# i really don't know how to get $s = xyz$ for pumping lemma for this language

Let $L=\{a^i b^j c^k d^l : i, j, k, l > 0, 3(i+j) \geq 2(k+l)\}$. Proof that this language is not a regular language.

I have no clue, cause i can't find any example for $3(i+j) \geq 2(k+l)$ or something like this.

The reversal of your language is $$L^R = \{ d^l c^k b^j a^i \mid 2(k+l) \le 3(i+j), i,j,k,l > 0 \}$$ and for this language we can do it with the Pumping lemma. Suppose it is regular with pumping constant $N$, then consider $w = d^N c^N b^N a^N$, for which we have a decomposition $uvw = w$ with $|uv| \le N$ and $|v| > 0$, in particular $uv \in d^{\ast}$... can you complete it from here on your own?
• no i can't. is it $d^n c^n b^{2n} a^{2n}$ and then $d^{n-i} (d^i)^m c^n b^{2n} a^{2n}$. i'm i going right? – Suparerk Yoscharoen May 6 '18 at 15:20