# Any barrier result for this kind of (non-relativising) technique?

We think that non-deterministic machines are more powerful than deterministic machines, by giving an oracle access to $$P\subseteq L\subseteq NP$$, it seems reasonable to expect there's some $$L$$ that is so weak compared to $$SAT$$ and thus gives "relatively more" power to $$P$$ than to $$NP$$.

Suppose we're given a monotonically harder sequence of oracle access, $$L_0 (\in P)\leq_P L_1 \leq_P L_2\leq_P\dots\leq_P L_k= SAT,$$ We'll have $$P=P^{L_0}\subseteq P^{L_1}\subseteq ... \subseteq P^{L_k}=NP\\ NP=NP^{L_0}\subseteq NP^{L_1}\subseteq ... \subseteq NP^{L_k}=\Sigma_2P$$

Suppose we can find, in the middle $$L_i$$ that does not give extra power to $$NP$$, i.e. $$NP^{L_i}\subseteq NP$$, but $$L_i$$ is already hard enough so that, in such relativised world we could resolve $$P^{L_i}\neq NP^{L_i}$$, then we'll have $$P\subseteq P^{L_i}\subsetneq NP^{L_i}=NP$$. So my question is,

Do we have any barrier result against this type of technique when resolving $$P$$ vs $$NP$$ ? It seems neither a relativizing technique nor a natural proof.