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# Example 1

### PFR Proof

The Prover interpolates polynomial $$h$$ such that $$seq\_{\mathbb{K}}(h)=\omega^t,\omega^{t+1},..,\omega^{n-1},0,..,0$$. Here $$t=2$$ and $$n=5$$, therefore $$seq\_{\mathbb{K}}(h)=\omega^2,\omega^3,\omega^4,0,..,0$$. This means that $${h(k):\hspace{1mm}k\in\mathbb{K}={1,43,39,48,73,62,132,65,80}}={\omega^2,\omega^3,\omega^4,0,..,0}={42,125,135,0,0,0,0,0,0}$$Therefore $$h(1)=42$$, $$h(43)=125$$ , $$h(39)=135$$, $$h(48)=h(73)=h(62)=h(132)=h(65)=h(80)=0$$. So $$h(x)=42L\_1(x)+125L\_2(x)+135L\_3(x)$$ where $$L\_1(x)=\frac{(x-43)(x-39)(x-48)(x-73)(x-62)(x-132)(x-65)(x-80)}{(-42)(-38)(-47)(-72)(-61)(-131)(-64)(-79)}=161x^8+161x^7+161x^6+161x^5+161x^4+161x^3+161x^2+161x+161$$,

$$L\_2(x)=45x^8+125x^7+126x^6+169x^5+27x^4+75x^3+148x^2+29x+161$$,

$$L\_3(x)=125x^8+169x^7+75x^6+29x^5+45x^4+126x^3+27x^2+148x+161$$,

Therefore $$h(x)=121x^8+133x^7+57x^6+127x^5+103x^4+24x^3+128x^2+140x+114$$. The Prover sends $$\pi\_1=(h\_0,h\_1,...,h\_8)$$ where $$h\_i$$ is coefficients of polynomial $$h(x)$$ to the Verifier.

#### 3-5-1-2- Step 2

The Prover and the Verifier do $$Geometric\hspace{1mm}Sequence\hspace{1mm}Test$$ on $$h$$ (to verify correct construction of polynomial $$h$$). In $$Geometric\hspace{1mm} Sequence\hspace{1mm} Test$$ want to prove that $$Seq\_{\mathbb{K}}(h)=a\_1,a\_1r,..,a\_1r^{c\_1-1},a\_2, a\_2r,..., a\_2r^{c\_2-1},...,a\_v,a\_vr,...,a\_vr^{c\_v-1}$$.

#### 3-5-1-2- Step 2-A- Geometric Sequence Test

Since $$h(\gamma^0)=\omega^2$$, $$h(\gamma^1)=\omega^3$$ ,$$h(\gamma^2)=\omega^4$$ and $$h(\gamma^3)=...=h(\gamma^8)=0$$, then Geometric Sequence Test with $$a\_1=\omega^2$$, $$a\_2=0$$, $$r=\omega$$, $$c\_1=n-t=3$$ and $$c\_2=m-(n-t)=9-3=6$$ run. Let $$p\_i=\sum\_{j\<i}c\_j$$ for $$i\in {1,2,..,v}$$. Here $$v$$ is the number of $$c\_i$$ that are non-zero. Therefore $$v=2$$ , $$p\_1=0$$ and $$p\_2=c\_1=3$$.

The Prover and the Verifier run $$Zero\hspace{1mm}Over\hspace{1mm}\mathbb{K}$$ (see the next section) to check that for all $$k\in\mathbb{K}$$, have $$(h(\gamma k)-rh(k))\prod\_{i=1}^{v}(k-\gamma^{p\_i+c\_i-1})=0$$. Here, $$Zero\hspace{1mm}Over\hspace{1mm}\mathbb{K}$$ run to check that for all $$k\in\mathbb{K}={1,\gamma,\gamma^2,..,\gamma^8}$$, have $$(h(\gamma k)-\omega h(k))(k-\gamma^2)(k-\omega^8)=0$$.\
$$Note$$: It is clear that if the construction of $$h$$ is correct, it applies this equality, because if $$k=1$$, $$h(\gamma k)=h(\gamma)=\omega h(1)$$ then $$h(\gamma k)-\omega h(k)=0$$. If $$k=\gamma$$, $$h(\gamma k)=h(\gamma^2)=\omega h(\gamma)$$, then $$h(\gamma k)-\omega h(k)=0$$. If $$k=\gamma^2$$, then $$k-\gamma^2=0$$. If $$k=\omega^3,...,\omega^7$$, then $$h(\gamma k)=\omega h(k)=0$$ and if $$k=\gamma^8$$, then $$k-\gamma^8=0$$.

#### 3-5-1-2-Step 2-B- Zero over K

In $$Zero\hspace{1mm}Over\hspace{1mm}\mathbb{K}$$, want to prove that for all $$k\in\mathbb{K}$$, $$F(k)=0$$ where $$F(x)=G(x,f\_{j\_1}(\alpha\_1x),f\_{j\_2}(\alpha\_2x),...,f\_{j\_{t}}(\alpha\_{t}x))$$. Here $$F(x)=(h(\gamma x)-\omega h(x))(x-\gamma^2)(x-\gamma^8)$$, therefore since $$t=2$$, $$F(x)=G(x,f\_{j\_1}(\alpha\_1x),f\_{j\_2}(\alpha\_2x))=(h(\gamma x)-\omega h(x))(x-\gamma^2)(x-\gamma^8)$$, so $$h\_1=f\_{j\_1}=h$$, $$h\_2=f\_{j\_2}=h$$, $$\alpha\_1=\gamma$$ and $$\alpha\_2=1$$.

2-2-1 The Prover selects polynomials $$r\_i\in \mathbb{F}^{<2}\[x]$$, $$i\in {1,2,..,t}$$. Then computes masks $$m\_i(x)=r\_i(\alpha\_i^{-1}x)Z\_{\mathbb{K}}(\alpha\_i^{-1}x)$$ and computes $$h'\_i(x)=h\_i(x)+m\_i(x)$$ for $$i\in{1,2,..,t}$$.

Here, the Prover selects $$r\_1,r\_2$$. For example, let $$r\_1(x)=2+x$$ and $$r\_2(x)=2x$$. Therefore $$m\_1(x)=r\_1(\alpha\_1^{-1}x)Z\_{\mathbb{K}}(\alpha\_1^{-1}x)$$ where $$\alpha\_1=\gamma=48$$ and $$Z\_{\mathbb{K}}(x)=\prod\_{x\in\mathbb{K}}(x-k)=(x-1)(x-43)(x-39)(x-48)(x-73)(x-62)(x-132)(x-65) (x-80)=x^9+180$$. Therefore $$m\_1(x)=r\_1(80x)Z\_{\mathbb{K}}(80x)=80x^{10}+2x^9+101x+179$$\
$$m\_2(x)=r\_2(\alpha\_2^{-1}x)Z\_{\mathbb{K}}(\alpha\_2^{-1}x)=r\_2(x)Z\_{\mathbb{K}}(x)=2x(x^9+180)=2x^{10}+179x$$. Then computes $$h'\_1(x)=h\_1(x)+m\_1(x)=h(x)+m\_1(x)=80x^{10}+2x^9+121x^8+133x^7+57x^6+127x^5+103x^4+$$\
$$24x^3+128x^2+60x+112$$ and $$h'\_2(x)=h\_2(x)+m\_2(x)=h(x)+m\_2(x)=2x^{10}+121x^8+133x^7+$$\
$$57x^6+127x^5+103x^4+24x^3+128x^2+138x+114$$.

2-2-2- The Prover computes $$F'(x)=G(x,h'*1(\alpha\_1x),h'*2(\alpha\_2x),...,h'*t(\alpha\_tx))$$ and $$q\_1(x)=\frac{F'(x)}{Z*{\mathbb{K}}(x)}$$ then the Prover sends $$\pi*{PFR*{1+i}}=(m\_{i1},...,m\_{ideg(m\_i)})$$ where $$m\_{ij}$$s are coefficients of polynomial $$m\_i(x)$$, $$\pi\_{PFR\_{1+t+i}}=(r\_{i1},...,r\_{ideg(r\_i)})$$ where $$r\_{ij}$$s are coefficients of polynomial $$r\_i(x)$$ and $$\pi\_{PFR\_{2t+2}}=(q\_{11},...,q\_{1deg(q\_1)})$$ where $$q\_{1j}$$s are coefficients of polynomial $$q\_1(x)$$

Here $$F'(x)=G(x,h'\_1(43x),h'\_2(x))=(h'\_1(43x)-\omega h'*2(x))(x-\gamma^2)(x-\gamma^8)=64x^{12}+169x^{11}+168x^{10}$$\
$$+51x^9+117x^3+12x^2+13x+130$$\
and then $$q\_1(x)=\frac{F'(x)}{Z*{\mathbb{K}}(x)}=64x^3+169x^2+168x+51$$.

The Prover sends $$\pi\_{PFR\_{2}}=(179,101,0,0,0,0,0,0,0,2,80)$$, $$\pi\_{PFR\_{3}}=(0,179,0,0,0,0,0,0,0,0,2)$$ , $$\pi\_{PFR\_4}=(2,1)$$, $$\pi\_{PFR\_5}=(0,2)$$ and $$\pi\_{PFR\_6}=(51,168,169,64)$$ to the Verifier.

2-2-3- The Verifier derives $$h'\_i=h\_i+m\_i$$ through additive homomorphism. Then samples $$\beta\_1$$, $$\beta\_2$$ and $$c$$ of $$\mathbb{F}^{\*}-\mathbb{K}$$ and sends $$c$$ to the Prover. Suppose $$\beta\_1=65$$, $$\beta\_2=21$$ and $$c=171$$. The Verifier sends $$c=171$$ to the Prover.

2-2-4- The Prover computes $$q\_2=r\_1+cr\_2+...+c^{t-1}r\_t$$ and the Verifier derives concrete oracle $$q\_2$$ through additive homomorphism. Here, the Prover computes $$q\_2(x)=r\_1(x)+cr\_2(x)=2+x+171(2x)=162x+2$$.

2-2-5- Let $$M(x)=m\_1(\alpha\_1 x)+cm\_2(\alpha\_2 x)+...+c^{t-1}m\_t(\alpha\_t x)$$. The Verifier computes $$Z\_{\mathbb{K}}(\beta\_1)$$, $$Z\_{\mathbb{K}}(\beta\_2)$$, queries $$q\_1(\beta\_1)$$and $$q\_2(\beta\_2)$$ and for $$i\in {1,2,..,t}$$, queries $$h'*i(\alpha\_i\beta\_1)$$ and $$m\_i(\alpha\_i\beta\_2)$$to compute $$F'(\beta\_1)$$ and $$M(\beta\_2)$$. The Verifier asserts two identities $$M(\beta\_2)-q\_2(\beta\_2)Z*{\mathbb{K}}(\beta\_2)=0$$ and $$F'(\beta\_1)-q\_1(\beta\_1)Z\_{\mathbb{K}}(\beta\_1)=0$$.

Here, $$M(x)=m\_1(\alpha\_1 x)+cm\_2(\alpha\_2 x)=m\_1(\gamma x)+171m\_2(x)=162x^{10}+2x^9+19x+179$$. The Verifier computes $$Z\_{\mathbb{K}}(\beta\_1)=Z\_{\mathbb{K}}(65)=65^9+180=0$$ and $$Z\_{\mathbb{K}}(\beta\_2)=Z\_{\mathbb{K}}(21)=21^9+180=30$$ and queries $$q\_1(\beta\_1)=86$$ and $$q\_2(\beta\_2)=146$$. Also queries values $$h'\_1(\alpha\_1\beta\_1)=h'\_1(\gamma\times 65)=h'\_1(80)=0$$ , $$h'*2(\alpha\_2\beta\_1)=h'*2(65)=0$$, $$m\_1(\alpha\_1\beta\_2)=m\_1(\gamma\times 21)=m\_1(179)=147$$ and $$m\_2(\alpha\_2\beta\_2)=m\_2(21)=174$$. Then checks$$M(\beta\_2)-q\_2(\beta\_2)Z*{\mathbb{K}}(\beta\_2)=0$$ , that means $$36-30\times 146=36-36\equiv 0\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, and $$F'(\beta\_1)-q\_1(\beta\_1)Z*{\mathbb{K}}(\beta\_1)=0$$, that means $$0-86\times 0\equiv 0\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$.

3- The Prover and the Verifier run $$Subset\hspace{1mm}over\hspace{1mm}\mathbb{K}$$ between $$row\_A$$ and $$h$$ to prove that $$row\_A(\mathbb{K})\subset h(\mathbb{K})$$ where $$row\_A(\mathbb{K})={row\_A(k):\hspace{1mm}k\in\mathbb{K}}$$and $$h(\mathbb{K})={h(k)\hspace{1mm}:\hspace{1mm}k\in\mathbb{K}}$$.

#### 3-5-1-2- Step 3- Subset over $$\mathbb{K}$$

#### .................

4- The Prover and the Verifier run $$Discrete-log\hspace{1mm} Comparison$$ between $$row\_A$$ and $$col\_A$$ as following:

#### 3-5-1-2- Step 4- Discrete-log Comparison

Let parameters $$(\Delta \in \mathbb{F}, n = |\mathbb{H}|)$$ such that $$ord(\Delta) = 2n$$ and $$\Delta^2=\omega$$.\
Here, $$\omega=59$$, $$\Delta=56$$ and $$ord(\Delta)=2n=10$$.

3-5-1-2- Step 4-Substep A- The Prover interpolates polynomial $$s$$ so that agrees with $$\frac{row\_A}{col\_A}$$ on $$\mathbb{K}$$. Then send them to the Verifier.

Here, $$s(1)=\frac{row\_A(1)}{col\_A(1)}=\frac{42}{59}\equiv 59\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, $$s(43)=\frac{row\_A(43)}{col\_A(43)}\equiv125\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, $$s(39)=\frac{row\_A(39)}{col\_A(39)}=\frac{135}{125}\equiv59\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, $$s(48)=\frac{row\_A(48)}{col\_A(48)}=\frac{48}{45}\equiv 170\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$,\
$$s(73)=\frac{row\_A(73)}{col\_A(73)}=\frac{36}{22}\equiv 51\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, $$s(62)=\frac{row\_A(62)}{col\_A(62)}=\frac{151}{166}\equiv 2\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$,\
$$s(132)=\frac{row\_A(132)}{col\_A(132)}=\frac{21}{46}\equiv 28\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, $$s(65)=\frac{row\_A(65)}{col\_A(65)}=\frac{131}{127}\equiv 135\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$ and\
$$s(80)=\frac{row\_A(80)}{col\_A(80)}=\frac{6}{40}\equiv 154\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$. Therefore, $$s(x)=59L\_1(x)+125L\_2(x)+59L\_3(x)+170L\_4(x)+51L\_5(x)+2L\_6(x)+28L\_7(x)+135L\_8(x)+$$\
$$154L\_9(x)$$.

where $$L\_1(x)$$, $$L\_2(x)$$ and $$L\_3(x)$$ are computed in step $$1$$ and the rest of $$L\_i(x)s$$ are\
$$L\_4(x)=126x^8+75x^7+161x^6+126x^5+75x^4+161x^3+126x^2+75x+161$$,\
$$L\_5(x)=169x^8+29x^7+126x^6+148x^5+125x^4+75x^3+45x^2+27x+161$$,\
$$L\_6(x)=27x^8+45x^7+75x^6+125x^5+148x^4+126x^3+29x^2+169x+161$$,\
$$L\_7(x)=75x^8+126x^7+161x^6+75x^5+126x^4+161x^3+75x^2+126x+161$$,\
$$L\_8(x)=148x^8+27x^7+126x^6+45x^5+29x^4+75x^3+169x^2+125x+161$$and\
$$L\_9(x)=29x^8+148x^7+75x^6+27x^5+169x^4+126x^3+125x^2+45x+161$$.

Therefore $$s(x)=41x^8+101x^7+34x^6+38x^5+x^4+25x^3+152x^2+123x+87$$. The Prover sends $$s(x)$$ to the Verifier.

3-5-1-2- Step 4-Substep B- For $$b\in{row\_A, col\_A, s}$$, the Prover interpolates polynomial $$b'$$ so that for all $$k\in \mathbb{K}$$, $$b'(k)=\Delta^{\log\_{\omega}^{b(k)}}$$.

Here, if $$b=row\_A$$, $$row'*A(k)=\Delta^{\log*{\omega}^{row\_A(k)}}=56^{\log\_{59}^{row\_A(k)}}$$that means $$row'*A(1)=56^{\log*{59}^{42}}=56^2\equiv 59\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, $$row'*A(43)=56^{\log*{59}^{125}}=56^3\equiv 46\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$,\
$$row'*A(39)=56^{\log*{59}^{135}}=56^4\equiv 42\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, $$row'*A(48)=56^{\log*{59}^{48}} \equiv 49\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$,\
$$row'*A(73)=56^{\log*{59}^{36}} \equiv 114\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, $$row'*A(62)=56^{\log*{59}^{151}} \equiv \hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$.\\

Therefore $$row'\_A(x)=59L\_1(x)+46L\_2(x)+42L\_3(x)+49L\_4(x)+114L\_5(x)+...$$.

if $$b=col\_A$$, $$col'*A(k)=\Delta^{\log*{\omega}^{col\_A(k)}}=56^{\log\_{59}^{col\_A(k)}}$$that means $$col'*A(1)=56^{\log*{59}^{59}}=56^1\equiv 56\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, $$col'*A(48)=56^{\log*{59}^{1}}=56^5\equiv 180\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$ and $$col'*A(132)=56^{\log*{59}^{125}}=56^3\equiv 46\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$. Therefore $$col'\_A(x)=56L\_1(x)+180L\_2(x)+46L\_3(x)=96x^2+47x+94$$.

if $$b=s$$, $$s'(k)=\Delta^{\log\_{\omega}^{s(k)}}=56^{\log\_{59}^{s(k)}}$$that means $$s'(1)=56^{\log\_{59}^{59}}=56^1\equiv 56\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, $$s'(48)=56^{\log\_{59}^{125}}=56^3\equiv 46\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$ and $$s'(132)=56^{\log\_{59}^{59}}=56^1\equiv 56\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$. Therefore $$s'(x)=56L\_1(x)+46L\_2(x)+56L\_3(x)=21x^2+103x+113$$.

Then the Prover sends $$row'\_A$$, $$col'\_A$$ and $$s'$$ to the Verifier.

3-5-1-2- Step 4-Substep C- The Prover interpolates polynomial $$h$$ so that $$seq\_{\mathbf{K}}(h)=1,\Delta,\Delta^2,..,\Delta^{n-1},0,0,..,0$$.

Here $$seq\_{\mathbf{K}}(h)=1, 56, 59, 46, 42$$.

### AHP Proof

Since in this example $$n\_i=1$$, therefore input and output are not vector. So, we denote input by $$x$$ and output by $$y$$ instead of $$X$$ and $$Y$$.

$$Proof (\mathbb{F}\_{181}, \mathbb{H}={1,59,42,125,135}, \mathbb{K}={1,49,48,180,132,133}, A, B, C, x=4,W=(20,31),y=82)$$

1- The Prover puts $$x=4$$ in $$Com\_{AHP\_X}^1$$ . We consider $$z=(1,x,w\_1,w\_2,y)$$ where $$w\_1=5x$$, $$w\_2=w\_1+11$$ and $$y=26w\_2$$. Therefore $$z=(1,x,W,y)=(1,4,20,31,82)$$. The Prover calculates $$z\_A=Az=\begin{bmatrix}0&0&0&0&0\0&0&0&0&0\0&1&0&0&0\1&0&0&0&0\0&0&0&1&0\ \end{bmatrix}\begin{bmatrix}1\4\20\31\82\\\end{bmatrix}=\begin{bmatrix}0\0\4\1\31\\\end{bmatrix}$$, $$z\_B=Bz=\begin{bmatrix}0&0&0&0&0\0&0&0&0&0\5&0&0&0&0\11&0&1&0&0\26&0&0&0&0\\\end{bmatrix}\begin{bmatrix}1\4\20\31\82\end{bmatrix}=\begin{bmatrix}0\0\5\31\26\\\end{bmatrix}$$and $$z\_C=Cz=\begin{bmatrix}0&0&0&0&0\0&0&0&0&0\0&0&1&0&0\0&0&0&1&0\0&0&0&0&1\\\end{bmatrix}\begin{bmatrix}1\4\20\31\82\\\end{bmatrix}=\begin{bmatrix}0\0\20\31\82\\\end{bmatrix}$$

2- The Prover calculates the polynomial $$z\_A(x)$$using indexing $$z\_A$$ by elements of $$\mathbb{H}$$ that mean $$z\_A(x)$$ is the polynomial where $$z\_A(1)=0$$, $$z\_A(59)=0$$, $$z\_A(42)=4$$, $$z\_A(125)=1$$ and $$z\_A(135)=31$$.

Then calculates polynomial $$\hat{z}\_A(x)$$ using the polynomial $$z\_A(x)$$ such that $$\hat{z}\_A(x)\in \mathbb{F}^{<|\mathbb{H}|+b}\[x]$$ that agree with $$z\_A(x)$$ on $$\mathbb{H}$$ . Note that values of up to $$b$$ locations in this polynomial reveals no information about the witness $$w$$provided the locations are in $$\mathbb{F}-\mathbb{H}$$.\
Here, for simplicity, let $$b=2$$. The Prover calculates $$\hat{z}\_A(x)$$ such that agree with $$z\_A(x)$$ on $$\mathbb{H}$$ and also $$\hat{z}\_A(150)=5$$, $$\hat{z}\_A(80)=47$$.

Therefore, we have $$\hat{z}\_A(x)=4L\_3(x)+L\_4(x)+31L\_5(x)+5L\_6(x)+47L\_7(x)$$ where $$L\_3(x)=133x^6+155x^5+117x^4+27x^3+48x^2+73x+171$$, $$L\_4(x)=4x^6+123x^5+25x^4+48x^3+27x^2+113x+22$$, $$L\_5(x)=75x^6+115x^5+27x^4+25x^3+117x^2+154x+30$$, $$L\_6(x)=132x^6+119x^5+49x+62$$ and $$L\_7(x)=97x^6+111x^5+84x+70$$.\
So $$\hat{z}\_A(x)=116x^6+165x^5+63x^4+26x^3+45x^2+141x+168$$.

Similarly, calculates polynomial $$\hat{z}\_B(x)$$ so that $$\hat{z}\_B(x)\in \mathbb{F}^{<|\mathbb{H}|+b}\[x]$$ that agree with $$z\_B(x)$$ on $$\mathbb{H}$$ that mean $$\hat{z}\_B(1)=0$$, $$\hat{z}\_B(59)=0$$, $$\hat{z}\_B(42)=5$$, $$\hat{z}\_B(125)=31$$, $$\hat{z}\_B(135)=26$$ and also $$b=2$$ random locations $$\hat{z}\_B(150)=15$$ and $$\hat{z}\_B(80)=170$$. So, $$\hat{z}\_B(x)=5L\_3(x)+31L\_4(x)+26L\_5(x)+15L\_6(x)+170L\_7(x)=32x^6+178x^5+71x^4+101x^3+137x^2+81x+124$$

Similarly, calculates polynomial $$\hat{z}\_C(x)$$ such that $$\hat{z}\_C(x)\in \mathbb{F}^{<|\mathbb{H}|+b}\[x]$$ that agree with $$z\_C(x)$$ on $$\mathbb{H}$$ that mean $$\hat{z}\_C(1)=0$$, $$\hat{z}\_C(59)=0$$, $$\hat{z}\_C(42)=20$$, $$\hat{z}\_C(125)=31$$, $$\hat{z}\_C(135)=82$$ and also $$b=2$$ random locations $$\hat{z}\_C(150)=1$$ and $$\hat{z}\_C(80)=100$$. So $$\hat{z}\_C(x)=20L\_3(x)+31L\_4(x)+82L\_5(x)+L\_6(x)+100L\_7(x)=123x^6+50x^5+80x^4+96x^3+169x^2+157x+49$$

The Prover calculates polynomial $$\hat{W}(x)\in \mathbb{F}^{\<n\_g+b}\[x]$$ that agree with $$\bar{W}(x)$$ on $$\mathbb{H}\[>|x|+1]$$ where

$$\bar{W}:\mathbb{H}\[>|x|+1]={42,125,135}\to \mathbb{F}$$

$$\bar{W}(h)=\frac{W(h)-\hat{x}(h)}{v\_{\mathbb{H}\[\leq |x|+1]}(h)}$$

and $$v\_{\mathbb{H}\[\leq |x|+1]}(h)$$ is vanishing polynomial on $$\mathbb{H}\[\leq |x|+1]={1,59}$$, therefore $$v\_{\mathbb{H}\[\leq |x|+1]}(h)=(h-1)(h-59)$$. Also $$\hat{x}(h)$$ is the polynomial such that $$\hat{x}(1)=1$$ and $$\hat{x}(59)=4$$, therefore $$\hat{x}(h)=128h+54$$. Therefore,

$$\bar{W}(42)=\frac{W(42)-\hat{x}(42)}{(42-1)(42-59)}=\frac{20-0}{(41)(-17)}\equiv 108\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$, $$\bar{W}(125)=\frac{W(125)-\hat{x}(125)}{(125-1)(125-59)}=\frac{31-126}{(124)(66)}\equiv 160\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$ and $$\bar{W}(135)=\frac{W(135)-\hat{x}(135)}{(135-1)(135-59)}=\frac{82-139}{(134)(76)}\equiv 78\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$.

Now, calculates $$\hat{W}(x)\in\mathbb{F}^{\<n\_g+b=3+2}\[x]$$ so that $$\hat{W}(42)=108$$, $$\hat{W}(125)=160$$ , $$\hat{W}(135)=78$$ and also $$b=2$$ random locations $$\hat{W}(150)=42$$ and $$\hat{W}(80)=180$$. Therefore, $$\hat{W}(x)=108L\_1(x)+160L\_2(x)+78L\_3(x)+42L\_4(x)+180L\_5(x)$$ where $$L\_1(x)=\frac{(x-125)(x-135)(x-150)(x-80)}{(-83)(-93)(-108)(-38)}\equiv 152x^4+92x^3+73x^2+43x+112\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$$,\
$$L\_2(x)=156x^4+39x^3+84x^2+86x+171$$, $$L\_3(x)=161x^4+157x^3+131x^2+159x+6$$,\
$$L\_4(x)=60x^4+67x^3+118x^2+50x+20$$ and $$L\_5(x)=14x^4+7x^3+137x^2+24x+54$$. Therefore,\
$$\hat{W}(x)=149x^4+97x^3+161x^2+121x+166$$.

3- The Prover finds polynomial $$h\_0(x)$$ so that $$\hat{z}\_A(x)\hat{z}\_B(x)-\hat{z}*C(x)=h\_0(x)v*{\mathbb{H}}(x)$$. Since $$\hat{z}\_A(x)\hat{z}*B(x)-\hat{z}*C(x)=$$\
$$92x^{12}+45x^{11}+164x^{10}+x^9+20x^8+61x^7+152x^6+49x^5+180x^4+161x^3+28x^2+165x+149$$ and $$v*{\mathbb{H}}(x)=\prod*{h\in\mathbb{H}}(x-h)=(x-1)(x-59)(x-42)(x-125)(x-135)=x^5+180$$, The Prover finds\
$$h\_0(x)=92x^7+45x^6+164x^5+x^4+20x^3+153x^2+16x+32$$.

4- The Prover samples a fully random $$s(x)\in\mathbb{F}^{<2|\mathbb{H}|+b-1=11}\[x]$$. Consider $$s(x)=5x^{10}+101x^8+17x^7+x^5+20x^4+3x+115$$. Then, the Prover computes sum $$\sigma\_1=\sum\_{k\in \mathbb{H}}s(k)=$$\
$$s(1)+s(59)+s(42)+s(125)+s(135)=81+47+141+46+109\equiv 62\hspace{1mm}(\hspace{1mm}\textrm{mod}\hspace{1mm}181)$$.

5- The Prover sends $$Com\_{AHP\_X}^2=\sum\_{i=0}^{deg\_{\hat{w}(x)}}\hat{w}*i\hspace{1mm}ck(i)=30$$,\
$$Com*{AHP\_X}^{3}=\sum\_{i=0}^{deg\_{\hat{z}*A(x)}}\hat{z}*{A\_i}ck(i)=160$$, $$Com\_{AHP\_X}^{4}=\sum\_{i=0}^{deg\_{\hat{z}*B(x)}}\hat{z}*{B\_i}ck(i)=69$$,\
$$Com\_{AHP\_X}^{5}=\sum\_{i=0}^{deg\_{\hat{z}*C(x)}}\hat{z}*{C\_i}ck(i)=11$$, $$Com\_{AHP\_X}^{6}=\sum\_{i=0}^{deg\_{h\_0(x)}}h\_{0\_i}ck(i)=18$$ and\
$$Com\_{AHP\_X}^{7}=\sum\_{i=0}^{deg\_{s(x)}}s\_i\hspace{1mm}ck(i)=178$$.

6- The Verifier chooses random numbers $$\alpha$$, $$\eta\_A$$, $$\eta\_B$$, $$\eta\_C$$ and sends them to the Prover. ( Note that the Prover can choose $$\alpha=hash(s(0))$$, $$\eta\_A=hash(s(1))$$, $$\eta\_B=hash(s(2))$$, $$\eta\_C=hash(s(3))$$. Let $$\alpha=10$$, $$\eta\_A=2$$, $$\eta\_B=30$$ and $$\eta\_C=100$$.

7- The Prover finds polynomials $$g\_1(x)$$ and $$h\_1(x)$$ such that

$$s(x)+r(\alpha,x)\sum\_{M}\eta\_M\hat{z}*M(x)-(\sum*{M}\eta\_Mr\_M(\alpha,x))\hat{z}(x)=h\_1(x)v\_{\mathbb{H}}(x)+xg\_1(x)+\frac{\sigma\_1}{|\mathbb{H}|}$$ $$(1)$$

where $$r(x,y)=u\_{\mathbb{H}}(x,y)=\frac{v\_{\mathbb{H}}(x)-v\_{\mathbb{H}}(y)}{x-y}$$ , $$v\_{\mathbb{H}}(x)=\prod\_{h\in \mathbb{H}}(x-h)=x^{|\mathbb{H}|}-1$$. Therefore $$r(x,y)=\frac{x^5-y^5}{x-y}$$ . Also $$r\_M(x,y)=\sum\_{k\in \mathbb{H}}r(x,k)\hat{M}(k,y)$$ for $$M\in {A,B,C}$$.

Now, since $$\sum\_M\eta\_M\hat{z}\_M(x)=\eta\_A\hat{z}\_A(x)+\eta\_B\hat{z}\_B(x)+\eta\_C\hat{z} \_C(x)=98x^6+172x^5+120x^4+12x^3+104x^2+131x+87$$and $$r(\alpha,x)=r(10,x)=\frac{10^5-x^5}{10-x}$$, so the second term of the left of equation $$(1)$$ is $$r(\alpha,x)\sum\_M\eta\_M\hat{z}\_M(x)=98x^{10}+66x^9+56x^8+29x^7+32x^6+153x^5+56x^4+136x^3+123x^2+42x+114$$

Also, $$\hat{z}(x)=\hat{W}(x)v\_{\mathbb{H}\[\leq |x|+1]}(x)+\hat{x}(x)=(149x^4+97x^3+161x^2+121x+166)(x-1)(x-59)+128x+54=149x^6+26x^5+55x^4+166x^3+52x^2+22x+74$$ that agree with $$z$$ on $$\mathbb{H}$$. Also, $$r\_A(10,x)=\sum\_{k\in \mathbb{H}}r(10,k)\hat{A}(k,x)$$ where $$\hat{A}(x,y)$$ is a polynomial such that $$\hat{A}(42,59)=1$$, $$\hat{A}(125,1)=1$$, $$\hat{A}(135,125)=1$$ and $$\hat{A}(x,y)=0$$ for the rest of points in $$\mathbb{H}\times\mathbb{H}$$. So, $$\hat{A}(x,y)$$ is a bivariate polynomial that passes from these 25 points. This polynomial can obtain as following:\
$$\hat{A}(x,y)=\sum\_{k\in \mathbb{K}}u\_{\mathbb{H}}(x,\hat{row}*{AHP\_A}(k))u*{\mathbb{H}}(y,\hat{col}*{AHP\_A}(k))\hat{val}*{AHP\_A}(k)=\sum\_{k\in \mathbb{K}}\frac{x^5-\hat{row}*{AHP\_A}(k)^5}{x-\hat{row}*{AHP\_A}(k)}\frac{y^5-\hat{col}*{AHP\_A}(k)^5}{y-\hat{col}*{AHP\_A}(k)}\hat{val}*{AHP\_A}(k)$$\
Therefore, $$\hat{A}(x,y)=5\frac{x^5-1}{x-42}\frac{y^5-1}{y-59}+5\frac{x^5-1}{x-125}\frac{y^5-1}{y-1}+132\frac{x^5-1}{x-135}\frac{y^5-1}{y-125}$$. So $$r\_A(10,x)=\sum*{k\in\mathbb{H}}r(10,k)\hat{A}(k,x)=70\hat{A}(1,x)+13\hat{A}(59,x)+150\hat{A}(42,x)+26\hat{A}(125,x)+147\hat{A}(135,x)$$where $$\hat{A}(1,x)=\hat{A}(59,x)=0$$, $$\hat{A}(42,x)=5\times82\times\frac{x^5-1}{x-59}=48(x^4+59x^3+42x^2+125x+135)=48x^4+117x^3+25x^2+27x+145$$, $$\hat{A}(125,x)=5\times 29\times\frac{x^5-1}{x-1}=145(x^4+x^3+x^2+x+1)=145x^4+145x^3+145x^2+145x+145$$, $$\hat{A}(135,x)=132\times 114\times\frac{x^5-1}{x-125}=25(x^4+125x^3+59x^2+135x+42)=25x^4+48x^3+27x^2+117x+145$$

Now, calculates $$\hat{B}(x,y)$$ similarly as following:\
$$\hat{B}(x,y)=\sum\_{k\in \mathbb{K}}u\_{\mathbb{H}}(x,\hat{row}*{AHP\_B}(k))u*{\mathbb{H}}(y,\hat{col}*{AHP\_B}(k))\hat{val}*{AHP\_B}(k)=\sum\_{k\in \mathbb{K}}\frac{x^5-\hat{row}*{AHP\_B}(k)^5}{x-\hat{row}*{AHP\_B}(k)}\frac{y^5-\hat{col}*{AHP\_B}(k)^5}{y-\hat{col}*{AHP\_B}(k)}\hat{val}*{AHP\_B}(k)$$\
\
Therefore, $$\hat{B}(x,y)=117\frac{x^5-1}{x-42}\frac{y^5-1}{y-1}+55\frac{x^5-1}{x-125}\frac{y^5-1}{y-1}+29\frac{x^5-1}{x-125}\frac{y^5-1}{y-42}+68\frac{x^5-1}{x-135}\frac{y^5-1}{y-1}$$. So, $$r\_B(10,x)=\sum*{k\in\mathbb{H}}r(10,k)\hat{B}(k,x)=70\hat{B}(1,x)+13\hat{B}(59,x)+150\hat{B}(42,x)+26\hat{B}(125,x)+147\hat{B}(135,x)$$\
where $$\hat{B}(1,x)=\hat{B}(59,x)=0$$, $$\hat{B}(42,x)=117\times 82\times \frac{x^5-1}{x-1}=1\times(x^4+x^3+x^2+x+1)=x^4+x^3+x^2+x+1$$, $$\hat{B}(125,x)=55\times 29\times\frac{x^5-1}{x-1}+29\times 29\times\frac{x^5-1}{x-42}=147(x^4+x^3+x^2+x+1)+117(x^4+42x^3+135x^2+59x+125)=83x^4+174x^3+14x^2+172x+111$$and $$\hat{B}(135,x)=68\times 114\times \frac{x^5-1}{x-1}=150(x^4+x^3+x^2+x+1)=150x^4+150x^3+150x^2+150x+150$$

Now, calculates $$\hat{C}(x,y)$$ similarly as following:\
$$\hat{C}(x,y)=\sum\_{k\in \mathbb{K}}u\_{\mathbb{H}}(x,\hat{row}*{AHP\_C}(k))u*{\mathbb{H}}(y,\hat{col}*{AHP\_C}(k))\hat{val}*{AHP\_C}(k)=\sum\_{k\in \mathbb{K}}\frac{x^5-\hat{row}^5\_{AHP\_C}(k)}{x-\hat{row}*{AHP\_C}(k)}\frac{y^5-\hat{col}^5*{AHP\_C}(k)}{y-\hat{col}*{AHP\_C}(k)}\hat{val}*{AHP\_C}(k)$$\
\
Therefore, $$\hat{C}(x,y)=114\frac{x^5-1}{x-42}\frac{y^5-1}{y-42}+82\frac{x^5-1}{x-125}\frac{y^5-1}{y-125}+5\frac{x^5-1}{x-135}\frac{y^5-1}{y-135}$$. So, $$r\_C(10,x)=70\hat{C}(1,x)+13\hat{C}(59,x)+150\hat{C}(42,x)+26\hat{C}(125,x)+147\hat{C}(135,x)$$ where $$\hat{C}(1,x)=\hat{C}(59,x)=0$$, $$\hat{C}(42,x)=114\times 82\times\frac{x^5-1}{x-42}=117(x^4+42x^3+135x^2+59x+125)=117x^4+27x^3+48x^2+25x+145$$, $$\hat{C}(125,x)=82\times 29\times\frac{x^5-1}{x-125}=25(x^4+125x^3+59x^2+135x+42)=25x^4+48x^3+27x^2+117x+145$$and $$\hat{C}(135,x)=5\times 114\times\frac{x^5-1}{x-135}=27(x^4+135x^3+125x^2+42x+59)=27x^4+25x^3+117x^2+48x+145$$

Therefore, the third term of the left of equation $$(1)$$ is

$$(\sum\_M \eta\_M r\_M(\alpha,x))\hat{z}(x)=(2r\_A(10,x)+30r\_B(10,x)+100r\_C(10,x))\hat{z}(x)=(23x^4+72x^3+144x^2+10x+19)(149x^6+26x^5+55x^4+166x^3+52x^2+22x+74)=169x^{10}+104x^9+158x^8+161x^7+86x^6+57x^5+85x^4+43x^3+99x^2+72x+139$$

Therefore, the left of equation $$(1)$$ is $$s(x)+r(\alpha,x)\sum\_M \eta\_M \hat{z}M(x)-(\sum\_M \eta\_M r\_M(\alpha,x))\hat{z}(x)=115x^{10}+143x^9+180x^8+66x^7+127x^6+97x^5+172x^4+93x^3+24x^2+154x+90$$

Now, the Prover finds polynomials $$g\_1(x)$$ and $$h\_1(x)$$ such that $$h\_1(x)v\_{\mathbb{H}}(x)+xg\_1(x)+\frac{\sigma\_1}{|\mathbb{H}|}=115x^{10}+143x^9+180x^8+66x^7+127x^6+97x^5+172x^4+93x^3+24x^2+154x+90$$

that means, the Prover finds polynomials $$g\_1(x)$$ and $$h\_1(x)$$ such that $$h\_1(x)(x^5+180)+xg\_1(x)+121=115x^{10}+143x^9+180x^8+66x^7+127x^6+97x^5+172x^4+93x^3+24x^2+154x+90$$

In this case, polynomials $$g\_1(x)= 134x^3+92x^2+90x+100$$ and $$h\_1(x)=115x^5+143x^4+180x^3+66x^2+127x+31$$ are obtained. The Prover sends ,\
$$Com\_{AHP\_X}^{8}=\sum\_{i=0}^{deg\_{g\_1(x)}}g\_{1\_i}ck(i)=129$$ and $$Com\_{AHP\_X}^{9}=\sum\_{i=0}^{deg\_{h\_1(x)}}h\_{1\_i}ck(i)=33$$ to the Verifier where $$g\_{1\_i}$$ is coefficient of $$x^i$$ of polynomial $$g\_1(x)$$ and $$h\_{1\_i}$$ is coefficient of $$x^i$$ of polynomial $$h\_1(x)$$.

8- The Verifier selects $$\beta\_1\in \mathbb{F}-\mathbb{H}$$ and sends it to the Prover. (The Prover can selects $$\beta\_1=hash(s(8))$$). Let $$\beta\_1=22$$.

9- The Prover calculates $$\sigma\_2=\sum\_{k\in\mathbb{H}}r(\alpha,k)\sum\_{M}\eta\_M\hat{M}(k,\beta\_1)=r(10,1)\sum\_{M}\eta\_M\hat{M}(1,22)+r(10,59)\sum\_{M}\eta\_M\hat{M}(59,22)+r(10,42)\sum\_{M}\eta\_M\hat{M}(42,22)+r(10,125)\sum\_{M}\eta\_M\hat{M}(125,22)+r(10,135)\sum\_{M}\eta\_M\hat{M}(135,22)$$\
that is $$\sigma\_2=70\times 0+13\times 0+150\times 135+26\times 88+147\times 22=70$$.

Then, the Prover finds $$g\_2(x)$$ and $$h\_2(x)$$ such that $$r(\alpha,x)\sum\_M \eta\_M\hat{M}(x,\beta\_1)=h\_2(x)v\_{\mathbb{H}}(x)+xg\_2(x)+\frac{\sigma\_2}{|\mathbb{H}|}$$

where $$r(\alpha,x)\sum\_M\eta\_M \hat{M}(x,\beta\_1)=r(10,x)(2\hat{A}(x,22)+30\hat{B}(x,22)+100\hat{C}(x,22))$$ where$$\hat{A}(x,22)=5\frac{x^5-1}{x-42}\frac{22^5-1}{22-59}+5\frac{x^5-1}{x-125}\frac{22^5-1}{22-1}+132\frac{x^5-1}{x-135}\frac{22^5-1}{22-125}=159\frac{x^5-1}{x-42}+56\frac{x^5-1}{x-125}+114\frac{x^5-1}{x-135}=159(x^4+42x^3+135x^2+59x+125)+56(x^4+125x^3+59x^2+135x+42)+114(x^4+135x^3+125x^2+42x+59)=148x^4+108x^3+104x^2+9x+174$$,$$\hat{B}(x,22)=117\frac{x^5-1}{x-42}\frac{22^5-1}{22-1}+55\frac{x^5-1}{x-125}\frac{22^5-1}{22-1}+29\frac{x^5-1}{x-125}\frac{22^5-1}{22-42}+68\frac{x^5-1}{x-135}\frac{22^5-1}{22-1}=146x^4+37x^3+95x^2+100x+165$$and $$\hat{C}(x,22)=114\frac{x^5-1}{x-42}\frac{22^5-1}{22-42}+82\frac{x^5-1}{x-125}\frac{22^5-1}{22-125}+5\frac{x^5-1}{x-135}\frac{22^5-1}{22-135}=6(x^4+42x^3+135x^2+59x+125)+5(x^4+125x^3+59x^2+135x+42)+177(x^4+135x^3+125x^2+42x+59)=7x^4+156x^3+62x^2+137x$$

Therefore, $$r(\alpha,x)\sum\_M \eta\_M \hat{M}(x,\beta\_1)=\frac{10^5-x^5}{10-x}(127x^4+93x^3+27x^2+66x+49)=(x^4+10x^3+100x^2+95x+45)(127x^4+93x^3+27x^2+66x+49)=127x^8+96x^7+82x^6+162x^5+40x^4+84x^3+77x^2+23x+33$$

Hence, the Prover finds $$g\_2(x)=40x^3+30x^2+173x+105$$ and $$h\_2(x)=127x^3+96x^2+82x+162$$ such that $$h\_2(x)v\_{\mathbb{H}}(x)+xg\_2(x)+\frac{\sigma\_2}{|\mathbb{H}|}=h\_2(x)(x^5+180)+xg\_2(x)+\frac{70}{5}=127x^8+96x^7+82x^6+162x^5+40x^4+84x^3+77x^2+23x+33$$

The Prover sends , $$Com\_{AHP\_X}^{10}=\sum\_{i=0}^{deg\_{g\_2(x)}}g\_{2\_i}ck(i)=100$$ and\
$$Com\_{AHP\_X}^{11}=\sum\_{i=0}^{deg\_{h\_2(x)}}h\_{2\_i}ck(i)=179$$ where $$g\_{2\_i}$$ is coefficient of $$x^i$$ of polynomial $$g\_2(x)$$ and $$h\_{2\_i}$$ is coefficient of $$x^i$$ of polynomial $$h\_2(x)$$.

10- The Verifier selects $$\beta\_2\in \mathbb{F}-\mathbb{H}$$ and sends it to the Prover. For example $$\beta\_2=80$$.

11- The Prover calculates $$\sigma\_3=\sum\_{k\in\mathbb{K}}(\sum\_M \eta\_M\frac{v\_{\mathbb{H}}(\beta\_2)v\_{\mathbb{H}}(\beta\_1)\hat{val'*M}(k)}{(\beta\_2-\hat{row'*M}(k))(\beta\_1-\hat{col'*M}(k))})=(2\frac{72\times 18\times 5}{38\times 144}+30\frac{72\times 18\times 117}{38\times 21}+100\frac{72\times 18\times 114}{38\times 161})+(2\frac{72\times 18\times 5}{136\times 21}+30\frac{72\times 18\times 55}{136\times 21}+100\frac{72\times 18\times 82}{136\times 78})+(2\frac{72\times 18\times 132}{126\times 78}+30\frac{72\times 18\times 29}{136\times 161}+100\frac{72\times 18\times 5}{126\times 68})+(30\frac{72\times 18\times 68}{126\times 21})=84$$ Then, the Prover finds polynomials $$g\_3(x)$$ and $$h\_3(x)$$ such that $$h\_3(x)v*{\mathbb{K}}(x)=a(x)-b(x)(xg\_3(x)+\frac{\sigma\_3}{|\mathbb{K}|})$$ where $$a(x)=\sum*{M\in {A,B,C}} \eta\_M v*{\mathbb{H}}(\beta\_2)v\_{\mathbb{H}}(\beta\_1)\hat{val}*{AHP\_M}(x)\prod*{N\in{A,B,C}-{M}}(\beta\_2-\hat{row}*{AHP\_N}(x))(\beta\_1-\hat{col}*{AHP\_N}(x))$$and $$b(x)=\prod\_{M\in{A,B,C}}(\beta\_2-\hat{row}*{AHP\_M}(x))(\beta\_1-\hat{col}*{AHP\_M}(x))$$.

Therefore, since $$v\_{\mathbb{H}}(\beta\_1)=22^5+180=18$$ and $$v\_{\mathbb{H}}(\beta\_2)=80^5+180=72$$ ,

$$a(x)=2\times 72\times 18\hat{val}*{AHP\_A}(x)(80-\hat{row}*{AHP\_B}(x))(22-\hat{col}*{AHP\_B}(x))(80-\hat{row}*C(x))(22-\hat{col}*{AHP\_C}(x))+30\times72\times18\hat{val}*{AHP\_B}(x)(80-\hat{row}*{AHP\_A}(x))(22-\hat{col}*{AHP\_A}(x))(80-\hat{row}*{AHP\_C}(x))(22-\hat{col}*{AHP\_C}(x))+100\times72\times 18\hat{val}*{AHP\_C}(x)(80-\hat{row}*{AHP\_A}(x))(22-\hat{col}*{AHP\_A}(x))(80-\hat{row}*{AHP\_B}(x))(22-\hat{col}\_{AHP\_B}(x))=63x^{25 }+ 56x^{24} + 45x^{23} + 27x^{22} + 81x^{21} + 80x^{20} + 66x^{19} + 152x^{18} + 97x^{17} + 55x^{16} + 170x^{15} + 142x^{14} + 3x^{13} + 96x^{12} + 11x^{11} + 69x^{10} + 43x^9 + 10x^8 + 75x^7 + 21x^6 + 109x^5 + 31x^4 + 90x^3 + 159x^2 + 119x + 79$$

and

$$b(x)=(80-\hat{row}*{AHP\_A}(x))(22-\hat{col}*{AHP\_A}(x))(80-\hat{row}*{AHP\_B}(x))(22-\hat{col}*{AHP\_B}(x))(80-\hat{row}*{AHP\_C}(x))(22-\hat{col}*{AHP\_C}(x))=23x^{30} + 176x^{29} + 75x^{28} + 63x^{27} + 174x^{26} + 13x^{25} + 43x^{24} + 169x^{23} + 35x^{22 }+ 66x^{21} + 160x^{20} + 99x^{19} + 150x^{18} + 31x^{17} + 69x^{16} + 83x^{15} + 93x^{14} + 179x^{13} + 30x^{12} + 111x^{11} + 58x^{10} + 131x^9 + 18x^8 + 36x^7 + 56x^6 + 59x^5 + 15x^4 + 171x^3 + 47x^2 + 110x + 142$$

Therefore,

$$g\_3(x)= 110x^4 + 123x^3 + 161x^2 + 111x + 134$$

\
and

$$h\_3(x)=4x^{29} + 74x^{28} + 65x^{27} + 16x^{26} + 139x^{25} + 135x^{24} + 92x^{23} + 139x^{22} + 60x^{21} + 75x^{20} + 75x^{19} + 99x^{18} + 98x^{17} + 71x^{16} + 15x^{15} + 53x^{14} + 138x^{13} + 128x^{12} + 18x^{11} + 147x^{10} + 111x^9 + 37x^8 + 18x^7 + 97x^6 + 143x^5 + 136x^4 + 53x^3 + 50x^2 + 177x + 99$$

The Prover sends , $$Com\_{AHP\_X}^{12}=\sum\_{i=0}^{deg\_{g\_3(x)}}g\_{3\_i}ck(i)=169$$ and $$Com\_{AHP\_X}^{13}=\sum\_{i=0}^{deg\_{h\_3(x)}}h\_{3\_i}ck(i)=166$$\
where $$g\_{3\_i}$$ is coefficient of $$x^i$$ of polynomial $$g\_3(x)$$ and $$h\_{3\_i}$$ is coefficient of $$x^i$$ of polynomial $$h\_3(x)$$.

12- The Prover sends $$\pi\_{AHP}^1=62$$, $$\pi\_{AHP}^2=(166,121,161,97,149)$$, $$\pi\_{AHP}^3=(168,141,45,26,63,165,116)$$, $$\pi\_{AHP}^4=(124,81,137,101,71,178,32)$$, $$\pi\_{AHP}^5=(49,157,169,96,80,50,123)$$, $$\pi\_{AHP}^6=(32,16,153,20,1,164,45,92)$$ and $$\pi\_{AHP}^7=(115,3,0,0,20,1,0,17,101,0,5)$$\
to the Verifier that are value of $$\sigma\_1$$, coefficients of polynomials $$\hat{W}(x)$$, $$\hat{z}\_A(x)$$, $$\hat{z}\_B(x)$$, $$\hat{z}\_C(x)$$,$$h\_0(x)$$ and $$s(x)$$, respectively.

13- The Prover sends $$\pi\_{AHP}^8=(100,90,92,134)$$ and $$\pi\_{AHP}^{9}=(31,127,66,180,143,115)$$ to the Verifier that are coefficients of polynomials $$g\_1(x)$$ and $$h\_1(x)$$, respectively.

14-The Prover sends $$\pi\_{AHP}^{10}=70$$, $$\pi\_{AHP}^{11}=(105,173,30,40)$$ and $$\pi\_{AHP}^{12}=(162,82,96,127)$$ that are value of $$\sigma\_2$$ and coefficients of polynomials $$g\_2(x)$$ and $$h\_2(x)$$, respectively.

15- The Prover sends $$\pi\_{AHP}^{13}=84$$, $$\pi\_{AHP}^{14}=(134,111,161,123,110)$$ and\
$$\pi\_{AHP}^{15}=(99,177,50,53,136,143,97,18,37,111,147,18,128,138,53,15,71,98,99,75,75,60,139,92 ,$$$$135,139,16,65,74,4)$$\
that are value of $$\sigma\_3$$ and coefficients of polynomials $$g\_3(x)$$ and $$h\_3(x)$$, respectively.

16- The Prover chooses random values $$\eta\_{\hat{w}}$$, $$\eta\_{\hat{z}*A}$$, $$\eta*{\hat{z}*B}$$, $$\eta*{\hat{z}*C}$$, $$\eta*{h\_0}$$, $$\eta\_s$$, $$\eta\_{g\_1}$$, $$\eta\_{h\_1}$$, $$\eta\_{g\_2}$$, $$\eta\_{h\_2}$$, $$\eta\_{g\_3}$$ and $$\eta\_{h\_3}$$ of $$\mathbb{F}$$. For example, $$\eta\_{\hat{w}}=1$$, $$\eta\_{\hat{z}*A}=4$$, $$\eta*{\hat{z}*B}=10$$, $$\eta*{\hat{z}*C}=8$$, $$\eta*{h\_0}=32$$, $$\eta\_s=45$$, $$\eta\_{g\_1}=92$$, $$\eta\_{h\_1}=11$$, $$\eta\_{g\_2}=1$$, $$\eta\_{h\_2}=5$$, $$\eta\_{g\_3}=25$$ and $$\eta\_{h\_3}=63$$.

17- The Prover calculates the linear combination$$p(x)=\eta\_{\hat{w}}\hat{w}(x)+\eta\_{\hat{z}*A}\hat{z}*A(x)+\eta*{\hat{z}*B}\hat{z}*B(x)+\eta*{\hat{z}*C}\hat{z}*C(x)+\eta*{h\_0}h\_0(x)+\eta\_ss(x)+\eta*{g\_1}g\_1(x)$$\
$$+\eta*{h\_1}h\_1(x)+\eta*{g\_2}g\_2(x)+\eta\_{h\_2}h\_2(x)+\eta\_{g\_3}g\_3(x)+\eta\_{h\_3}h\_3(x)$$.

The Prover obtains

$$p(x)=71x^{29} + 137x^{28} + 113x^{27} + 103x^{26} + 69x^{25} + 179x^{24} + 4x^{23} + 69x^{22} + 160x^{21} + 19x^{20} + 19x^{19} + 83x^{18} + 20x^{17} + 129x^{16} + 40x^{15} + 81x^{14} + 6x^{13} + 100x^{12} + 48x^{11} + 74x^{10} + 115x^9 + 179x^8 + 137x^7 + 88x^6 + 126x^5 + 8x^4 + 124x^3 + 37x^2 + 72x + 114$$

18- The Prover calculates $$p(x)$$ in $$x=x'$$ (value of $$x'$$ is received from the Verifier), then puts it in $$\pi\_{AHP}^{16}$$ . Therefore $$\pi\_{AHP}^{16}=p(x')=y'$$. For example, if $$x'=2$$, then $$\pi\_{AHP}^{16}=p(2)=119$$.

19- The Prover computes $$\pi\_{AHP}^{17}=PC.Eval(ck,p(x),d\_p,r\_p,x')$$ where $$d\_p$$ is degree bound of $$p(x)$$ and $$r\_p$$ is a random value.\
For example, if the polynomial commitment scheme $$KZG$$ is used, then the Prover builds polynomial\
$$q(x)=\frac{p(x)-y'}{x-x'}=\frac{p(x)-119}{x-2}=71x^{28} + 98x^{27} + 128x^{26} + 178x^{25} + 63x^{24} + 124x^{23} + 71x^{22} + 30x^{21} + 39x^{20} + 97x^{19} + 32x^{18} + 147x^{17} + 133x^{16} + 33x^{15} + 106x^{14} + 112x^{13} + 49x^{12} + 17x^{11} + 82x^{10} + 57x^9 + 48x^8 + 94x^7 + 144x^6 + 14x^5 + 154x^4 + 135x^3 + 32x^2 + 101x + 93$$\
and calculates $$\pi\_{AHP}^{17}=gq(\tau)$$ by using $$ck$$ as following:\
$$\pi\_{AHP}^{17}=\sum\_{i=0}^{deg\_{q(x)}}q\_i\hspace{1mm}ck(i)=149$$ where $$q\_i$$ is the coefficient of $$x^i$$ of $$q(x)$$.

## Proof JSON file Example 1

IoT\_Manufacturer\_Name = "zkIoT"\
IoT\_Device\_Name = "MultiSensor"\
Device\_Hardware\_Version = "1.0"\
Firmware\_Version = "1.0"\
Device Picture= <>\
Lines = \[200, 350, 4000-4010]

```
    {
    "commitmentId": 64-bit,
    "class": 32-bit Integer,
    "input": 4
    "output": 82    
       
    "P_AHP1": 62
    "P_AHP2": [166,121,161,97,149],
    "P_AHP3": [168,141,45,26,63,165,116],
    "P_AHP4": [124,81,137,101,71,178,32],  
    "P_AHP5": [49,157,169,96,80,50,123], 
    "P_AHP6": [32,16,153,20,1,164,45,92],
    "P_AHP7": [115,3,0,0,20,1,0,17,101,0,5],
    "P_AHP8": [100,90,92,134],
    "P_AHP9": [31,127,66,180,143,115],
    "P_AHP10": 70
    "P_AHP11": [105,173,30,40],
    "P_AHP12": [162,82,96,127],
    "P_AHP13": 84
    "P_AHP14": [134,111,161,123,110],
    "P_AHP15": [99,177,50,53,136,143,97,18,37,111,147,18,128,138,53,15,71,98,99,75,75,60,139,92,135,139,16,65,74,4]
    "P_AHP16": 119
    "P_AHP17": 149
      
    "Com_AHP1_x": 4,
    "Com_AHP2_x": 30,
    "Com_AHP3_x": 160,
    "Com_AHP4_x": 69,
    "Com_AHP5_x": 11,
    "Com_AHP6_x": 18,  
    "Com_AHP7_x": 178,
    "Com_AHP8_x": 129,
    "Com_AHP9_x": 33,
    "Com_AHP10_x": 100,
    "Com_AHP11_x": 179,
    "Com_AHP12_x": 169,
    "Com_AHP13_x": 166
}
```

##