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Protocol Description

Preliminaries

We take \sec as our security parameter, and unless explicitly noted all algorithms and adversaries are probabilistic (interactive) Turing machines that run in polynomial time in this security parameter. We use \negl to denote a function that is negligible in \sec.

Cryptographic Groups

We let \group denote a cyclic group of prime order p. The identity of a group is written as \zero. We refer to the scalars of elements in \group as elements in a scalar field \field of size p. Group elements are written in capital letters while scalars are written in lowercase or Greek letters. Vectors of scalars or group elements are written in boldface, i.e. $\mathbf{a} \in \field^n$ and \mathbf{G} \in \group^n. Group operations are written additively and the multiplication of a group element G by a scalar a is written $[a] G$.

We will often use the notation \langle \mathbf{a}, \mathbf{b} \rangle to describe the inner product of two like-length vectors of scalars $\mathbf{a}, \mathbf{b} \in \field^n$. We also use this notation to represent the linear combination of group elements such as \langle \mathbf{a}, \mathbf{G} \rangle with \mathbf{a} \in \field^n, \mathbf{G} \in \group^n, computed in practice by a multiscalar multiplication.

We use \mathbf{0}^n to describe a vector of length n that contains only zeroes in \field.

Discrete Log Relation Problem. The advantage metric


\adv^\dlrel_{\group,n}(\a, \sec) = \textnormal{Pr} \left[ \mathsf{G}^\dlrel_{\group,n}(\a, \sec) \right]

is defined with respect the following game.


\begin{array}{ll}
&\underline{\bold{Game} \, \mathsf{G}^\dlrel_{\group,n}(\a, \sec):} \\
&\mathbf{G} \gets \group^n_\sec \\
&\mathbf{a} \gets \a(\mathbf{G}) \\
&\textnormal{Return} \, \left( \langle \mathbf{a}, \mathbf{G} \rangle = \zero \land \mathbf{a} \neq \mathbf{0}^n \right)
\end{array}

Given an n-length vector \mathbf{G} \in \group^n of group elements, the discrete log relation problem asks for \mathbf{g} \in \field^n such that \mathbf{g} \neq \mathbf{0}^n and yet $\innerprod{\mathbf{g}}{\mathbf{G}} = \zero$, which we refer to as a non-trivial discrete log relation. The hardness of this problem is tightly implied by the hardness of the discrete log problem in the group as shown in Lemma 3 of [JT20]. Formally, we use the game \dlgame defined above to capture this problem.

Interactive Proofs

Interactive proofs are a triple of algorithms $\ip = (\setup, \prover, \verifier). The algorithm \setup(1^\sec)$ produces as its output some public parameters commonly referred to by \pp. The prover \prover and verifier \verifier are interactive machines (with access to \pp) and we denote by \langle \prover(x), \verifier(y) \rangle an algorithm that executes a two-party protocol between them on inputs x, y. The output of this protocol, a transcript of their interaction, contains all of the messages sent between \prover and \verifier. At the end of the protocol, the verifier outputs a decision bit.

Zero knowledge Arguments of Knowledge

Proofs of knowledge are interactive proofs where the prover aims to convince the verifier that they know a witness w such that (x, w) \in \relation for a statement x and polynomial-time decidable relation \relation. We will work with arguments of knowledge which assume computationally-bounded provers.

We will analyze arguments of knowledge through the lens of four security notions.

Completeness: If the prover possesses a valid witness, can they always convince the verifier? It is useful to understand this property as it can have implications for the other security notions.
Soundness: Can a cheating prover falsely convince the verifier of the correctness of a statement that is not actually correct? We refer to the probability that a cheating prover can falsely convince the verifier as the soundness error.
Knowledge soundness: When the verifier is convinced the statement is correct, does the prover actually possess ("know") a valid witness? We refer to the probability that a cheating prover falsely convinces the verifier of this knowledge as the knowledge error.
Zero knowledge: Does the verifier learn anything besides that which can be inferred from the correctness of the statement and the prover's knowledge of a valid witness?

First, we will visit the simple definition of completeness.

Perfect Completeness. An interactive argument (\setup, \prover, \verifier) has perfect completeness if for all polynomial-time decidable relations \relation and for all non-uniform polynomial-time adversaries \a


Pr \left[
    (x, w) \notin \relation \lor
    \langle \prover(\pp, x, w), \verifier(\pp, x) \rangle \, \textnormal{accepts}
    \, \middle| \,
    \begin{array}{ll}
    &\pp \gets \setup(1^\sec) \\
    &(x, w) \gets \a(\pp)
    \end{array}
\right] = 1

Soundness

Complicating our analysis is that although our protocol is described as an interactive argument, it is realized in practice as a non-interactive argument through the use of the Fiat-Shamir transformation.

Public coin. We say that an interactive argument is public coin when all of the messages sent by the verifier are each sampled with fresh randomness.

Fiat-Shamir transformation. In this transformation an interactive, public coin argument can be made non-interactive in the random oracle model by replacing the verifier algorithm with a cryptographically strong hash function that produces sufficiently random looking output.

This transformation means that in the concrete protocol a cheating prover can easily "rewind" the verifier by forking the transcript and sending new messages to the verifier. Studying the concrete security of our construction after applying this transformation is important. Fortunately, we are able to follow a framework of analysis by Ghoshal and Tessaro ([GT20]) that has been applied to constructions similar to ours.

We will study our protocol through the notion of state-restoration soundness. In this model the (cheating) prover is allowed to rewind the verifier to any previous state it was in. The prover wins if they are able to produce an accepting transcript.

State-Restoration Soundness. Let \ip be an interactive argument with $r = r(\sec)$ verifier challenges and let the $i$th challenge be sampled from \ch_i. The advantage metric


\adv^\srs_\ip(\prover, \sec) = \textnormal{Pr} \left[ \srs^\ip_\prover(\sec) \right]

of a state restoration prover \prover is defined with respect to the following game.


\begin{array}{ll}
\begin{array}{ll}
&\underline{\bold{Game} \, \srs_\ip^\prover(\sec):} \\
&\textnormal{win} \gets \tt{false}; \\
&\tr \gets \epsilon \\
&\pp \gets \ip.\setup(1^\sec) \\
&(x, \textsf{st}_\prover) \gets \prover_\sec(\pp) \\
&\textnormal{Run} \, \prover^{\oracle_\srs}_\sec(\textsf{st}_\prover) \\
&\textnormal{Return win}
\end{array} &
\begin{array}{ll}
&\underline{\bold{Oracle} \, \oracle_\srs(\tau = (a_1, c_1, ..., a_{i - 1}, c_{i - 1}), a_i):} \\
& \textnormal{If} \, \tau \in \tr \, \textnormal{then} \\
& \, \, \textnormal{If} \, i \leq r \, \textnormal{then} \\
& \, \, \, \, c_i \gets \ch_i; \tr \gets \tr || (\tau, a_i, c_i); \textnormal{Return} \, c_i \\
& \, \, \textnormal{Else if} \, i = r + 1 \, \textnormal{then} \\
& \, \, \, \, d \gets \ip.\verifier (\pp, x, (\tau, a_i)); \tr \gets (\tau, a_i) \\
& \, \, \, \, \textnormal{If} \, d = 1 \, \textnormal{then win} \gets \tt{true} \\
& \, \, \, \, \textnormal{Return} \, d \\
&\textnormal{Return} \bottom
\end{array}
\end{array}

As shown in [GT20] (Theorem 1) state restoration soundness is tightly related to soundness after applying the Fiat-Shamir transformation.

Knowledge Soundness

We will show that our protocol satisfies a strengthened notion of knowledge soundness known as witness extended emulation. Informally, this notion states that for any successful prover algorithm there exists an efficient emulator that can extract a witness from it by rewinding it and supplying it with fresh randomness.

However, we must slightly adjust our definition of witness extended emulation to account for the fact that our provers are state restoration provers and can rewind the verifier. Further, to avoid the need for rewinding the state restoration prover during witness extraction we study our protocol in the algebraic group model.

Algebraic Group Model (AGM). An adversary \alg{\prover} is said to be algebraic if whenever it outputs a group element X it also outputs a representation \mathbf{x} \in \field^n such that \langle \mathbf{x}, \mathbf{G} \rangle = X where \mathbf{G} \in \group^n is the vector of group elements that \alg{\prover} has seen so far. Notationally, we write \rep{X} to describe a group element X enhanced with this representation. We also write \repv{X}{G}{i} to identify the component of the representation of X that corresponds with \mathbf{G}_i. In other words,


X = \sum\limits_{i=0}^{n - 1} \left[ \repv{X}{G}{i} \right] \mathbf{G}_i

The algebraic group model allows us to perform so-called "online" extraction for some protocols: the extractor can obtain the witness from the representations themselves for a single (accepting) transcript.

State Restoration Witness Extended Emulation Let \ip be an interactive argument for relation \relation with r = r(\sec) challenges. We define for all non-uniform algebraic provers \alg{\prover}, extractors \extractor, and computationally unbounded distinguishers \distinguisher the advantage metric


\adv^\srwee_{\ip, \relation}(\alg{\prover}, \distinguisher, \extractor, \sec) = \textnormal{Pr} \left[ \weereal^{\prover,\distinguisher}_{\ip,\relation}(\sec) \right] - \textnormal{Pr} \left[ \weeideal^{\extractor,\prover,\distinguisher}_{\ip,\relation}(\sec) \right]

is defined with the respect to the following games.


\begin{array}{ll}
\begin{array}{ll}
&\underline{\bold{Game} \, \weereal_{\ip,\relation}^{\alg{\prover},\distinguisher}(\sec):} \\
&\tr \gets \epsilon \\
&\pp \gets \ip.\setup(1^\sec) \\
&(x, \state{\prover}) \gets \alg{\prover}(\pp) \\
&\textnormal{Run} \, \alg{\prover}^{\oracle_\real}(\state{\prover}) \\
&b \gets \distinguisher(\tr) \\
&\textnormal{Return} \, b = 1 \\
&\underline{\bold{Game} \, \weeideal_{\ip,\relation}^{\extractor,\alg{\prover},\distinguisher}(\sec):} \\
&\tr \gets \epsilon \\
&\pp \gets \ip.\setup(1^\sec) \\
&(x, \state{\prover}) \gets \alg{\prover}(\pp) \\
&\state{\extractor} \gets (1^\sec, \pp, x) \\
&\textnormal{Run} \, \alg{\prover}^{\oracle_\ideal}(\state{\prover}) \\
&w \gets \extractor(\state{\extractor}, \bottom) \\
&b \gets \distinguisher(\tr) \\
&\textnormal{Return} \, (b = 1) \\
&\, \, \land (\textnormal{Acc}(\tr) \implies (x, w) \in \relation) \\
\end{array} &
\begin{array}{ll}
&\underline{\bold{Oracle} \, \oracle_\real(\tau = (a_1, c_1, ..., a_{i - 1}, c_{i - 1}), a_i):} \\
& \textnormal{If} \, \tau \in \tr \, \textnormal{then} \\
& \, \, \textnormal{If} \, i \leq r \, \textnormal{then} \\
& \, \, \, \, c_i \gets \ch_i; \tr \gets \tr || (\tau, a_i, c_i); \textnormal{Return} \, c_i \\
& \, \, \textnormal{Else if} \, i = r + 1 \, \textnormal{then} \\
& \, \, \, \, d \gets \ip.\verifier (\pp, x, (\tau, a_i)); \tr \gets (\tau, a_i) \\
& \, \, \, \, \textnormal{If} \, d = 1 \, \textnormal{then win} \gets \tt{true} \\
& \, \, \, \, \textnormal{Return} \, d \\
& \, \, \textnormal{Return} \, \bottom \\
\\
\\
&\underline{\bold{Oracle} \, \oracle_\ideal(\tau, a):} \\
& \textnormal{If} \, \tau \in \tr \, \textnormal{then} \\
& \, \, (r, \state{\extractor}) \gets \extractor(\state{\extractor}, \left[(\tau, a)\right]) \\
& \, \, \tr \gets \tr || (\tau, a, r) \\
& \, \, \textnormal{Return} \, r \\
&\textnormal{Return} \, \bottom
\end{array}
\end{array}

Zero Knowledge

We say that an argument of knowledge is zero knowledge if the verifier also does not learn anything from their interaction besides that which can be learned from the existence of a valid w. More formally,

Perfect Special Honest-Verifier Zero Knowledge. A public coin interactive argument (\setup, \prover, \verifier) has perfect special honest-verifier zero knowledge (PSHVZK) if for all polynomial-time decidable relations \relation and for all (x, w) \in \relation and for all non-uniform polynomial-time adversaries \a_1, \a_2 there exists a probabilistic polynomial-time simulator \sim such that


\begin{array}{rl}
&Pr \left[ \a_1(\sigma, x, \tr) = 1 \, \middle| \,
\begin{array}{ll}
&\pp \gets \setup(1^\lambda); \\
&(x, w, \rho) \gets \a_2(\pp); \\
&tr \gets \langle \prover(\pp, x, w), \verifier(\pp, x, \rho) \rangle
\end{array}
 \right] \\
 \\
=&Pr \left[ \a_1(\sigma, x, \tr) = 1 \, \middle| \,
\begin{array}{ll}
&\pp \gets \setup(1^\lambda); \\
&(x, w, \rho) \gets \a_2(\pp); \\
&tr \gets \sim(\pp, x, \rho)
\end{array}
\right]
\end{array}

where \rho is the internal randomness of the verifier.

In this (common) definition of zero-knowledge the verifier is expected to act "honestly" and send challenges that correspond only with their internal randomness; they cannot adaptively respond to the prover based on the prover's messages. We use a strengthened form of this definition that forces the simulator to output a transcript with the same (adversarially provided) challenges that the verifier algorithm sends to the prover.

Protocol

Let \omega \in \field be a n = 2^k primitive root of unity forming the domain D = (\omega^0, \omega^1, ..., \omega^{n - 1}) with t(X) = X^n - 1 the vanishing polynomial over this domain. Let n_g, n_a, n_e be positive integers with n_a, n_e \lt n and n_g \geq 4. We present an interactive argument \halo = (\setup, \prover, \verifier) for the relation


\relation = \left\{
\begin{array}{ll}
&\left(
\begin{array}{ll}
\left(
g(X, C_0, ..., C_{n_a - 1}, a_0(X), ..., a_{n_a - 1}\left(X, C_0, ..., C_{n_a - 1}, a_0(X), ..., a_{n_a - 2}(X) \right))
\right); \\
\left(
a_0(X), a_1(X, C_0, a_0(X)), ..., a_{n_a - 1}\left(X, C_0, ..., C_{n_a - 1}, a_0(X), ..., a_{n_a - 2}(X) \right)
\right)
\end{array}
\right) : \\
\\
& g(\omega^i, \cdots) = 0 \, \, \, \, \forall i \in [0, 2^k)
\end{array}
\right\}

where a_0, a_1, ..., a_{n_a - 1} are (multivariate) polynomials with degree n - 1 in X and g has degree n_g(n - 1) at most in any indeterminates X, C_0, C_1, ....

\setup(\sec) returns \pp = (\group, \field, \mathbf{G} \in \group^n, U, W \in \group).

For all i \in [0, n_a):

Let \mathbf{p_i} be the exhaustive set of integers j (modulo n) such that a_i(\omega^j X, \cdots) appears as a term in g(X, \cdots).
Let \mathbf{q} be a list of distinct sets of integers containing \mathbf{p_i} and the set \mathbf{q_0} = \{0\}.
Let \sigma(i) = \mathbf{q}_j when \mathbf{q}_j = \mathbf{p_i}.

Let n_q \leq n_a denote the size of \mathbf{q}, and let n_e denote the size of every \mathbf{p_i} without loss of generality.

In the following protocol, we take it for granted that each polynomial a_i(X, \cdots) is defined such that n_e + 1 blinding factors are freshly sampled by the prover and are each present as an evaluation of a_i(X, \cdots) over the domain D. In all of the following, the verifier's challenges cannot be zero or an element in D, and some additional limitations are placed on specific challenges as well.

\prover and \verifier proceed in the following n_a rounds of interaction, where in round j (starting at 0)

\prover sets a'_j(X) = a_j(X, c_0, c_1, ..., c_{j - 1}, a_0(X, \cdots), ..., a_{j - 1}(X, \cdots, c_{j - 1}))
\prover sends a hiding commitment A_j = \innerprod{\mathbf{a'}}{\mathbf{G}} + [a^*_j] W where \mathbf{a'} are the coefficients of the univariate polynomial a'_j(X) and a^*_j is some fresh (random and independently sampled) blinding factor. Similar notation is used throughout this protocol description.
\verifier responds with a challenge c_j.

\prover sets g'(X) = g(X, c_0, c_1, ..., c_{n_a - 1}, \cdots).
\prover sends a commitment R = \innerprod{\mathbf{r}}{\mathbf{G}} + [r^*] W where \mathbf{r} \in \field^n are the coefficients of a randomly sampled univariate polynomial r(X) of degree n - 1.
\prover computes univariate polynomial h(X) = \frac{g'(X)}{t(X)} of degree n_g(n - 1) - n.
\prover computes at most n - 1 degree polynomials h_0(X), h_1(X), ..., h_{n_g - 2}(X) such that h(X) = \sum\limits_{i=0}^{n_g - 2} X^{ni} h_i(X).
\prover sends commitments H_i = \innerprod{\mathbf{h_i}}{\mathbf{G}} + [h^*_i] W for all i where \mathbf{h_i} denotes the vector of coefficients for h_i(X), and h^*_i are fresh blinding factors.
\verifier responds with challenge x and computes H' = \sum\limits_{i=0}^{n_g - 2} [x^{ni}] H_i.
\prover sets h'(X) = \sum\limits_{i=0}^{n_g - 2} x^{ni} h_i(X), and synthetic blinding factor h'^* = \sum\limits_{i=0}^{n_g - 2} x^{ni} h^*_i.
\prover sends r = r(x) and for all i \in [0, n_a) sends \mathbf{a_i} such that (\mathbf{a_i})_j = a'_i(\omega^{(\mathbf{p_i})_j} x) for all j \in [0, n_e - 1].
For all i \in [0, n_a) \prover and \verifier set s_i(X) to be the lowest degree univariate polynomial defined such that s_i(\omega^{(\mathbf{p_i})_j} x) = (\mathbf{a_i})_j for all j \in [0, n_e - 1).
\verifier responds with challenges x_1, x_2 and initializes Q_0, Q_1, ..., Q_{n_q - 1} = \zero.

Starting at i=0 and ending at n_a - 1 \verifier sets Q_{\sigma(i)} := [x_1] Q_{\sigma(i)} + A_i.
\verifier finally sets Q_0 := [x_1^2] Q_0 + [x_1] H' + R.

\prover initializes q_0(X), q_1(X), ..., q_{n_q - 1}(X) = 0 and blinding factors q^*_0, q^*_1, ..., q^*_{n_q-1} = 0.

Starting at i=0 and ending at n_a - 1 \prover sets q_{\sigma(i)} := x_1 q_{\sigma(i)} + a'(X) and q^*_{\sigma(i)} := x_1 q^*_{\sigma(i)} + a^*_i.
\prover finally sets q_0(X) := x_1^2 q_0(X) + x_1 h'(X) + r(X), and its corresponding synthetic blinding factor q^*_0 := x_1^2 q^*_0 + x_1 h'^* + r^*.

\prover and \verifier initialize r_0(X), r_1(X), ..., r_{n_q - 1}(X) = 0.

Starting at i = 0 and ending at n_a - 1 \prover and \verifier set r_{\sigma(i)}(X) := x_1 r_{\sigma(i)}(X) + s_i(X).
Finally \prover and \verifier set r_0 := x_1^2 r_0 + x_1 h + r and where h is computed by \verifier as \frac{g'(x)}{t(x)} using the values r, \mathbf{a} provided by \prover.

\prover sends Q' = \innerprod{\mathbf{q'}}{\mathbf{G}} + [q^{\prime *}] W where q^{\prime *} is the synthetic blinding factor calculated in step 12, and \mathbf{q'} defines the coefficients of the polynomial

q'(X) = \sum\limits_{i=0}^{n_q - 1}

x_2^{n_q - 1 - i}
  \left(
  \frac
  {q_i(X) - r_i(X)}
  {\prod\limits_{j=0}^{n_e - 1}
    \left(
      X - \omega^{\left(
        \mathbf{q_i}
      \right)_j} x
    \right)
  }
  \right)

\verifier responds with challenge x_3.
\prover sends \mathbf{u} \in \field^{n_q} such that \mathbf{u}_i = q_i(x_3) for all i \in [0, n_q).
\verifier responds with challenge x_4.
\verifier sets P = [x_4^{n_q}]Q' + \sum\limits_{i=0}^{n_q - 1} [x_4^{n_q - 1 - i}] Q_i and v =


x_4^{n_q} \cdot
\sum\limits_{i=0}^{n_q - 1}
\left(
x_2^{n_q - 1 - i}
  \left(
  \frac
  { \mathbf{u}_i - r_i(x_3) }
  {\prod\limits_{j=0}^{n_e - 1}
    \left(
      x_3 - \omega^{\left(
        \mathbf{q_i}
      \right)_j} x
    \right)
  }
  \right)
\right)
+
\sum\limits_{i=0}^{n_q - 1} x_4^{n_q - 1 - i} \mathbf{u}_i

\prover sets p(X) = x_4^{n_q} \cdot q'(X) + \sum\limits_{i=0}^{n_q - 1} x_4^{n_q - 1 - i} \cdot q_i(X) and the corresponding synthetic blinding factor p^* = x_4^{n_q} \cdot q'^* + \sum\limits_{i=0}^{n_q - 1} x_4^{n_q - 1 - i} \cdot q^*_i.
\prover samples a random polynomial s(X) of degree n - 1 with a root at x_3 and sends a commitment S = \innerprod{\mathbf{s}}{\mathbf{G}} + [s^{*}] W where \mathbf{s} defines the coefficients of s(X) and s^{*} is a fresh blinding factor.
\verifier responds with challenges \xi, z.
\verifier sets P' = P - [v] \mathbf{G}_0 + [\xi] S.
\prover sets p'(X) = p(X) - p(x_3) + \xi s(X) and p^{\prime *} = s^* \cdot \xi + p^* (where p(x_3) should correspond with the verifier's computed value v).
Initialize \mathbf{p'} as the coefficients of p'(X) and \mathbf{G'} = \mathbf{G} and \mathbf{b} = (x_3^0, x_3^1, ..., x_3^{n - 1}). \prover and \verifier will interact in the following k rounds, where in the j$th round starting in round $j=0 and ending in round j=k-1:

\prover sends L_j = \innerprod{\mathbf{p'}_\hi}{\mathbf{G'}_\lo} + [z \innerprod{\mathbf{p'}_\hi}{\mathbf{b}_\lo}] U + [\ell_j^*] W and R_j = \innerprod{\mathbf{p'}_\lo}{\mathbf{G'}_\hi} + [z \innerprod{\mathbf{p'}_\lo}{\mathbf{b}_\hi}] U + [r_j^*] W where \ell_j^* and r_j^* are fresh blinding factors.
\verifier responds with challenge u_j chosen such that 1 + u_{k-1-j} x_3^{2^j} is nonzero.
\prover and \verifier set \mathbf{G'} := \mathbf{G'}_\lo + u_j \mathbf{G'}_\hi and \mathbf{b} := \mathbf{b}_\lo + u_j \mathbf{b}_\hi.
\prover sets \mathbf{p'} := \mathbf{p'}_\lo + u_j^{-1} \mathbf{p'}_\hi.

\prover sends c = \mathbf{p'}_0 and synthetic blinding factor f = p^{\prime *} + \sum_{j=0}^{k - 1}\left(\ell_j^* \cdot u_j^{-1} + r_j^* \cdot u_j\right).
\verifier accepts only if \sum_{j=0}^{k - 1} [u_j^{-1}] L_j + P' + \sum_{j=0}^{k - 1} [u_j] R_j = [c] \mathbf{G'}_0 + [c \mathbf{b}_0 z] U + [f] W.

Zero-knowledge and Completeness

We claim that this protocol is perfectly complete. This can be verified by inspection of the protocol; given a valid witness a_i(X, \cdots) \forall i the prover succeeds in convincing the verifier with probability 1.

We claim that this protocol is perfect special honest-verifier zero knowledge. We do this by showing that a simulator \sim exists which can produce an accepting transcript that is equally distributed with a valid prover's interaction with a verifier with the same public coins. The simulator will act as an honest prover would, with the following exceptions:

In step 1 of the protocol \sim chooses random degree n - 1 polynomials (in X) a_i(X, \cdots) \forall i.
In step 5 of the protocol \sim chooses a random n - 1 degree polynomials h_0(X), h_1(X), ..., h_{n_g - 2}(X).
In step 14 of the protocol \sim chooses a random n - 1 degree polynomial q'(X).
In step 20 of the protocol \sim uses its foreknowledge of the verifier's choice of \xi to produce a degree n - 1 polynomial s(X) conditioned only such that p(X) - v + \xi s(X) has a root at x_3.

First, let us consider why this simulator always succeeds in producing an accepting transcript. \sim lacks a valid witness and simply commits to random polynomials whenever knowledge of a valid witness would be required by the honest prover. The verifier places no conditions on the scalar values in the transcript. \sim must only guarantee that the check in step 26 of the protocol succeeds. It does so by using its knowledge of the challenge \xi to produce a polynomial which interferes with p'(X) to ensure it has a root at x_3. The transcript will thus always be accepting due to perfect completeness.

In order to see why \sim produces transcripts distributed identically to the honest prover, we will look at each piece of the transcript and compare the distributions. First, note that \sim (just as the honest prover) uses a freshly random blinding factor for every group element in the transcript, and so we need only consider the scalars in the transcript. \sim acts just as the prover does except in the mentioned cases so we will analyze each case:

\sim and an honest prover reveal n_e openings of each polynomial a_i(X, \cdots), and at most one additional opening of each a_i(X, \cdots) in step 16. However, the honest prover blinds their polynomials a_i(X, \cdots) (in X) with n_e + 1 random evaluations over the domain D. Thus, the openings of a_i(X, \cdots) at the challenge x (which is prohibited from being 0 or in the domain D by the protocol) are distributed identically between \sim and an honest prover.
Neither \sim nor the honest prover reveal h(x) as it is computed by the verifier. However, the honest prover may reveal h'(x_3) --- which has a non-trivial relationship with h(X) --- were it not for the fact that the honest prover also commits to a random degree n - 1 polynomial r(X) in step 3, producing a commitment R and ensuring that in step 12 when the prover sets q_0(X) := x_1^2 q_0(X) + x_1 h'(X) + r(X) the distribution of q_0(x_3) is uniformly random. Thus, h'(x_3) is never revealed by the honest prover nor by \sim.
The expected value of q'(x_3) is computed by the verifier (in step 18) and so the simulator's actual choice of q'(X) is irrelevant.
p(X) - v + \xi s(X) is conditioned on having a root at x_3, but otherwise no conditions are placed on s(X) and so the distribution of the degree n - 1 polynomial p(X) - v + \xi s(X) is uniformly random whether or not s(X) has a root at x_3. Thus, the distribution of c produced in step 25 is identical between \sim and an honest prover. The synthetic blinding factor f also revealed in step 25 is a trivial function of the prover's other blinding factors and so is distributed identically between \sim and an honest prover.

Notes:

In an earlier version of our protocol, the prover would open each individual commitment H_0, H_1, ... at x as part of the multipoint opening argument, and the verifier would confirm that a linear combination of these openings (with powers of x^n) agreed to the expected value of h(x). This was done because it's more efficient in recursive proofs. However, it was unclear to us what the expected distribution of the openings of these commitments H_0, H_1, ... was and so proving that the argument was zero-knowledge is difficult. Instead, we changed the argument so that the verifier computes a linear combination of the commitments and that linear combination is opened at x. This avoided leaking h_i(x).
As mentioned, in step 3 the prover commits to a random polynomial as a way of ensuring that h'(x_3) is not revealed in the multiopen argument. This is done because it's unclear what the distribution of h'(x_3) would be.
Technically it's also possible for us to prove zero-knowledge with a simulator that uses its foreknowledge of the challenge x to commit to an h(X) which agrees at x to the value it will be expected to. This would obviate the need for the random polynomial s(X) in the protocol. This may make the analysis of zero-knowledge for the remainder of the protocol a little bit tricky though, so we didn't go this route.
Group element blinding factors are technically not necessary after step 23 in which the polynomial is completely randomized. However, it's simpler in practice for us to ensure that every group element in the protocol is randomly blinded to make edge cases involving the point at infinity harder.
It is crucial that the verifier cannot challenge the prover to open polynomials at points in D as otherwise the transcript of an honest prover will be forced to contain what could be portions of the prover's witness. We therefore restrict the space of challenges to include all elements of the field except D and, for simplicity, we also prohibit the challenge of 0.

Witness-extended Emulation

Let \protocol = \protocol[\group] be the interactive argument described above for relation \relation and some group \group with scalar field \field. We can always construct an extractor \extractor such that for any non-uniform algebraic prover \alg{\prover} making at most q queries to its oracle, there exists a non-uniform adversary \dlreladv with the property that for any computationally unbounded distinguisher \distinguisher


\adv^\srwee_{\protocol, \relation}(\alg{\prover}, \distinguisher, \extractor, \sec) \leq q\epsilon + \adv^\dlrel_{\group,n+2}(\dlreladv, \sec)

where \epsilon \leq \frac{n_g \cdot (n - 1)}{|\ch|}.

Proof. We will prove this by invoking Theorem 1 of [GT20]. First, we note that the challenge space for all rounds is the same, i.e. \forall i \ \ch = \ch_i. Theorem 1 requires us to define:

\badch(\tr') \in \ch for all partial transcripts \tr' = (\pp, x, [a_0], c_0, \ldots, [a_i]) such that |\badch(\tr')| / |\ch| \leq \epsilon.
an extractor function e that takes as input an accepting extended transcript \tr and either returns a valid witness or fails.
a function \pfail(\protocol, \alg{\prover}, e, \relation) returning a probability.

We say that an accepting extended transcript \tr contains "bad challenges" if and only if there exists a partial extended transcript \tr', a challenge c_i \in \badch(\tr'), and some sequence of prover messages and challenges ([a_{i+1}], c_{i+1}, \ldots [a_j]) such that \tr = \tr' \,||\, (c_i, [a_{i+1}], c_{i+1}, \ldots [a_j]).

Theorem 1 requires that e, when given an accepting extended transcript \tr that does not contain "bad challenges", returns a valid witness for that transcript except with probability bounded above by \pfail(\protocol, \alg{\prover}, e, \relation).

Our strategy is as follows: we will define e, establish an upper bound on \pfail with respect to an adversary \dlreladv that plays the \dlrel_{\group,n+2} game, substitute these into Theorem 1, and then walk through the protocol to determine the upper bound of the size of \badch(\tr'). The adversary \dlreladv plays the \dlrel_{\group,n+2} game as follows: given the inputs U, W \in \mathbb{G}, \mathbf{G} \in \mathbb{G}^n, the adversary \dlreladv simulates the game \srwee_{\protocol, \relation} to \alg{\prover} using the inputs from the \dlrel_{\group,n+2} game as public parameters. If \alg{\prover} manages to produce an accepting extended transcript \tr, \dlreladv invokes a function h on \tr and returns its output. We shall define h in such a way that for an accepting extended transcript \tr that does not contain "bad challenges", e(\tr) always returns a valid witness whenever h(\tr) does not return a non-trivial discrete log relation. This means that the probability \pfail(\protocol, \alg{\prover}, e, \relation) is no greater than \adv^\dlrel_{\group,n+2}(\dlreladv, \sec), establishing our claim.

Helpful substitutions

We will perform some substitutions to aid in exposition. First, let us define the polynomial


\kappa(X) = \prod_{j=0}^{k - 1} (1 + u_{k - 1 - j} X^{2^j})

so that we can write \mathbf{b}_0 = \kappa(x_3). The coefficient vector \mathbf{s} of \kappa(X) is defined such that

\mathbf{s}_i = \prod\limits_{j=0}^{k-1} u_{k - 1 - j}^{f(i, j)}

where f(i, j) returns 1 when the j$th bit of $i is set, and 0 otherwise. We can also write \mathbf{G'}_0 = \innerprod{\mathbf{s}}{\mathbf{G}}.

Description of function `h`

Recall that an accepting transcript \tr is such that


\sum_{i=0}^{k - 1} [u_j^{-1}] \rep{L_j} + \rep{P'} + \sum_{i=0}^{k - 1} [u_j] \rep{R_j} = [c] \mathbf{G'}_0 + [c z \mathbf{b}_0] U + [f] W

By inspection of the representations of group elements with respect to \mathbf{G}, U, W (recall that \alg{\prover} is algebraic and so \dlreladv has them), we obtain the n equalities


\sum_{i=0}^{k - 1} u_j^{-1} \repv{L_j}{G}{i} + \repv{P'}{G}{i} + \sum_{i=0}^{k - 1} u_j \repv{R_j}{G}{i} = c \mathbf{s}_i \forall i \in [0, n)

and the equalities


\sum_{i=0}^{k - 1} u_j^{-1} \repr{L_j}{U} + \repr{P'}{U} + \sum_{i=0}^{k - 1} u_j \repr{R_j}{U} = c z \kappa(x_3)


\sum_{i=0}^{k - 1} u_j^{-1} \repr{L_j}{W} + \repr{P'}{W} + \sum_{i=0}^{k - 1} u_j \repr{R_j}{W} = f

We define the linear-time function h that returns the representation of


\begin{array}{rll}
\sum\limits_{i=0}^{n - 1} &\left[ \sum_{i=0}^{k - 1} u_j^{-1} \repv{L_j}{G}{i} + \repv{P'}{G}{i} + \sum_{i=0}^{k - 1} u_j \repv{R_j}{G}{i} - c \mathbf{s}_i \right] & \mathbf{G}_i \\[1ex]
+ &\left[ \sum_{i=0}^{k - 1} u_j^{-1} \repr{L_j}{U} + \repr{P'}{U} + \sum_{i=0}^{k - 1} u_j \repr{R_j}{U} - c z \kappa(x_3) \right] & U \\[1ex]
+ &\left[ \sum_{i=0}^{k - 1} u_j^{-1} \repr{L_j}{W} + \repr{P'}{W} + \sum_{i=0}^{k - 1} u_j \repr{R_j}{W} - f \right] & W
\end{array}

which is always a discrete log relation. If any of the equalities above are not satisfied, then this discrete log relation is non-trivial. This is the function invoked by \dlreladv.

The extractor function `e`

The extractor function e simply returns a_i(X) from the representation \rep{A_i} for i \in [0, n_a). Due to the restrictions we will place on the space of bad challenges in each round, we are guaranteed to obtain polynomials such that g(X, C_0, C_1, \cdots, a_0(X), a_1(X), \cdots) vanishes over D whenever the discrete log relation returned by the adversary's function h is trivial. This trivially gives us that the extractor function e succeeds with probability bounded above by \pfail as required.

Defining `\badch(\tr')`

Recall from before that the following n equalities hold:


\sum_{i=0}^{k - 1} u_j^{-1} \repv{L_j}{G}{i} + \repv{P'}{G}{i} + \sum_{i=0}^{k - 1} u_j \repv{R_j}{G}{i} = c \mathbf{s}_i \forall i \in [0, n)

as well as the equality


\sum_{i=0}^{k - 1} u_j^{-1} \repr{L_j}{U} + \repr{P'}{U} + \sum_{i=0}^{k - 1} u_j \repr{R_j}{U} = c z \kappa(x_3)

For convenience let us introduce the following notation


\begin{array}{ll}
\mv{G}{i}{m} &= \sum_{i=0}^{m - 1} u_j^{-1} \repv{L_j}{G}{i} + \repv{P'}{G}{i} + \sum_{i=0}^{m - 1} u_j \repv{R_j}{G}{i} \\[1ex]
\m{U}{m} &= \sum_{i=0}^{m - 1} u_j^{-1} \repr{L_j}{U} + \repr{P'}{U} + \sum_{i=0}^{m - 1} u_j \repr{R_j}{U}
\end{array}

so that we can rewrite the above (after expanding for \kappa(x_3)) as


\mv{G}{i}{k} = c \mathbf{s}_i \forall i \in [0, n)


\m{U}{k} = c z \prod_{j=0}^{k - 1} (1 + u_{k - 1 - j} x_3^{2^j})

We can combine these equations by multiplying both sides of each instance of the first equation by \mathbf{s}_i^{-1} (because \mathbf{s}_i is never zero) and substituting for c in the second equation, yielding the following n equalities:


\m{U}{k} = \mv{G}{i}{k} \cdot \mathbf{s}_i^{-1} z \prod_{j=0}^{k - 1} (1 + u_{k - 1 - j} x_3^{2^j}) \forall i \in [0, n)

Lemma 1. If \m{U}{k} = \mv{G}{i}{k} \cdot \mathbf{s}_i^{-1} z \prod_{j=0}^{k - 1} (1 + u_{k - 1 - j} x_3^{2^j}) \forall i \in [0, n) then it follows that \repr{P'}{U} = z \sum\limits_{i=0}^{2^k - 1} x_3^i \repv{P'}{G}{i} for all transcripts that do not contain bad challenges.

Proof. It will be useful to introduce yet another abstraction defined starting with

\z{k}{m}{i} = \mv{G}{i}{m}


> and then recursively defined for all integers $r$ such that $0 \lt r \leq k$
> $$
\z{k - r}{m}{i} = \z{k - r + 1}{m}{i} + x_3^{2^{k - r}} \z{k - r + 1}{m}{i + 2^{k - r}}

This allows us to rewrite our above equalities as

\m{U}{k} = \z{k}{k}{i} \cdot \mathbf{s}i^{-1} z \prod{j=0}^{k - 1} (1 + u_{k - 1 - j} x_3^{2^j}) \forall i \in [0, n)


>
> We will now show that for all integers $r$ such that $0 \lt r \leq k$ that whenever the following holds for $r$
> $$
\m{U}{r} = \z{r}{r}{i} \cdot \mathbf{s}_i^{-1} z \prod_{j=0}^{r - 1} (1 + u_{k - 1 - j} x_3^{2^j}) \forall i \in [0, 2^r)

that the same also holds for

\m{U}{r - 1} = \z{r - 1}{r - 1}{i} \cdot \mathbf{s}i^{-1} z \prod{j=0}^{r - 2} (1 + u_{k - 2 - j} x_3^{2^j}) \forall i \in [0, 2^{r-1})


>
> For all integers $r$ such that $0 \lt r \leq k$ we have that $\mathbf{s}_{i + 2^{r - 1}} = u_{r - 1} \mathbf{s}_i \forall i \in [0, 2^{r - 1})$ by the definition of $\mathbf{s}$. This gives us $\mathbf{s}_{i+2^{r - 1}}^{-1} = \mathbf{s}_i^{-1} u_{r - 1}^{-1} \forall i \in [0, 2^{r - 1})$ as no value in $\mathbf{s}$ nor any challenge $u_r$ are zeroes. We can use this to relate one half of the equalities with the other half as so:
> $$
\begin{array}{rl}
\m{U}{r} &= \z{r}{r}{i} \cdot \mathbf{s}_i^{-1} z \prod_{j=0}^{r - 1} (1 + u_{k - 1 - j} x_3^{2^j}) \\
&= \z{r}{r}{i + 2^{r - 1}} \cdot \mathbf{s}_i^{-1} u_{r - 1}^{-1} z \prod_{j=0}^{r - 1} (1 + u_{k - 1 - j} x_3^{2^j}) \\
&\forall i \in [0, 2^{r - 1})
\end{array}

Notice that \z{r}{r}{i} can be rewritten as u_{r - 1}^{-1} \repv{L_{r - 1}}{G}{i} + \z{r}{r - 1}{i} + u_{r - 1} \repv{R_{r - 1}}{G}{i} for all i \in [0, 2^{r}). Thus we can rewrite the above as

\begin{array}{rl} \m{U}{r} &= \left( u_{r - 1}^{-1} \repv{L_{r - 1}}{G}{i} + \z{r}{r - 1}{i} + u_{r - 1} \repv{R_{r - 1}}{G}{i} \right) \ &\cdot ; \mathbf{s}i^{-1} z \prod{j=0}^{r - 1} (1 + u_{k - 1 - j} x_3^{2^j}) \ &= \left( u_{r - 1}^{-1} \repv{L_{r - 1}}{G}{i + 2^{r - 1}} + \z{r}{r - 1}{i + 2^{r - 1}} + u_{r - 1} \repv{R_{r - 1}}{G}{i + 2^{r - 1}} \right) \ &\cdot ; \mathbf{s}i^{-1} u{r - 1}^{-1} z \prod_{j=0}^{r - 1} (1 + u_{k - 1 - j} x_3^{2^j}) \ &\forall i \in [0, 2^{r - 1}) \end{array}


>
> Now let us rewrite these equalities substituting $u_{r - 1}$ with formal indeterminate $X$.
>
> $$
\begin{array}{rl}
& X^{-1} \repr{L_{r - 1}}{U} + \m{U}{r - 1} + X \repr{R_{r - 1}}{U} \\
&= \left( X^{-1} \repv{L_{r - 1}}{G}{i} + \z{r}{r - 1}{i} + X \repv{R_{r - 1}}{G}{i} \right) \\
&\cdot \; \mathbf{s}_i^{-1} z \prod_{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) (1 + x_3^{2^{r - 1}} X) \\
&= \left( X^{-1} \repv{L_{r - 1}}{G}{i + 2^{r - 1}} + \z{r}{r - 1}{i + 2^{r - 1}} + X \repv{R_{r - 1}}{G}{i + 2^{r - 1}} \right) \\
&\cdot \; \mathbf{s}_i^{-1} z \prod_{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) (X^{-1} + x_3^{2^{r - 1}}) \\
&\forall i \in [0, 2^{r - 1})
\end{array}

Now let us rescale everything by X^2 to remove negative exponents.

\begin{array}{rl} & X \repr{L_{r - 1}}{U} + X^2 \m{U}{r - 1} + X^3 \repr{R_{r - 1}}{U} \ &= \left( X^{-1} \repv{L_{r - 1}}{G}{i} + \z{r}{r - 1}{i} + X \repv{R_{r - 1}}{G}{i} \right) \ &\cdot ; \mathbf{s}i^{-1} z \prod{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) (X^2 + x_3^{2^{r - 1}} X^3) \ &= \left( X^{-1} \repv{L_{r - 1}}{G}{i + 2^{r - 1}} + \z{r}{r - 1}{i + 2^{r - 1}} + X \repv{R_{r - 1}}{G}{i + 2^{r - 1}} \right) \ &\cdot ; \mathbf{s}i^{-1} z \prod{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) (X + x_3^{2^{r - 1}} X^2) \ &\forall i \in [0, 2^{r - 1}) \end{array}


>
> This gives us $2^{r - 1}$ triples of maximal degree-$4$ polynomials in $X$ that agree at $u_{r - 1}$ despite having coefficients determined prior to the choice of $u_{r - 1}$. The probability that two of these polynomials would agree at $u_{r - 1}$ and yet be distinct would be $\frac{4}{|\ch|}$ by the Schwartz-Zippel lemma and so by the union bound the probability that the three of these polynomials agree and yet any of them is distinct from another is $\frac{8}{|\ch|}$. By the union bound again the probability that any of the $2^{r - 1}$ triples have multiple distinct polynomials is $\frac{2^{r - 1}\cdot8}{|\ch|}$. By restricting the challenge space for $u_{r - 1}$ accordingly we obtain $|\badch(\trprefix{\tr'}{u_r})|/|\ch| \leq \frac{2^{r - 1}\cdot8}{|\ch|}$ for integers $0 \lt r \leq k$ and thus $|\badch(\trprefix{\tr'}{u_k})|/|\ch| \leq \frac{4n}{|\ch|} \leq \epsilon$.
> 
> We can now conclude an equality of polynomials, and thus of coefficients. Consider the coefficients of the constant terms first, which gives us the $2^{r - 1}$ equalities
> $$
0 = 0 = \mathbf{s}_i^{-1} z \left( \prod_{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) \right) \cdot \repv{L_{r - 1}}{G}{i + 2^{r - 1}} \forall i \in [0, 2^{r - 1})

No value of \mathbf{s} is zero, z is never chosen to be 0 and each u_j is chosen so that 1 + u_{k - 1 - j} x_3^{2^j} is nonzero, so we can then conclude

0 = \repv{L_{r - 1}}{G}{i + 2^{r - 1}} \forall i \in [0, 2^{r - 1})


>
> An identical process can be followed with respect to the coefficients of the $X^4$ term in the equalities to establish $0 = \repv{R_{r - 1}}{G}{i} \forall i \in [0, 2^{r - 1})$ contingent on $x_3$ being nonzero, which it always is. Substituting these in our equalities yields us something simpler
> 
> $$
\begin{array}{rl}
& X \repr{L_{r - 1}}{U} + X^2 \m{U}{r - 1} + X^3 \repr{R_{r - 1}}{U} \\
&= \left( X^{-1} \repv{L_{r - 1}}{G}{i} + \z{r}{r - 1}{i} \right) \\
&\cdot \; \mathbf{s}_i^{-1} z \prod_{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) (X^2 + x_3^{2^{r - 1}} X^3) \\
&= \left( \z{r}{r - 1}{i + 2^{r - 1}} + X \repv{R_{r - 1}}{G}{i + 2^{r - 1}} \right) \\
&\cdot \; \mathbf{s}_i^{-1} z \prod_{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) (X + x_3^{2^{r - 1}} X^2) \\
&\forall i \in [0, 2^{r - 1})
\end{array}

Now we will consider the coefficients in X, which yield the equalities

\begin{array}{rl} \repr{L_{r - 1}}{U} &= \mathbf{s}i^{-1} z \prod{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) \cdot \repv{L_{r - 1}}{G}{i} \ &= \mathbf{s}i^{-1} z \prod{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) \cdot \z{r}{r - 1}{i + 2^{r - 1}} \ &\forall i \in [0, 2^{r - 1}) \end{array}


>
> which for similar reasoning as before yields the equalities
> $$
\repv{L_{r - 1}}{G}{i} = \z{r}{r - 1}{i + 2^{r - 1}} \forall i \in [0, 2^{r - 1})

Finally we will consider the coefficients in X^2 which yield the equalities

\begin{array}{rl} \m{U}{r - 1} &= \mathbf{s}i^{-1} z \prod{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) \cdot \left( \z{r}{r - 1}{i} + \repv{L_{r - 1}}{G}{i} x_3^{2^{r - 1}} \right) \ &\forall i \in [0, 2^{r - 1}) \end{array}


>
> which by substitution gives us $\forall i \in [0, 2^{r - 1})$
> $$
\m{U}{r - 1} = \mathbf{s}_i^{-1} z \prod_{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) \cdot \left( \z{r}{r - 1}{i} + \z{r}{r - 1}{i + 2^{r - 1}} x_3^{2^{r - 1}} \right)

Notice that by the definition of \z{r - 1}{m}{i} we can rewrite this as

\m{U}{r - 1} = \z{r - 1}{r - 1}{i} \cdot \mathbf{s}i^{-1} z \prod{j=0}^{r - 2} (1 + u_{k - 1 - j} x_3^{2^j}) \forall i \in [0, 2^{r - 1})


>
> which is precisely in the form we set out to demonstrate.
>
> We now proceed by induction from the case $r = k$ (which we know holds) to reach $r = 0$, which gives us

\m{U}{0} = \z{0}{0}{0} \cdot \mathbf{s}_0^{-1} z


>
> and because $\m{U}{0} = \repr{P'}{U}$ and $\z{0}{0}{0} = \sum_{i=0}^{2^k - 1} x_3^i \repv{P'}{G}{i}$, we obtain $\repr{P'}{U} = z \sum_{i=0}^{2^k - 1} x_3^i \repv{P'}{G}{i}$, which completes the proof.

Having established that $\repr{P'}{U} = z \sum_{i=0}^{2^k - 1} x_3^i \repv{P'}{G}{i}$, and given that $x_3$ and $\repv{P'}{G}{i}$ are fixed in advance of the choice of $z$, we have that at most one value of $z \in \ch$ (which is nonzero) exists such that $\repr{P'}{U} = z \sum_{i=0}^{2^k - 1} x_3^i \repv{P'}{G}{i}$ and yet $\repr{P'}{U} \neq 0$. By restricting $|\badch(\trprefix{\tr'}{z})|/|\ch| \leq \frac{1}{|\ch|} \leq \epsilon$ accordingly we obtain $\repr{P'}{U} = 0$ and therefore that the polynomial defined by $\repr{P'}{\mathbf{G}}$ has a root at $x_3$.

By construction $P' = P - [v] \mathbf{G}_0 + [\xi] S$, giving us that the polynomial defined by $\repr{P + [\xi] S}{\mathbf{G}}$ evaluates to $v$ at the point $x_3$. We have that $v, P, S$ are fixed prior to the choice of $\xi$, and so either the polynomial defined by $\repr{S}{\mathbf{G}}$ has a root at $x_3$ (which implies the polynomial defined by $\repr{P}{\mathbf{G}}$ evaluates to $v$ at the point $x_3$) or else $\xi$ is the single solution in $\ch$ for which $\repr{P + [\xi] S}{\mathbf{G}}$ evaluates to $v$ at the point $x_3$ while $\repr{P}{\mathbf{G}}$ itself does not. We avoid the latter case by restricting $|\badch(\trprefix{\tr'}{\xi})|/|\ch| \leq \frac{1}{|\ch|} \leq \epsilon$ accordingly and can thus conclude that the polynomial defined by $\repr{P}{\mathbf{G}}$ evaluates to $v$ at $x_3$.

The remaining work deals strictly with the representations of group elements sent previously by the prover and their relationship with $P$ as well as the challenges chosen in each round of the protocol. We will simplify things first by using $p(X)$ to represent the polynomial defined by $\repr{P}{\mathbf{G}}$, as it is the case that this $p(X)$ corresponds exactly with the like-named polynomial in the protocol itself. We will make similar substitutions for the other group elements (and their corresponding polynomials) to aid in exposition, as the remainder of this proof is mainly tedious application of the Schwartz-Zippel lemma to upper bound the bad challenge space size for each of the remaining challenges in the protocol.

Recall that $P = Q' + x_4 \sum\limits_{i=0}^{n_q - 1} [x_4^i] Q_i$, and so by substitution we have $p(X) = q'(X) + x_4 \sum\limits_{i=0}^{n_q - 1} x_4^i q_i(X)$. Recall also that

v = \sum\limits_{i=0}^{n_q - 1} \left( x_2^i \left( \frac { \mathbf{u}i - r_i(x_3) } {\prod\limits{j=0}^{n_e - 1} \left( x_3 - \omega^{\left( \mathbf{q_i} \right)j} x \right) } \right) \right) + x_4 \sum\limits{i=0}^{n_q - 1} x_4 \mathbf{u}_i



We have already established that $p(x_3) = v$. Notice that the coefficients in the above expressions for $v$ and $P$ are fixed prior to the choice of $x_4 \in \ch$. By the Schwartz-Zippel lemma we have that only at most $n_q + 1$ possible choices of $x_4$ exist such that these expressions are satisfied and yet $q_i(x_3) \neq \mathbf{u}_i$ for any $i$ or

q'(x_3) \neq \sum\limits_{i=0}^{n_q - 1} \left( x_2^i \left( \frac { \mathbf{u}i - r_i(x_3) } {\prod\limits{j=0}^{n_e - 1} \left( x_3 - \omega^{\left( \mathbf{q_i} \right)_j} x \right) } \right) \right)



By restricting $|\badch(\trprefix{\tr'}{x_4})|/|\ch| \leq \frac{n_q + 1}{|\ch|} \leq \epsilon$ we can conclude that all of the aforementioned inequalities are untrue. Now we can substitute $\mathbf{u}_i$ with $q_i(x_3)$ for all $i$ to obtain

q'(x_3) = \sum\limits_{i=0}^{n_q - 1} \left( x_2^i \left( \frac { q_i(x_3) - r_i(x_3) } {\prod\limits_{j=0}^{n_e - 1} \left( x_3 - \omega^{\left( \mathbf{q_i} \right)_j} x \right) } \right) \right)



Suppose that $q'(X)$ (which is the polynomial defined by $\repr{Q'}{\mathbf{G}}$, and is of degree at most $n - 1$) does _not_ take the form

$$\sum\limits_{i=0}^{n_q - 1}

x_2^i
  \left(
  \frac
  {q_i(X) - r_i(X)}
  {\prod\limits_{j=0}^{n_e - 1}
    \left(
      X - \omega^{\left(
        \mathbf{q_i}
      \right)_j} x
    \right)
  }
  \right)

and yet q'(X) agrees with this expression at x_3 as we've established above. By the Schwartz-Zippel lemma this can only happen for at most n - 1 choices of x_3 \in \ch and so by restricting |\badch(\trprefix{\tr'}{x_3})|/|\ch| \leq \frac{n - 1}{|\ch|} \leq \epsilon we obtain that

q'(X) = \sum\limits_{i=0}^{n_q - 1}

x_2^i
  \left(
  \frac
  {q_i(X) - r_i(X)}
  {\prod\limits_{j=0}^{n_e - 1}
    \left(
      X - \omega^{\left(
        \mathbf{q_i}
      \right)_j} x
    \right)
  }
  \right)

Next we will extract the coefficients of this polynomial in x_2 (which are themselves polynomials in formal indeterminate X) by again applying the Schwartz-Zippel lemma with respect to x_2; again, this leads to the restriction |\badch(\trprefix{\tr'}{x_2})|/|\ch| \leq \frac{n_q}{|\ch|} \leq \epsilon and we obtain the following polynomials of degree at most n - 1 for all i \in [0, n_q - 1)


\frac
  {q_i(X) - r_i(X)}
  {\prod\limits_{j=0}^{n_e - 1}
    \left(
      X - \omega^{\left(
        \mathbf{q_i}
      \right)_j} x
    \right)
  }

Having established that these are each non-rational polynomials of degree at most n - 1 we can then say (by the factor theorem) that for each i \in [0, n_q - 1] and j \in [0, n_e - 1] we have that q_i(X) - r_i(X) has a root at \omega^{\left(\mathbf{q_i}\right)_j} x. Note that we can interpret each q_i(X) as the restriction of a bivariate polynomial at the point x_1 whose degree with respect to x_1 is at most n_a + 1 and whose coefficients consist of various polynomials a'_i(X) (from the representation \repr{A'_i}{\mathbf{G}}) as well as h'(X) (from the representation \repr{H'_i}{\mathbf{G}}) and r(X) (from the representation \repr{R}{\mathbf{G}}). By similarly applying the Schwartz-Zippel lemma and restricting the challenge space with |\badch(\trprefix{\tr'}{x_1})|/|\ch| \leq \frac{n_a + 1}{|\ch|} \leq \epsilon we obtain (by construction of each q'_i(X) and r_i(X) in steps 12 and 13 of the protocol) that the prover's claimed value of r in step 9 is equal to r(x); that the value h computed by the verifier in step 13 is equal to h'(x); and that for all i \in [0, n_q - 1] the prover's claimed values (\mathbf{a_i})_j = a'_i(\omega^{(\mathbf{p_i})_j} x) for all j \in [0, n_e - 1].

By construction of h'(X) (from the representation \repr{H'}{\mathbf{G}}) in step 7 we know that h'(x) = h(x) where by h(X) we refer to the polynomial of degree at most (n_g - 1) \cdot (n - 1) whose coefficients correspond to the concatenated representations of each \repr{H_i}{\mathbf{G}}. As before, suppose that h(X) does not take the form g'(X) / t(X). Then because h(X) is determined prior to the choice of x then by the Schwartz-Zippel lemma we know that it would only agree with g'(X) / t(X) at (n_g - 1) \cdot (n - 1) points at most if the polynomials were not equal. By restricting again |\badch(\trprefix{\tr'}{x})|/|\ch| \leq \frac{(n_g - 1) \cdot (n - 1)}{|\ch|} \leq \epsilon we obtain h(X) = g'(X) / t(X) and because h(X) is a non-rational polynomial by the factor theorem we obtain that g'(X) vanishes over the domain D.

We now have that g'(X) vanishes over D but wish to show that g(X, C_0, C_1, \cdots) vanishes over D at all points to complete the proof. This just involves a sequence of applying the same technique to each of the challenges; since the polynomial g(\cdots) has degree at most n_g \cdot (n - 1) in any indeterminate by definition, and because each polynomial a_i(X, C_0, C_1, ..., C_{i - 1}, \cdots) is determined prior to the choice of concrete challenge c_i by similarly bounding |\badch(\trprefix{\tr'}{c_i})|/|\ch| \leq \frac{n_g \cdot (n - 1)}{|\ch|} \leq \epsilon we ensure that g(X, C_0, C_1, \cdots) vanishes over D, completing the proof.

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