Question: L = {a^j b^k c^k d^k : j, k 1} {b^j c^k d^l : j, k, l 0 }. The language L is not context-free

L = {a^j b^k c^k d^k : j, k 1} {b^j c^k d^l : j, k, l 0 }. The language L is not context-free but satisfies the pumping lemma for context-free languages. Prove that L satisfies the pumping lemma for CFLs. Please show full proof.

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