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Since $G_{\mathrm{Enc}} = D \cdot \mathrm{ek}^{-1} + H_0 \cdot (-r \cdot \mathrm{ek}^{-1})$, the device proves knowledge of $(\mathrm{ek}^{-1}, -r \cdot \mathrm{ek}^{-1})$ in this relation. Pick blindings $(\mathrm{ek}^{-1})_r, (-r \cdot \mathrm{ek}^{-1})_r$:
The blinding $r$ hides $\mathrm{ek}$ in $D$. This Sigma proof binds $\mathrm{ek}^{-1}$ to the true inverse of $\mathrm{ek}$ in $D$, and the $\mathrm{ek}^{-1}$ response is reused as the bridging element in all per-leg ciphertext checks below.
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The blinding $r$ hides $\mathrm{ek}$ in $D$. The device proves two things about $D$:
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-**$D$ opening.** Knowledge of $(\mathrm{ek}, r)$ in $D = G_{\mathrm{Enc}} \cdot \mathrm{ek} + H_0 \cdot r$, reusing the account partials' $\mathrm{ek}_r$ (step 1) so the $\mathrm{ek}$ response $s_{\mathrm{ek}}$ is shared. Pick a blinding $r_r$:
The shared $s_{\mathrm{ek}}$ binds the $\mathrm{ek}$ inside $D$ to the account's $\mathrm{ek}$; without it the device could open $D$ with an arbitrary key.
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-**Inverse relation.** Since $G_{\mathrm{Enc}} = D \cdot \mathrm{ek}^{-1} + H_0 \cdot (-r \cdot \mathrm{ek}^{-1})$, knowledge of $(\mathrm{ek}^{-1}, -r \cdot \mathrm{ek}^{-1})$ in it. Pick blindings $(\mathrm{ek}^{-1})_r, (-r \cdot \mathrm{ek}^{-1})_r$:
Given $D$ opens to the account's $\mathrm{ek}$, this forces $\mathrm{ek} \cdot \mathrm{ek}^{-1} = 1$, so $\mathrm{ek}^{-1}$ is the true inverse. The $\mathrm{ek}^{-1}$ response is reused as the bridging element in all per-leg ciphertext checks below.
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3.**Per-leg amount partial** (when the asset-id is revealed in this leg, or the balance changes). Pick fresh $k_{\mathrm{amt}} \in \mathbb{Z}_p$ and a blinding $k_{\mathrm{amt},r}$, publish
@@ -295,9 +299,10 @@ with $r_{\mathrm{old},D} + r_{\mathrm{old},H} = b_0$ (so old partials sum to $\m
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5. Device hashes its transcript — containing $\mathrm{nonce}$, $P_{\mathrm{old},D}, P_{\mathrm{new},D}, D$, the per-leg partials $P_{\mathrm{ct}_{\mathrm{amt}},D}, P_{\mathrm{ct\_at},D}$, the leg encryptions, and all $T$ values above — to derive challenge $c$.
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6. Device responses:
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- $s_{\mathrm{sk}} = \mathrm{sk}_r + \mathrm{sk} \cdot c$ (shared across both account partials)
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- $s_{\mathrm{ek}} = \mathrm{ek}_r + \mathrm{ek} \cdot c$ (shared across both account partials)
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- $s_{\mathrm{ek}} = \mathrm{ek}_r + \mathrm{ek} \cdot c$ (shared across both account partials and the $D$ opening)
- $s_{\mathrm{ek}^{-1}} = (\mathrm{ek}^{-1})_r + \mathrm{ek}^{-1} \cdot c$ (shared across the D-relation, the per-leg amount partial, and the per-leg asset-id)
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- $s_r = r_r + r \cdot c$ (for the $D$ opening)
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- $s_{\mathrm{ek}^{-1}} = (\mathrm{ek}^{-1})_r + \mathrm{ek}^{-1} \cdot c$ (shared across the inverse relation, the per-leg amount partial, and the per-leg asset-id)
- Per leg with a $\mathrm{ct}_{\mathrm{amt}}$ partial: $s_{k_{\mathrm{amt}}} = k_{\mathrm{amt},r} + k_{\mathrm{amt}} \cdot c$
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- Per leg with hidden asset-id: $s_{k_{\mathrm{at}}} = k_{\mathrm{at},r} + k_{\mathrm{at}} \cdot c$
@@ -307,7 +312,8 @@ with $r_{\mathrm{old},D} + r_{\mathrm{old},H} = b_0$ (so old partials sum to $\m
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The verifier recomputes $c$ from the device's transcript and checks each Sigma relation. Each check has the standard structure $\text{(linear combination of responses)} \stackrel{?}{=} T + \text{(committed value)}.c$:
- Per leg with a $\mathrm{ct}_{\mathrm{amt}}$ partial (asset-id revealed in this leg, or balance changes): $s_{\mathrm{ek}^{-1}} \cdot \mathrm{eph\_pk}_{\mathrm{amt}} + s_{k_{\mathrm{amt}}} \cdot H_0 \stackrel{?}{=} T_{P_{\mathrm{ct}_{\mathrm{amt}},D}} + P_{\mathrm{ct}_{\mathrm{amt}},D} \cdot c$
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- Per leg with hidden asset-id: $s_{\mathrm{ek}^{-1}} \cdot \mathrm{eph\_pk}_{\mathrm{asset\_id}} + s_{k_{\mathrm{at}}} \cdot H_0 \stackrel{?}{=} T_{P_{\mathrm{ct\_at},D}} + P_{\mathrm{ct\_at},D} \cdot c$
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- Per leg with asset-id revealed elsewhere: $s_{\mathrm{ek}^{-1}} \cdot \mathrm{eph\_pk}_{\mathrm{asset\_id}} \stackrel{?}{=} T_{\mathrm{at},\mathrm{DL}} + (\mathrm{CT}_{\mathrm{at}} - \mathrm{at} \cdot H) \cdot c$
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