## 7 Security Analysis
### Formal Security Analysis
Since the probability that \(A\) correctly guesses the bit \(c\) is equal to the probability that A successfully breaks the real scheme, it follows from the definition of semantic security that  
\[ADV_{A}=|2 ADV_{G_{0}}^{A}-1| \tag{1}\]  
\(G_{1}\): Within the game, adversary \(A\) carries out an eavesdropping attack through the EXECUTE query. The adversary A can obtain the messages \((M_{1},M_{2},PK_{D_{i}^{X}},E_{i},T_{1})\) and \((M_{3},M_{4},V_{t},T_{2})\) exchanged during the initial authentication phase, as well as the messages \((TID_{i},W_{i},M_{5},T_{5})\) and \((M_{6},X_{m},T_{5})\) exchanged during the handover authentication phase. Subsequently, \(A\) may also perform REVEAL and TEST queries to determine whether the returned bit \(c\) corresponds to a real session key or a random function. However, the values \(E_{i}=T_{a_{i}}(x),V_{t}=T_{a_{i}}(x),W_{i}=T_{c_{i}}(m)\) contain random parameters \((a_{i},a_{t},c_{i},c_{m})\), which \(A\) cannot obtain. Therefore, \(A\) is unable to extract any useful information. In this round of the game, \(A\)'s winning probability does not increase. Thus, \(A\)'s advantage in winning game \(G_{0}\) is equal to that in \(G_{1}\):  
\[ADV^{A}_{G_{1}}=ADV^{A}_{G_{0}} \tag{2}\]  
\(G_{2}\): In this game, the adversary \(A\) simulates an active attack by performing hash and SEND queries, attempting to forge messages that deceive the protocol participants into believing they are legitimate authentication messages. The game terminates only if \(A\) successfully constructs a valid message. During the initial authentication phase, [PERSON] attempts to forge the message \(\{M_{1},M_{2},PK_{v_{i}},E_{i},T_{1}\}\), where \(M_{1}=PDID^{X}_{i}\oplus h(EID^{X}_{i}||B_{it})\), \(M_{2}=h(PDID^{X}_{i}||B_{it}||E_{i}||T_{1})\). To obtain the parameters \(PDID^{X}_{i}\) and \(B_{it}\), \(A\) must solve the CMDLP problem to compute \(B_{it}\) or, alternatively, find a hash collision to forge the message. Similarly, in the cross-domain handover authentication phase, [PERSON] also needs to solve the CMDLP problem or find a hash collision. If hash collisions and CMDLP-solving capabilities are excluded in game \(G_{2}\), [PERSON]'s advantage in winning \(G_{2}\) is the same as in \(G_{1}\). Based on the birthday paradox, the likelihood of a hash collision can be estimated as \(\frac{q_{h}^{2}}{|Hash|}\). Therefore, \(A\)'s advantage in winning game \(G_{2}\) is given by  
\[|ADV^{A}_{G_{2}}-ADV^{A}_{G_{1}}|\leq\frac{q_{h}^{2}}{2|Hash|}+ADV^{CMDLP}_{A} \tag{3}\]  
At this point, [PERSON] has completed all queries. In order to win game \(G_{2}\), [PERSON] performs the TEST query and then randomly guesses the bit \(c\). Therefore,  
\[ADV^{A}_{G_{2}}=1/2 \tag{4}\]  
By using Equations (1)-(4), the following equation can be derived:  
\[\frac{1}{2}ADV_{A}=|ADV^{A}_{G_{1}}-1/2|=|ADV^{A}_{G_{2}}-ADV^{A}_{G_{1}}|\leq \frac{q_{h}^{2}}{2|Hash|}+ADV^{CMDLP}_{A} \tag{5}\]  
By multiplying both sides of Equation (5) by 2, we obtain  
\[ADV_{A}\leq\frac{q_{h}^{2}}{|Hash|}+ADV^{CMDLP}_{A} \tag{6}\]  
Theorem 1 is thus proved, and the advantage of adversary \(A\) is negligible. Therefore, the session key in the proposed protocol is semantically secure under the ROR model.