1\documentclass{article}
  2\usepackage[utf8]{inputenc}
  3\usepackage{amsfonts}
  4\usepackage{amsthm}
  5\usepackage{amsmath}
  6\usepackage{enumerate}
  7\usepackage{hyperref}
  8\usepackage{amssymb}
  9\usepackage{tikz} % diagram
 10
 11\begin{filecontents}[overwrite]{pinocchio-groth16.bib}
 12@misc{cryptoeprint:2013/279,
 13  author = {Bryan Parno and Craig Gentry and Jon Howell and Mariana Raykova},
 14  title = {Pinocchio: Nearly Practical Verifiable Computation},
 15  howpublished = {Cryptology {ePrint} Archive, Paper 2013/279},
 16  year = {2013},
 17  url = {https://eprint.iacr.org/2013/279}
 18}
 19@misc{cryptoeprint:2016/260,
 20  author = {Jens Groth},
 21  title = {On the Size of Pairing-based Non-interactive Arguments},
 22  howpublished = {Cryptology {ePrint} Archive, Paper 2016/260},
 23  year = {2016},
 24  url = {https://eprint.iacr.org/2016/260}
 25}
 26
 27@misc{pairings,
 28  note = {\url{https://github.com/arnaucube/math/blob/master/weil-pairing.pdf}},
 29  url = {https://github.com/arnaucube/math/blob/master/weil-pairing.pdf}
 30}
 31\end{filecontents}
 32\nocite{*} % TODO review
 33
 34
 35\newtheorem{thm}{Theorem}[section]
 36\newtheorem{lemma}[thm]{Lemma}
 37\newtheorem{prop}[thm]{Proposition}
 38\newtheorem{cor}[thm]{Corollary}
 39\newtheorem{defn}[thm]{Definition}
 40
 41
 42\title{Pinocchio and Groth16 notes}
 43\author{arnaucube}
 44\date{2026}
 45
 46\begin{document}
 47
 48\maketitle
 49
 50\begin{abstract}
 51    Notes taken while studying the schemes Pinocchio (\cite{cryptoeprint:2013/279}) and Groth16 (\cite{cryptoeprint:2016/260}).
 52
 53    Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
 54
 55    The proofs may slightly differ from the ones from the books or papers, since I try to extend them for a deeper understanding.
 56\end{abstract}
 57
 58\tableofcontents
 59\newpage
 60
 61
 62\section{Introduction}
 63
 64% TODO.
 65% what we want to prove, how it connects with R1CS & QAP
 66
 67
 68\section{Notation}
 69
 70Let $g \in \mathbb{G}$ be the generator of the group $\mathbb{G}$, ie. $\mathbb{G}=<g>$, then denote with $[a]$ by $g^a \in \mathbb{G}$.
 71
 72\section{R1CS}
 73% TODO Add R_{R1CS} relation definition.
 74
 75Rank-One Constraint System (R1CS) is the representation of the constraints of the circuit by 3 linear combinations, where there is one constraint per operation.
 76
 77$$
 78\begin{aligned}
 79(a_{11}w_1 + a_{12}w_2 + ... + a_{1n}w_n) \cdot (b_{11}w_1 + b_{12}w_2 + ... + b_{1n}w_n)\\- (c_{11}w_1 + c_{12}w_2 + ... + c_{1n}w_n) = 0
 80\\
 81(a_{21}w_1 + a_{22}w_2 + ... + a_{2n}w_n) \cdot (b_{21}w_1 + b_{22}w_2 + ... + b_{2n}w_n)\\- (c_{21}w_1 + c_{22}w_2 + ... + c_{2n}w_n) = 0
 82\\
 83(a_{31}w_1 + a_{32}w_2 + ... + a_{3n}w_n) \cdot (b_{31}w_1 + b_{32}w_2 + ... + b_{3n}w_n)\\- (c_{31}w_1 + c_{32}w_2 + ... + c_{3n}w_n) = 0
 84\\
 85\ldots
 86\\
 87(a_{m1}w_1 + a_{m2}w_2 + ... + a_{mn}w_n) \cdot (b_{m1}w_1 + b_{m2}w_2 + ... + b_{mn}w_n)\\- (c_{m1}w_1 + c_{m2}w_2 + ... + c_{mn}w_n) = 0
 88\end{aligned}
 89$$
 90
 91Where the R1CS linear combinations can be represented by three matrices $A$, $B$ and $C$:
 92$$
 93\begin{aligned}
 94A =
 95\begin{pmatrix}
 96a_{11} & a_{12} & ... & a_{1n}\\
 97a_{21} & a_{22} & ... & a_{2n}\\
 98a_{31} & a_{32} & ... & a_{3n}\\
 99...\\
100a_{m1} & a_{m2} & ... & a_{mn}
101\end{pmatrix}
102B =
103\begin{pmatrix}
104b_{11} & b_{12} & ... & b_{1n}\\
105b_{21} & b_{22} & ... & b_{2n}\\
106b_{31} & b_{32} & ... & b_{3n}\\
107...\\
108b_{m1} & b_{m2} & ... & b_{mn}
109\end{pmatrix}
110\\
111C =
112\begin{pmatrix}
113c_{11} & c_{12} & ... & c_{1n}\\
114c_{21} & c_{22} & ... & c_{2n}\\
115c_{31} & c_{32} & ... & c_{3n}\\
116...\\
117c_{m1} & c_{m2} & ... & c_{mn}
118\end{pmatrix}
119\end{aligned}
120$$
121
122Where
123$(A, B, C) = A w \circ B w - C w = 0$
124
125The process that converts the \emph{flat constraints} into \emph{R1CS} is the following:
126
127Lets assume that we have a constraint $5 w_3 = w_1 \cdot 2 w_2$ (being $w_1$, $w_2$ and $w_3$ the signals 1, 2 and 3 respectively)). We can represent this constraint in the matrices form by:
128
129$$
130\begin{aligned}
131A =
132\begin{pmatrix}
1331 & 0 & 0 & ... & 0\\
134\end{pmatrix}
135\\
136B =
137\begin{pmatrix}
1380 & 2 & 0 & ... & 0\\
139\end{pmatrix}
140\\
141C =
142\begin{pmatrix}
1430 & 0 & 5 & ... & 0\\
144\end{pmatrix}
145\end{aligned}
146$$
147
148
149Which in the R1CS equation form would look as:
150$$
151\begin{aligned}
152(1 \cdot w_1 + 0 \cdot w_2 + 0 \cdot w_3 + ... + 0 \cdot w_n) \cdot
153(0 \cdot w_1 + 2 \cdot w_2 + 0 \cdot w_3 + ... + 0 \cdot w_n) -\\
154(0 \cdot w_1 + 0 \cdot w_2 + 5 \cdot w_3 + ... + 0 \cdot w_n) = 0
155\end{aligned}
156$$
157
158If we simplify the previous liniar combination we get:
159$$
160\begin{aligned}
161(1 \cdot w_1) \cdot (2 \cdot w_2) - (5 \cdot w_3) = 0
162\\
163w_1 \cdot 2w_1 - 5 w_3 = 0
164\\
1655 w_3 = w_1 \cdot 2w_1
166\end{aligned}
167$$
168Where $5 w_3 = w_1 \cdot 2w_1$ is the constraint that we wanted to represent in the R1CS format.
169
170The nice thing from R1CS is that allows us to standaridze the constraints in an equation fashion, where all the signals are involved and we 'activate' these ones that are used in each constraint.
171Since we have now the constraints in a polynomial format, we can do work with it in the polynomial way.
172
173
174\section{QAP (Quadratic Arithmetic Programs)} \label{qap}
175
176Once we have the constraints of a circuit represented in \emph{R1CS}, we can do one more iteration and represent it in \emph{Quadratic Arithmetic Programs (QAP)}.
177%% TODO %%
178\emph{QAP} is a representation of the XXX in a linear combination of the \emph{R1CS}, represented by 3 polynomials $a(X)$, $b(X)$ and $c(X)$.
179The \emph{QAP} consists of the 3 polynomials $a(X)$, $b(X)$ and $c(X)$, plus a divisor polynomial $h(X) \in \mathbb{F}[X]^{<d}$.
180
181The main idea behind the \emph{QAP} is that they 'contain' all the constraints of the circuit, meaning that these polynomials need to be able to generate all the constraints.
182
183In order to 'collapse' the \emph{R1CS} polynomials into the 3 polynomials of the \emph{QAP}, we make use of \emph{polynomial interpolation}, also named \emph{Lagrange interpolation}. \emph{Lagrange interpolation} allows for a group of points, to find the smallest degree and unique polynomial that goes through all those points. It can be calculated by computing:
184
185$$L(X) = \sum_{j=0}^{n} y_j l_j(X)$$
186
187where $l_j(X)$ is:
188$$
189l_j(X) = \prod_{0\leq m \leq k ; m\neq j}
190\frac{X-x_m}{x_j - x_m}
191...
192\frac{(X-x_{j-1})}{(x_j - x_{j-1})}
193\frac{(X-x_{j+1})}{(x_j - x_{j+1})}
194...
195\frac{X-x_k}{x_j - x_k}
196$$
197
198The intuition behind the \emph{Lagrange interpolation} is quite simple. If we have two points (on a 2 dimensional plane) we can find a line that goes through both points (this can be done with a pen and a ruler on a paper), if we have 3 points we can find a curved line that goes through all the three points, if we do this through math trying to get the lowest degree curve we obtain a  2nd degree polynomial. This can be generalized using the *Lagrange interpolation*, allowing us for a group of $n$ points to find a $n-1$ degree polynomial that goes through all of them.
199
200In our case, \emph{Lagrange interpolation} is used to go from the \emph{R1CS} polynomials into the \emph{QAP} polynomials, which are defined by:
201
202$$
203\begin{aligned}
204    a(X) = (a_1(X)w_1 + a_2(X)w_2 + ... + a_n(X)w_n) = \sum_{i=1}^n a_i(X) w_i\\
205    b(X) = (b_1(X)w_1 + b_2(X)w_2 + ... + b_n(X)w_n) = \sum_{i=1}^n b_i(X) w_i\\
206    c(X) = (c_1(X)w_1 + c_2(X)w_2 + ... + c_n(X)w_n) = \sum_{i=1}^n c_i(X) w_i
207\end{aligned}
208$$
209
210The polynomials $a(X), b(X), c(X)$ are the QAP, such that
211$$a(X) \cdot b(X) = c(X)$$
212so that
213$$p(X) = a(X) \cdot b(X) - c(X) = 0$$
214
215that is,
216$$
217\begin{aligned}
218    p(X) = &(a_1(X)w_1 + a_2(X)w_2 + ... + a_n(X)w_n)\\
219	 &\cdot (b_1(X)w_1 + b_2(X)w_2 + ... + b_n(X)w_n)\\
220	 &- (c_1(X)w_1 + c_2(X)w_2 + ... + c_n(X)w_n)\\
221    = &(\sum_{i=1}^n a_i(X) w_i) \cdot (\sum_{i=1}^n b_i(X) w_i) - (\sum_{i=1}^n c_i(X) w_i) = 0
222\end{aligned}
223$$
224
225The main idea, is that $p(X)$ defines the constraints of our circuit, so that when evaluating $p(X)$ at the challenge, it will be equal to zero.
226
227
228\vspace{0.5cm}
229
230\section{QAP Relation}
231
232Want to ensure that
233$$
234(\sum_{i=1}^n a_i(X) w_i) \cdot (\sum_{i=1}^n b_i(X) w_i) - (\sum_{i=1}^n c_i(X) w_i) = 0
235$$
236
237for a given witness $(w_1, \ldots, w_n) \in \mathbb{F}^n$, which we divide between public inputs and private inputs:
238\begin{align*}
239    \text{public inputs:}~~ \phi &= (w_1, \ldots, w_l) \in \mathbb{F}^l, \\
240    \text{private inputs:}~~ w &= (w_{l+1}, \ldots, w_n) \in \mathbb{F}^{n-l},
241\end{align*}
242
243We denote the relation
244
245$$
246R_{QAP} =
247\left\{
248\begin{aligned}
249(\phi, w) \ \big| \;&
250\phi = (w_1, \ldots, w_l) \in \mathbb{F}^l,\\
251& w = (w_{l+1}, \ldots, w_n) \in \mathbb{F}^{n-l},\\
252& (\sum_{i=1}^n a_i(X) w_i) \cdot (\sum_{i=1}^n b_i(X) w_i)
253\equiv
254\sum_{i=1}^{n} c_i(X) w_i 
255~\bmod Z(X)
256\end{aligned}
257\right.
258$$
259
260where $Z(X)$ is the "zero polynomial" or "vanishing polynomial".
261
262Set
263$$a(X) = \sum_{i=1}^n a_i(X) w_i,~~~~ b(X) = \sum_{i=1}^n b_i(X) w_i),~~~~ c(X) = \sum_{i=1}^{n} c_i(X) w_i$$
264
265so that
266$$a(X) \cdot b(X) \equiv c(X) \bmod Z(X)$$
267
268which means that we can prove that
269$$\exists~ h(X) \in \mathbb{F}[X]^{<d}~~ \text{s.th.}~
270a(X) \cdot b(X) - c(X) = h(X) \cdot Z(X)$$
271
272
273\section{Simplified protocol}
274Protocol to prove $a(X)b(X)-c(X)=0$ in order to provide intuition on what we're trying to do.
275
276\paragraph{Setup.}
277The protocol requires a secret random value, which it is encrypted over the elliptic curve group. That is, for given $g_1 \in \mathbb{G}_1$ and $g_2 \in \mathbb{G}_2$ generators of curves $\mathbb{G}_1$ and $\mathbb{G}_2$ respectively, set $[\tau]_1 = g_1^\tau$ and $[\tau]_2 =g_2^\tau$.
278
279$$\{ [\tau^i]_1 \}_0^n = ( [\tau^0]_1, [\tau^1]_1, [\tau^2]_1, \ldots, [\tau^{n-1}]_1)$$
280$$\{ [\tau^i]_2 \}_0^n = ( [\tau^0]_2, [\tau^1]_2, [\tau^2]_2, \ldots, [\tau^{n-1}]_2)$$
281
282\paragraph{Prover.}
283\begin{enumerate}[i.]
284    \item commit to $a(X), b(X), c(X) \in \mathbb{F}[X]$
285
286	$$cm(a) = [a(\tau)]_1 = \sum_{i=0}^{\delta a} a_i \cdot [\tau^i]_1 \in \mathbb{G}_1$$
287
288	notice that it is equal to $\sum a_i \cdot g_1^{\tau^i}$ without knowing $\tau$.
289	
290	Same is done for $b(X)$ and $c(X)$:
291	$$cm(b) = [b(\tau)]_2 = \sum_{i=0}^{\delta b} b_i \cdot [\tau^i]_2 \in \mathbb{G}_2$$
292	$$cm(c) = [c(\tau)]_1 = \sum_{i=0}^{\delta c} c_i \cdot [\tau^i]_1 \in \mathbb{G}_1$$
293
294
295	For a challenge $z \in^R \mathbb{F}$, set by the verifier, open the commitments:
296
297	$$q_a(X) = \frac{a(X) - a(z)}{X-z}$$
298
299	Since $a(X)-\underbrace{a(z)}_{=y}$ should be zero at $X-z$, so $a(X) - a(z)$ has a root at $z$, hence it is divisible by $(X-z)$, therefore,
300	$$\exists ~q_a(X) ~~\text{such that}~ a(X)-a(z) = (X-z) \cdot q_a(X)$$
301	therefore
302	$$q_a(X) = \frac{a(X)-a(z)}{X-z}$$
303
304	Same is done for $b(X)$ and $c(X)$,
305	$$q_b(X) = \frac{b(X)-b(z)}{X-z}$$
306	$$q_c(X) = \frac{c(X)-c(z)}{X-z}$$
307
308	And then, evaluate them at the encrypted $\tau$:
309	$$[q_a(\tau)]_1 = \sum_{i=0}^{\delta q_a} q_{a_i} \cdot [\tau^i]_1 \in \mathbb{G}_1$$
310	similarly for $[q_b(\tau)]_2 \in \mathbb{G}_2$ and $[q_c(\tau)]_1 \in \mathbb{G}_1$.
311
312    \item for the relation that we want to prove, $a(X) \cdot b(X) - c(X)=0$,
313	$$\exists~ q(X) \in \mathbb{F}[X] ~~\text{s.th.}~~ a(X)b(X)-c(X)=q(X) \cdot z(X)$$
314	So the prover computes
315	$$q(X) = \frac{a(X) \cdot b(X) - c(X)}{z(X)}$$
316	and it's encrypted evaluation
317	$$[q(\tau)]_1 = \sum_{i=0}^{\delta q} q_i \cdot [\tau^i]_1 \in \mathbb{G}_1$$
318\end{enumerate}
319
320\paragraph{Verifier.}
321\begin{enumerate}[i.]
322    \item verify the commitments:\\
323	The prover has sent the commitments $[a(\tau)]_1,~ [b(\tau)]_2,~ [c(\tau)]_1$, and their respective evaluations at the challenge $z \in \mathbb{F}$: $a(z), b(z), c(z) \in \mathbb{F}$; together with their respective opening proofs: $[q_a(\tau)]_1, [q_b(\tau)]_2, [q_c(\tau)]_1$
324
325	For each commitment, the verifier checks:
326	$$e([q_a(\tau)]_1, [\tau]_2 - [z]_2) \stackrel{?}{=} e([a(\tau)]_1 - [a(z)]_1, [1]_2)$$
327	which is a way of checking the existence of $q_a(X)$ such that $q_a(X) \cdot (X-z) = a(X) - a(z)$.
328
329	If the checks passes, it means that $q_a \in \mathbb{F}[X]$ exists, and thus $a(X) - a(z)$ is divisible ("vanishes") at $(X-z)$, hence $a(X) - a(z)$ has a root at $z$.
330
331	Same is done for $b(X)$ and $c(X)$ commitments:
332	$$e([\tau]_1 - [z]_2, [q_b(\tau)]_1) \stackrel{?}{=} e([1]_1, [b(\tau)]_1 - [b(z)]_1)$$
333	$$e([q_c(\tau)]_1, [\tau]_2 - [z]_2) \stackrel{?}{=} e([c(\tau)]_1 - [c(z)]_1, [1]_2)$$
334
335    \item check the $a(X) \cdot b(X) - c(X) = 0$ relation:
336	$$e([a(\tau)]_1, [b(\tau)]_2) \stackrel{?}{=} e([q(\tau)]_1, [z(\tau)]_2) \cdot e([c(s)]_1, [1]_2)$$
337	which ensures that $\exists~ q(X) \in \mathbb{F}[X] ~\text{s.th.}~ a(X) b(X) - c(X) = q(X) z(X)$.
338\end{enumerate}
339
340\section{Pinocchio}
341\section{Groth16}
342
343\vspace{1cm}
344
345\section{Powers of tau (trusted setup)}
346\subsection{Main idea}
347% TODO. mention "SRS" concept
348
349\subsection{Participant contribution}
350In the context of SRS (Structured Reference String), the powers of tau consist of an array containing the scalar multiplication of the generator point by each power of tau, eg. $\tau^0 \cdot G,~\tau^1 \cdot G,~\tau^2 \cdot G,~\tau^3 \cdot G \ldots$
351
352For our case, we'll generate $n$ powers of tau on $\mathbb{G}_1$, and $m$ on $\mathbb{G}_2$. We will denote the combination of both $\mathbb{G}_1$ and $\mathbb{G}_2$ powers of tau by SRS.
353
354Each participant will obtain the previous-participant generated SRS, which is formed by
355
356\begin{align*}
357        \{ [\tau^0]_1, [\tau^1]_1, [\tau^2]_2, \ldots, [\tau^{n-1}]_1 \},\\
358        \{ [\tau^0]_2, [\tau^1]_2, [\tau^2]_2, \ldots, [\tau^{m-1}]_2 \}
359\end{align*}
360
361for simplicity, we will represent it as
362
363$$\{~[~\tau^i]_1~\}^{n-1}_{i=0},~\{~[~\tau^i]_2~\}^{m-1}_{j=0}$$
364
365
366Each participant generates their secret random tau $\tau_p$, which to avoid confusion we will denote as $p \in \mathbb{F}_r$. From it, the participant can compute $[p]_2 \in \mathbb{G}_2$.
367
368Then the participant proceeds to update the reference string (SRS), by multiplying each element by $p^i$:
369$$\{~ p^i \cdot [~ \tau^i]_1~\}^{n-1}_{i=0},~\{~ p^i \cdot [~\tau^i]_2~\}^{m-1}_{j=0}$$
370which, by the properties of scalar multiplication of a point, is equivalent to
371$$\{~ [~ (p \cdot \tau)^i]_1~\}^{n-1}_{i=0},~\{~ [~ (p \cdot \tau)^i]_2~\}^{m-1}_{j=0}$$
372and we denote $p \cdot \tau$ as $\tau'$, so we can write the previous expression as
373$$\{~[~\tau'^i]_1~\}^{n-1}_{i=0},~\{~[~\tau'^i]_2~\}^{m-1}_{j=0}$$
374
375which is
376\begin{align*}
377        \{ [(\tau \cdot p)^0]_1, [(\tau \cdot p)^1]_1, [(\tau \cdot p)^2]_2, \ldots, [(\tau \cdot p)^{n-1}]_1 \},\\
378        \{ [(\tau \cdot p)^0]_2, [(\tau \cdot p)^1]_2, [(\tau \cdot p)^2]_2, \ldots, [(\tau \cdot p)^{m-1}]_2 \}
379\end{align*}
380
381Notice that the contributor does not know the previous $\tau$, but can obtain $(\tau \cdot p)^i$.
382
383
384\subsection{Proof of correct computation}
385The participant will compute the proof, which consists of
386$$\pi = ([\tau']_1,~[p]_2)$$
387which is equivalent to $\pi = (\tau' \cdot G,~p \cdot H) = ((p \cdot \tau) \cdot G,~p \cdot H)$
388
389In the next section we will see how this is used to verify the contribution correct computation.
390
391
392\subsection{Contribution verification}
393This is the interesting part. We want to check that the new SRS is well computed from the previous SRS together with the new tau. We want to avoid accepting contributions that do not follow the powers of tau structure, or that contain empty elements or that do not derive from the previous SRS.
394
395Following the powers of tau structure means that
396$$\tau'^{i+1} = \tau' \cdot \tau'^i ~~\forall~i \in [1, n-1]$$
397
398And the relation between the new SRS and the previous SRS is
399$$\tau' = \tau \cdot p$$
400where $\tau$ is the secret element from the previous SRS and $p$ is the secret one from the new SRS, resulting in the combined $\tau'$.
401
402Now, we're interested into how we can check these relations in the context of the SRS elements. A new contribution can be verified as follows:
403
404\begin{enumerate}
405        \item Check that elements of the new SRS are non-empty, non-zero and in the correct prime order subgroups.
406\item Get the proof ($\pi = ([\tau']_1, [p]_2)$), and confirm that the 1st element is equal to the element in position 1 $[\tau'^1]_1$ of the new SRS (which corresponds to $\tau'$ powered to $1$).
407
408\item Check that $\tau'$ (the new SRS) is correctly related to $\tau$ (the previous SRS).
409
410    To check this, we rely on the pairing properties and we use the proof value generated by the contributor, which consists of $[p]_2$
411    $$e([\tau]_1, [p]_2) \stackrel{?}{=} e([\tau']_1, [1]_2)$$
412    
413    We can see how this holds, as $\tau' = p \cdot \tau$, and by the bilinear property of the pairing (see \cite{pairings}) we have $e(a \cdot G, b \cdot H) = e((a \cdot b) \cdot G, H)$, which in our notation is represented as
414    $$e([a]_1, [b]_2) = e([a \cdot b]_1, [1]_2)$$
415    by applying this to our case we can see that:
416    $$e([\tau]_1, [p]_2) = e([\tau \cdot p]_1, [1]_2) = e([\tau']_1, [1]_2)$$
417
418\item Check if the new SRS follows the powers of tau structure:
419    $$e([\tau'^i]_1, [\tau']_2) \stackrel{?}{=} e([\tau'^{i+1}]_1, [1]_2) ~~ \forall ~i \in [1, n-1]$$
420    $$e([\tau']_1, [\tau'^j]_2) \stackrel{?}{=} e([1]_1, [\tau'^{j+1}]_2) ~~ \forall ~i \in [1, m-1]$$
421    
422    Which, again, by the bilinarity property, we can see that
423    $$e([\tau'^i]_1, [\tau']_2) = e([\tau'^{i+1}]_1, [1]_2)$$
424    is enforcing that $\tau'^i \cdot \tau'^1 = \tau'^{i+1}$.
425    
426    And the same for
427    $$e([\tau']_1, [\tau'^j]_2) = e([1]_1, [\tau'^{j+1}]_2)$$
428    which checks that $\tau'^1 \cdot \tau'^j = \tau'^{j+1}$.
429    
430    With this check, we're ensuring that each power of tau of the SRS is consistent with the previous one.
431\end{enumerate}
432
433
434
435\bibliographystyle{unsrt}
436\bibliography{pinocchio-groth16.bib}
437
438\end{document}