0}]=\text{Pr}[\text{Win}_{1}]\] (2)
* \(Game_{2}\): In this game, we simulate collision attacks on the transcripts and hash results. According to the definition of the birthday paradox, the probability of a hash collision is less than \(\frac{q_{\text{H}}^{2}}{2^{\prime}\mu^{+}}\), and the collision probability of other transcripts is less than \(\frac{(q_{\text{S}}+q_{\text{E}})^{2}}{2n}\). Thus, we have: \[\text{Pr}[\text{Win}_{2}]-\text{Pr}[\text{Win}_{1}]\leq\frac{q_{\text{H}}^{2 }}{2^{\prime}\mu^{+}}+\frac{\left(q_{\text{S}}+q_{\text{E}}\right)^{2}}{2n}\] (3)
* \(Game_{3}\): This game simulates that \(\mathcal{A}\) executes \(Corrupt\left(IN^{i}_{VUI}\right)\) and \(CorruptRSU\left(IN^{i}_{RSU}\right)\) to gain the stored information \(<G,p,P,e_{v},C_{v},PID_{v},h(\cdot),PA_{v}>\) in the SCand \(<C_{v},sk_{r},ID_{v},PD_{r},t_{v}>\) in RSU, where \(C_{v}=h(ID_{v},PW_{v},BSK_{v})mod\ n\), \(PPID_{v}=PID_{v}\bigoplus h(PW_{v},ID_{v},BSK_{v})\), \(PA_{v}=A_{v}\bigoplus h(PW_{v},BSK_{v},ID_{v})\), where \(n\in(2^{4},2^{8})\) and \(BSK_{v}\) is the biometric key. In addition, \(sk_{r}=x_{r}\oplus h(R_{r}\|t_{r})\), \(PD_{r}=PSID_{r}\oplus h(t_{r}\|R_{r})\), and \(R_{r}=PUF(C_{r})\). If \(\mathcal{A}\) wants to gain these parameters, they must guess the biological key \(BSK_{v}\) or break PUF. Let us assume that the probability of A successfully breaking the PUF is denoted as \(Adv^{A}_{PUF}\). Thus, we have: \[\text{Pr}[\text{Win}_{3}]-\text{Pr}[\text{Win}_{2}]\leq q_{\text{S}}\left( \frac{1}{2^{\prime}\mu_{\text{bio}}}+\text{Adv}^{A}_{PUF}\right)\] (4)\(Game_{4}\): \(\mathcal{A}\) can obtain \(R_{0}=r\cdot P\), \(R_{2}=s\cdot P\) through the public channel, which are utilized for session key agreements based on the ECDLP. The simulation in the game assumes that \(\mathcal{A}\) calculates the session keys based on the transcripts. We have: \[\Pr[\text{Win}_{4}]-\Pr[\text{Win}_{3}]\leq\text{Adv}_{\text{ ECDLP}}^{\text{A}}\] (5) Since the session keys are generated independently and randomly, the advantage of guessing the value \(r\) is equivalent to guessing the session key itself.