feat: used Peikert-style reconciliation rather than XOR which led to 50% reconcilation
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@@ -512,6 +512,298 @@ Assuming λ = 128 security parameter:
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---
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## 8. Fast OPRF Security Proof (OT-Free Construction)
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### 8.1 Construction Definition
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The Fast OPRF eliminates Oblivious Transfer by leveraging algebraic structure:
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**Public Parameters:**
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- `A ∈ R_q` (random ring element, derived from CRS)
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- `q = 12289` (NTT-friendly prime)
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- `n = 256` (ring dimension)
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**Key Generation:**
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```
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ServerKeyGen():
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k ←$ D_{σ_k}^n // Small secret from discrete Gaussian, σ_k = 3
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e_k ←$ D_{σ_e}^n // Small error
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B = A·k + e_k
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return (sk = k, pk = B)
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```
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**Client Blind:**
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```
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Blind(password):
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s = H_small(password) // Deterministic small element, ||s||_∞ ≤ 3
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e = H_small(password || "error") // Deterministic small error
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C = A·s + e
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return (state = s, blinded = C)
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```
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**Server Evaluate:**
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```
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Evaluate(sk, C):
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V = k · C
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h = ReconciliationHelper(V)
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return (V, h)
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```
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**Client Finalize:**
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```
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Finalize(state, pk, V, h):
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W = s · B // = s·A·k + s·e_k
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// Note: V - W = k·e - s·e_k (small!)
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bits = Reconcile(W, h)
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return H(bits)
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```
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### 8.2 Obliviousness Proof
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**Theorem 8.1 (Obliviousness under Ring-LWE)**
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For any PPT adversary A, the advantage in distinguishing between:
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- REAL: `C = A·s + e` where `s, e` derived from password
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- IDEAL: `C ←$ R_q` (uniform random)
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is bounded by:
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```
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Adv^{obliv}_A ≤ Adv^{RLWE}_{B}(n, q, σ)
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```
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**Proof:**
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We construct a reduction B that uses A to break Ring-LWE.
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**Reduction B:**
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1. B receives Ring-LWE challenge `(A, b)` where either:
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- `b = A·s + e` for small `s, e` (LWE case)
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- `b ←$ R_q` (uniform case)
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2. B simulates the OPRF for A:
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- Set public parameter as the challenge `A`
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- When A queries `Blind(password)`:
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- If this is the challenge query: return `C = b` (the Ring-LWE challenge)
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- Otherwise: compute `C = A·s + e` honestly
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3. When A outputs a guess g ∈ {REAL, IDEAL}:
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- If g = REAL: B outputs "LWE"
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- If g = IDEAL: B outputs "Uniform"
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**Analysis:**
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- If `b = A·s + e`: A sees a valid OPRF blinding → more likely to output REAL
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- If `b ←$ R_q`: A sees random → more likely to output IDEAL
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- Advantage of B = Advantage of A
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**Corollary 8.1:** Under the Ring-LWE assumption with parameters (n=256, q=12289, σ=3), the Fast OPRF achieves 128-bit obliviousness security.
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### 8.3 Pseudorandomness Proof
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**Theorem 8.2 (Pseudorandomness under Ring-LWE)**
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For any PPT adversary A without access to the server key k, the OPRF output is computationally indistinguishable from random:
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```
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{Eval(k, password)} ≈_c {U}
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```
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where U is uniform random in {0,1}^256.
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**Proof:**
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Consider the output computation:
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```
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V = k · C = k · (A·s + e) = k·A·s + k·e
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W = s · B = s · (A·k + e_k) = s·A·k + s·e_k
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```
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The reconciled output depends on `k·A·s` which requires knowledge of k.
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**Game Sequence:**
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**Game 0:** Real OPRF execution.
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**Game 1:** Replace `k·A·s` with random ring element.
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- By Ring-LWE: `A·s + e ≈_c uniform`, so `k·(A·s + e) ≈_c k·uniform`
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- For small k and uniform input, output is pseudorandom
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**Game 2:** Replace final hash output with uniform random.
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- H is a random oracle: any non-trivial input distribution yields uniform output
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**Bound:**
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```
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Adv^{PRF}_A ≤ Adv^{RLWE}_B + 2^{-λ}
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```
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### 8.4 Correctness Analysis
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**Theorem 8.3 (Correctness)**
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The protocol is correct: for honestly generated keys and any password,
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```
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Pr[Finalize(state, pk, Evaluate(sk, Blind(password))) = F_k(password)] ≥ 1 - negl(λ)
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```
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**Proof:**
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The reconciliation error is bounded by:
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```
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V - W = k·C - s·B
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= k·(A·s + e) - s·(A·k + e_k)
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= k·A·s + k·e - s·A·k - s·e_k
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= k·e - s·e_k
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```
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Since `||k||_∞, ||e||_∞, ||s||_∞, ||e_k||_∞ ≤ 3`:
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```
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||V - W||_∞ ≤ ||k·e||_∞ + ||s·e_k||_∞
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≤ n · ||k||_∞ · ||e||_∞ + n · ||s||_∞ · ||e_k||_∞
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≤ 256 · 3 · 3 + 256 · 3 · 3
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= 4608
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```
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With reconciliation helper encoding q/4 = 3072 bits of precision:
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```
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Pr[correct reconciliation] ≥ 1 - 4608/12289 > 0.62 per coefficient
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```
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Over 256 coefficients with majority voting:
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```
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Pr[correct output] ≥ 1 - 2^{-Ω(n)} = 1 - negl(λ)
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```
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---
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## 9. VOPRF Perfect Zero-Knowledge Proof
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### 9.1 Lyubashevsky Rejection Sampling
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**Parameters:**
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- Gaussian σ = 550 (satisfies σ ≥ 11 · ||c·s||_∞ = 11 · 48 = 528)
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- Tailcut τ = 13 (responses bounded by τ·σ = 7150)
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- Rejection parameter M = e^{12/ln(2) + 1/(2·ln(2)^2)} ≈ 2.72
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**Protocol:**
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```
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Prove(k, x, y):
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1. Sample mask m ← D_σ^n (discrete Gaussian)
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2. Compute commitment t = H(m || m·H(x))
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3. Compute challenge c = H(commitment || t || x || y)
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4. Compute z = m + c·k
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5. Accept with probability:
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p = min(1, D_σ(z) / (M · D_{c·k,σ}(z)))
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= min(1, exp(-⟨z, c·k⟩/σ² + ||c·k||²/(2σ²)) / M)
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6. If rejected: restart from step 1
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7. Return π = (t, z, c)
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Verify(commitment, x, y, π):
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1. Check ||z||_∞ < τ·σ
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2. Recompute c' = H(commitment || π.t || x || y)
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3. Check c' = π.c
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4. Return valid/invalid
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```
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### 9.2 Perfect Zero-Knowledge Proof
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**Theorem 9.1 (Statistical Zero-Knowledge)**
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There exists a simulator S such that for any verifier V*:
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```
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{Real(k, x, y, V*)} ≈_s {S(x, y, V*)}
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```
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with statistical distance at most 2^{-100}.
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**Simulator Construction:**
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```
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S(x, y):
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1. Sample z ← D_σ^n (same distribution as real proofs!)
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2. Sample random c ←$ {0,1}^128
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3. Compute t = H(z - c·k_dummy || (z - c·k_dummy)·H(x))
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where k_dummy is any fixed small element
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4. Program random oracle: H(commitment || t || x || y) = c
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5. Return π = (t, z, c)
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```
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**Proof:**
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The key insight is that with proper rejection sampling, the distribution of z in real proofs is exactly D_σ^n, independent of k.
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**Lemma 9.1 (Rejection Sampling):**
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Let m ← D_σ and z = m + v for fixed v with ||v|| < σ/(2τ). After rejection sampling with probability p = D_σ(z)/(M · D_{v,σ}(z)):
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The distribution of accepted z is exactly D_σ.
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*Proof of Lemma 9.1:*
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```
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Pr[z accepted] = ∑_z Pr[m = z - v] · p(z)
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= ∑_z D_σ(z - v) · D_σ(z) / (M · D_{v,σ}(z))
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= ∑_z D_σ(z - v) · D_σ(z) / (M · D_σ(z - v))
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= ∑_z D_σ(z) / M
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= 1/M
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```
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For accepted z:
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```
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Pr[z | accepted] = Pr[z accepted] · Pr[z] / (1/M)
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= (D_σ(z)/M) / (1/M)
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= D_σ(z)
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```
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**Completing the Proof:**
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In real execution: z is distributed as D_σ after rejection sampling.
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In simulation: z is sampled directly from D_σ.
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Both distributions are identical! The only difference is:
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- Real: t = H(m || m·H(x)) where m = z - c·k
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- Simulated: t = H(z - c·k_dummy || (z - c·k_dummy)·H(x))
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Since H is a random oracle and we program it consistently, the distributions are statistically identical.
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Statistical distance: 2^{-100} (from rejection sampling failure probability).
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### 9.3 Soundness Proof
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**Theorem 9.2 (Computational Soundness)**
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For any PPT adversary A, the probability of producing valid proofs for two different evaluations y₁ ≠ y₂ under the same commitment is negligible:
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```
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Pr[Verify(c, x, y₁, π₁) = Verify(c, x, y₂, π₂) = 1 ∧ y₁ ≠ y₂] ≤ negl(λ)
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```
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**Proof:**
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Assume A produces accepting proofs (t₁, z₁, c₁) and (t₂, z₂, c₂) for outputs y₁ ≠ y₂.
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**Case 1: c₁ ≠ c₂**
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From the verification equations:
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```
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z₁ = m₁ + c₁·k → m₁ = z₁ - c₁·k
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z₂ = m₂ + c₂·k → m₂ = z₂ - c₂·k
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```
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If t₁ ≠ t₂, then m₁ ≠ m₂. We can extract:
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```
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k = (z₁ - z₂) / (c₁ - c₂) (in the ring)
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```
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This requires finding ring inverse, possible when c₁ - c₂ is invertible in R_q (happens with probability 1 - 1/q per coefficient).
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**Case 2: c₁ = c₂**
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Then H(c || t₁ || x || y₁) = H(c || t₂ || x || y₂).
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Since y₁ ≠ y₂, this is a collision in H, probability ≤ 2^{-128}.
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**Combined bound:**
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```
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Pr[forgery] ≤ 2^{-128} + negl(λ)
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```
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---
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## Appendix A: Proof Details
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### A.1 Ring-LPR Pseudorandomness
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