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@ -1532,7 +1532,7 @@ Hence, the difference $\mathbf{h} = \mathbf{z}' - \mathbf{z}_{i^\star} \in \ZZ^{
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\begin{description}
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\begin{description}
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\item[$\textsf{Game}^{(d)}$~0:] This is the real anonymity experiment $\Expt^\textrm{anon$-d$}_\adv(\lambda)$ as described in Definition~\ref{def:anon}.
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\item[$\textsf{Game}^{(d)}$~0:] This is the real anonymity experiment $\Expt^\textrm{anon$-d$}_\adv(\lambda)$ as described in Definition~\ref{def:anon}.
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More precisely, the challenger starts by running the algorithm $\mathsf{Setup}(1^\lambda, 1^{\Ngs})$ to obtain $(\gspk, \mathcal{S}_\GM = \mathbf{T_A} \in \ZZ^{m \times m}, \mathcal{S}_\OA = \mathbf{T_B} \in \ZZ^{m \times m})$ along with state information $St$. The challenger next hands the public parameters $\gspk$ and the group manager key $\mathcal{S}_\GM$ to the adversary $\adv$.
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More precisely, the challenger starts by running the~$\mathsf{Setup}(1^\lambda, 1^{\Ngs})$ algorithm to obtain $(\gspk, \mathcal{S}_\GM = \mathbf{T_A} \in \ZZ^{m \times m}, \mathcal{S}_\OA = \mathbf{T_B} \in \ZZ^{m \times m})$ along with state information $St$. The challenger next hands the public parameters $\gspk$ and the group manager key $\mathcal{S}_\GM$ to the adversary $\adv$.
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On the following adversary signature opening queries on signatures $\Sigma = (\vk, \mathbf{c}_{\mathbf{v}_d}, \pi_K, sig)$, the challenger uses the opening authority key $\mathbf{T_A} \in \ZZ^{m \times m}$ he possesses to decrypt the GPV encryption of the signer identity $\mathbf{c}_{\mathbf{v}_d} \in \Zq^m \times \Zq^{2m}$.
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On the following adversary signature opening queries on signatures $\Sigma = (\vk, \mathbf{c}_{\mathbf{v}_d}, \pi_K, sig)$, the challenger uses the opening authority key $\mathbf{T_A} \in \ZZ^{m \times m}$ he possesses to decrypt the GPV encryption of the signer identity $\mathbf{c}_{\mathbf{v}_d} \in \Zq^m \times \Zq^{2m}$.
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At some point, the adversary $\adv$ requests a challenge by outputting a target message $M^\star \in \bit^*$ and two user key pairs
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At some point, the adversary $\adv$ requests a challenge by outputting a target message $M^\star \in \bit^*$ and two user key pairs
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\[ \bigl(\scr_i^\star = \mathbf{z}^\star_i \in \ZZ^{4m}, \crt_i^\star \in (\mathsf{id}^\star_i, \mathbf{d}^\star_i, \mathbf{s}^\star_i) \in \bit^\ell \times \ZZ^{2m} \times \ZZ^{2m} \bigr)_{i \in \bit} \]
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\[ \bigl(\scr_i^\star = \mathbf{z}^\star_i \in \ZZ^{4m}, \crt_i^\star \in (\mathsf{id}^\star_i, \mathbf{d}^\star_i, \mathbf{s}^\star_i) \in \bit^\ell \times \ZZ^{2m} \times \ZZ^{2m} \bigr)_{i \in \bit} \]
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@ -1776,7 +1776,8 @@ By inspection, it can be seen that the properties in~(\ref{eq:zk-equivalence}) a
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We now describe how to derive the protocol for proving the possession of a signature on a committed value, that is used in Section~\ref{commit-sig}.
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We now describe how to derive the protocol for proving the possession of a signature on a committed value, that is used in Section~\ref{commit-sig}.
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\begin{description}
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\begin{description}
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\item[Common Input:] Matrices $\mathbf{A}, \{\mathbf{A}_j\}_{j=0}^\ell, \mathbf{D} \in \ZZ_q^{n \times m}$; $\{\mathbf{D}_k\in \ZZ_q^{2n \times 2m}\}_{k=0}^N$;$\mathbf{B}\in \ZZ_q^{n \times m}$; $\mathbf{G}_1 \in \mathbb{Z}_q^{n \times 2m}$;
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\item[Common Input:] Matrices $\mathbf{A}, \{\mathbf{A}_j\}_{j=0}^\ell, \mathbf{D} \in \ZZ_q^{n \times m}$; $\{\mathbf{D}_k\in \ZZ_q^{2n \times 2m}\}_{k=0}^N$;$\mathbf{B}\in \ZZ_q^{n \times m}$; $\mathbf{G}_1 \in \mathbb{Z}_q^{n \times 2m}$;
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$\mathbf{G}_0 \in \mathbb{Z}_q^{n \times \ell}$; vectors
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$\mathbf{G}_0 \in \mathbb{Z}_q^{n \times \ell}$;\\
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vectors
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$ \{\mathbf{c}_{k,1}\}_{k=1}^N, \mathbf{c}_{\tau,1}, \mathbf{c}_{\mathbf{v}, 1}, \mathbf{c}_{s, 1} \in \ZZ_q^m$; $\{\mathbf{c}_{k,2}\}_{k=1}^N,\mathbf{c}_{\mathbf{v}, 2}, \mathbf{c}_{s,2} \in \ZZ_q^{2m}$; $\mathbf{c}_{\tau,2} \in \ZZ_q^\ell$; $\mathbf{u} \in \mathbb{Z}_q^n$.
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$ \{\mathbf{c}_{k,1}\}_{k=1}^N, \mathbf{c}_{\tau,1}, \mathbf{c}_{\mathbf{v}, 1}, \mathbf{c}_{s, 1} \in \ZZ_q^m$; $\{\mathbf{c}_{k,2}\}_{k=1}^N,\mathbf{c}_{\mathbf{v}, 2}, \mathbf{c}_{s,2} \in \ZZ_q^{2m}$; $\mathbf{c}_{\tau,2} \in \ZZ_q^\ell$; $\mathbf{u} \in \mathbb{Z}_q^n$.
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\smallskip
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\smallskip
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