# For every $\mathrm{NP}$ language $L$, is there a verifier such that, for all the certificates $u$ of other verifiers of $L$, it accepts $(x, u)$?

Let $$L$$ be an $$\mathrm{NP}$$ language. Then there exists a verifier $$V$$ of $$L$$ and a polynomial $$p\colon \mathbb{N} \to \mathbb{N}$$, such that for every $$x \in \Sigma^{*}$$, $$x \in L$$ if and only if there exists a certificate $$u \in \Sigma^{p(|x|)}$$ for $$x$$ satisfying $$V(x, u) = 1$$.

Suppose that $$\mathcal{V}_{q}$$ be all the verifiers of $$L$$ such that the length of the certificates of $$x$$ is $$q(|x|)$$. And for every $$V \in \mathcal{V}_{q}$$, let $$V_{x}$$ be all the certificates of $$x$$.

For every polynomial $$q\colon \mathbb{N} \to \mathbb{N}$$ satisfying $$\mathcal{V}_{q} \neq \varnothing$$, I want to know whether there is a verifier $$\bar{V} \in \mathcal{V}_{q}$$ such that for every $$x \in L$$, $$\bar{V}_{x} = \bigcup_{V \in \mathcal{V}_{q}} V_{x}.$$

Fix $$x \in L$$ of length $$|x|=n$$. Suppose that $$\mathcal{V}_q \neq \emptyset$$, and let $$m = q(n)$$. Choose some $$V^0 \in \mathcal{V}_q$$. For any $$s \in \{0,1\}^m$$, let $$V^s(y,u) = V^0(y,u \oplus s)$$ for inputs of length $$n$$ and witnesses of length $$m$$, $$V^s(y,u) = 0$$ for inputs of length $$n$$ and witnesses lengths different from $$m$$, and $$V^s(y,u) = V^0(y,u)$$ otherwise. Then $$V^s$$ is a verifier for $$L$$ for all $$s$$, and so $$\bigcup_{V \in \mathcal{V}_q} V_x = \{0,1\}^m$$.
This means that if a verifier $$\bar{V}$$ exists, then $$\bar{V}_x$$ is the set of all strings of length $$q(|x|)$$. This means that $$\bar{V}$$ need not consult its witness, showing that $$L \in \mathsf{P}$$.