Question: Proof using pumping lemma for Context free language L = {0^n 1^n 2^n | n >=1}? The string division should be used as uv^iwx^iy (means

Proof using pumping lemma for Context free language L = {0^n 1^n 2^n | n >=1}? The string division should be used as uv^iwx^iy (means divide string into five parts) and solve with different cases. I Need a general proof for this not hardcoded values.

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