## 4 Syntax about LR-SP-IBBE
### Security Descriptions for LR-SP-IBBE  
The security for the LR-SP-IBBE scheme could be achieved through an interactive game that was executed by one challenger \(\mathcal{B}\) and one adversary \(\mathcal{A}\). This scheme obtained the security against the chosen ciphertext attack. The security for LR-SP-IBBE was described through the game \(\mathrm{GM}_{\mathbb{R}}\). In \(\mathrm{GM}_{\mathbb{R}}\), the challenger maintained a table, \(\mathcal{L}=\{(\mathcal{H},\mathcal{I},\mathcal{K},\mathcal{SK},\mathcal{LK})\}\), where \(\mathcal{H}\), \(\mathcal{I}\), \(\mathcal{K}\), \(\mathcal{SK}\), and \(\mathcal{LK}\) are the handles' space, identities' space, symmetric keys' space, private keys' space, and the leakage amount's space, respectively. Assume \(\mathcal{H}=\mathbb{N}\) and \(\mathcal{LK}=\mathbb{N}\). \(\mathrm{GM}_{\mathbb{R}}\):  
**Initialization:** By running the initialization algorithm, the challenger produced the public parameters \(PP\) and the master private key \(MK\). \(\mathcal{B}\) gives \(PP\) to \(\mathcal{A}\), then \(\mathcal{B}\) keeps \(MK\) in secret. Stage 1: the adversary conducted the following inquires. \(\mathcal{O}-Create(ID)\). Given one identity \(ID\), \(\mathcal{B}\) searched for the item corresponding to \(ID\) in the list \(\mathcal{L}\). In the event that the corresponding item was found in the list \(\mathcal{L}\), the challenger ended the operation. Otherwise, \(\mathcal{B}\) obtained one private key \(SK_{ID}\) by running the private key generation algorithm and updated \(h\gets h+1\). The challenger put \((h,ID,K,SK_{ID},0)\) in \(\mathcal{L}\). \(\mathcal{O}-Leak(ID)\). Given the identity \(ID_{i}\), \(\mathcal{A}\) chose one arbitrary leak function \(f(\cdot)\). \(f(\cdot)\) input the symmetric key. \(\mathcal{B}\) sent the outcome of \(f(\cdot)\) to the adversary \(\mathcal{A}\). The limitation was that the output could not extend a bound. The information obtained from the output of this leakage function was related to the one encapsulated symmetric key \(K\). Specifically, \(\mathcal{B}\) sought an item for the handle \(h\) in the table \(\mathcal{L}\) provided that \((h,ID,K,SK_{ID},L)\) was found. \(\mathcal{B}\) determined whether \(L+|f(K)|\leq L_{k}\) or not, where \(L_{k}\) is the maximum value that allowed private key disclosure. If \(L+|f(K)|\leq L_{k}\), the challenger sent \(f(K)\) to \(\mathcal{A}\) and updated \((h,ID,K,SK_{ID},L)\) with \((h,ID,K,SK_{ID},L+|f(K)|)\). Otherwise, the challenger would output \(\bot\). \(\mathcal{O}-Reveal(h)\). \(\mathcal{A}\) inquired about one private key of the handle \(h\). \(\mathcal{B}\) looked for an item corresponding to the handle \(h\) in the table \(\mathcal{L}\). In the event that \((h,ID,K,SK_{ID},L)\) was found, \(\mathcal{B}\) gave \(SK_{ID}\) to the adversary. \(\mathcal{O}-KeyU\). \(\mathcal{A}\) inquired about the updating of the private key corresponding to the handle \(h\). \(\mathcal{B}\) looked for one item corresponding to the handle \(h\) in \(\mathcal{L}\). Provided that \((h,ID,K,SK_{ID},L)\) was found, \(\mathcal{B}\) ran the updating algorithm to obtain the new private key \(SK_{ID}\). The challenger then gave \(\mathcal{A}\) the new private key \(SK_{ID}\) and updated \((h,ID,K,SK_{ID},0)\) to \((h,ID,K,SK_{ID},L)\). \(\mathcal{O}-D1\). The adversary inquired about the plaintext of \((ID,CT)\), and the challenger looked up the list \(\mathcal{L}\) and found the private key \(SK_{ID}\). \(\mathcal{B}\) invoked the decryption algorithm \(\mathbf{D1}(PP,SK_{ID,k1},S,CT)\to CT^{\prime}\). If \(ID_{i}\in S\), the challenger calculated the part of plaintext \(CT^{\prime}\) and sent it to \(\mathcal{A}\). \(\mathcal{O}-D2\). \(\mathcal{A}\) inquired about the plaintext of \((ID,CT)\), and the challenger looked up the list \(\mathcal{L}\) and found the private key \(SK_{ID}\). \(\mathcal{B}\) invoked the decryption algorithm \(\mathbf{D2}(PP,SK_{ID,k2},S,\)\(CT^{\prime})\to M\). First, it divided \(CT\) into \((CM,Hd)\). If \(ID_{i}\in S\), the challenger used \(Hd\) to calculate the symmetric key \(K\).