I am trying to deduct how i can, using closure properties, deduct that since the following language is not context free $$L=\left\{abc^{i_1}bc^{i_2}...bc^{i_{2m}}def^{j_1}ef^{j_2}..ef^{j_{2n}}ghq^{k_1}hq^{k_2}...hq^{k_o}\right\}$$
Such that:
$m,n,o \geq 1;$
$m>n>o>0;$
$i_1,i_2,...,i_{2m} \geq 0;$
$j_1,j_2,...,j_{2n} \geq 0;$
$k_1,k_2,...,k_o \geq 0$
this language is also: $$L=\left\{0^i1^j2^k|1\le \:i<j<k\right\}$$
My attempt: basically, they both look very similar, but I am not sure the coming procedure is correct: If we take $L=\left\{abc^{i_1}bc^{i_2}...bc^{i_{2m}}def^{j_1}ef^{j_2}..ef^{j_{2n}}ghq^{k_1}hq^{k_2}...hq^{k_o}\right\}$, Unionize bc, ef and hq to obtain the following: $L=\left\{aB^{i_{2m}^*}dE^{j_{2n}^*}gH^{k_o^*}\right\}$, And then using assignment or homomorphism, defining $h:B^{i_{2m}^*} -> 0^{i_{2m}^*}$, $h:E^{i_{2n}^*} -> 1^{j_{2n}^*}$ , $h:H^{k_o^*} -> 2^{k^*_o}$ obtaining: $L=\left\{a0^{i_{2m}^*}d1^{j_{2n}^*}g2^{k^*_o}\right\}$. Since we can decompose string S using the pumping lemma into $S=uvxyz$ as we choose, We can only regard the $0^{i_{2m}^*}1^{j_{2n}^*}2^{k^*_o}$ part. Because of that, And knowing in advance that $L=\left\{abc^{i_1}bc^{i_2}...bc^{i_{2m}}def^{j_1}ef^{j_2}..ef^{j_{2n}}ghq^{k_1}hq^{k_2}...hq^{k_o}\right\}$ is not context free, we can deduct that $L=\left\{a0^{i^*_{2m}}d1^{j^*_{2n}}g2^{k^*_o}\right\}$ is not context free as well.
Would really appreciate your corrections or knowing if there exists a better and easier way to deduct that.
Thank you very much.
