Getting from one language to the other using closure properties(automata) 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: 

$m,n,o \geq 1;$

$m&gt;n&gt;o&gt;0;$ 

$i_1,i_2,...,i_{2m} \geq 0;$

$j_1,j_2,...,j_{2n} \geq 0;$

$k_1,k_2,...,k_o \geq 0$

this language is also: $$L=\left\{0^i1^j2^k|1\le \:i&lt;j&lt;k\right\}$$

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 $h:B^{i_{2m}^*} -&gt; 0^{i_{2m}^*}$, $h:E^{i_{2n}^*} -&gt; 1^{j_{2n}^*}$ , $h:H^{k_o^*} -&gt; 2^{k^*_o}$ 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.
If I understand your argument correctly, you are reducing languages in the wrong direction.


  If $L$ is not context-free, then $K$ is not context-free.


Is equivalent to


  If $K$ is context-free, then $L$ is context-free.


We have to reverse the construction, as we are using the closure properties of the context-free languages. In do not know of any useful closure properties of the non-context-free languages.

PS. The pumping argument seems out of place here.