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should work now
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katat committed Dec 6, 2023
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In this manner, the prover can keep calculating for the whole domain. Then they would response the verifier query with the evaluations across layers for checking if an instance $a_i$ corresponding to the query is in the $\mathcal{L}\left(\mathcal{R}_{\text{consistent\_layers }}\right)$
In this manner, the prover can keep calculating for the whole domain. Then they would response the verifier query with the evaluations across layers for checking if an instance $a_i$ corresponding to the query is in the $\mathcal{L}\left(\mathcal{R}_{\text{consistent-layers }}\right)$

### Consistency check

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Indeed, the value provided by prover, $p_2((2^2)^2)$, is consistent with the recursively accumulated sum from symmetric points on previous layers.

Therefore, $(a_2,w_2)$ is a pair in $\mathcal{R}_{\text {consistent\_layers }}$. So through a single query, this naive FRI convince the verifier that $p_0$ has an expected degree bound.
Therefore, $(a_2,w_2)$ is a pair in $\mathcal{R}_{\text {consistent-layers }}$. So through a single query, this naive FRI convince the verifier that $p_0$ has an expected degree bound.

For the next, we will see how to deal with a malicious prover by improving this naive version.

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