Is it possible to prove with pumping lemma that the languge $$L=\{w_1w_2 \mid w_1,w_2\in\{a,b,c\}^* \text{ and } \#_a(w_1)>\#_b(w_1) \text{ and } \#_b(w_2)>\#_c(w_2)\}$$ (where $\#_x(w)$ is the number of the letter $x$ in $w$) is not regular.
Using Myhill-Nerode, we could simply say for $k>0: [a_k]=\{a^k\}$ would describe an equivalence class, they are all disjoint pairwise, because i could say: $(i,j ≥ 1) \wedge (i>j) \implies (a^ib^i ∈ L) \wedge (a^{i-1}b^j ∉ L)$
Yes, the pumping lemma works just fine. Given pumping length $k$, pick the word $b^ka^{k+1}c^{4k}b$. This pattern gives us control over which part of the word is $w_1$, namely everything but the final $b$.
Now, any decomposition $uvw = b^ka^{k+1}c^{4k}b$ with $|uv| \leq k$ and $|v| > 0$ means that $v$ consists of between $1$ and $k$ many $b$'s, say $\ell$. Now $uv^2w = b^{k+\ell}a^{k+1}c^{4k}b$. If this word would belong to the language, then $w_2$ could only be $b$ - anything longer has too many $c$'s in it. But $b^{k+\ell}a^{k+1}c^{4k}$ doesn't have more $a$'s than $b$'s, so it can't be $w_1$ either. We conclude that $b^{k+\ell}a^{k+1}c^{4k}b$ doesn't belong to the language, and thus the pumping lemma tells us that it is not regular.