Using pumping lemma to show a language is not context free(Complicated)
Start with a long enough string $w$ in $L$ in which $m=p+2,n=p+1,o=p$ and


$i_1,...,i_{2m}=0$
$j_1,...,j_{2n}=0$
$k_1,...,k_{o}=0$


$w = a\; b^{2(p+2)}\; d\; e^{2(p+1)}\; g\; h^{p} $

Then apply pumping lemma (it should be easier ;-).

If you want to "reduce" $L$ to $L' = \left\{0^i1^j2^k|1\le \:i&lt;j&lt;k\right\}$ then you must use closure properties, in particular CFLs are:


closed under homomorphism
closed under reverse homomorphism
closed under reversal


The homomorphism that you can use is: $H(a) = H(c) = H(d) = H(f) = H(g) = H(q) = \epsilon, H(b) = \bar{2}, H(e) = \bar{1}, H(h) = \bar{0}$

Applying $H$ to the language gives $H(L) = \{ \bar{2}^{2m} \bar{1}^{2n} \bar{0}^{o} \mid m &gt; n &gt; o &gt; 0\}$

Then you can use a reverse homomorphism $\varphi(2)=\bar{2}\bar{2}, \varphi(1)=\bar{1}\bar{1}, \varphi(0) = \bar{0}$ and you get:

$\varphi^{-1}( H(L)) = \{ 2^m 1^n 0^o \mid m&gt;n&gt;o&gt;0 \}$

and finally apply closure under reversal:

$(\varphi^{-1}( H(L)))^R = \{ 0^o 1^n 2^m \mid 0&lt;o&lt;n&lt;m \} = L'$

If $L$ is CF then by closure properties $(\varphi^{-1}( h(L)))^R = L'$ should be CF, which is not the case.