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there is a lot of examples for pumpinglemma proof, but I did not figure out this, can anyone help ?

L= { a^n b^n c^m d^m : n >= 1, m >= 1 } U { a^n b^m c^m d^n : n >= 1, m >= 1 }

John Nick
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  • None of these words can be generated using regular grammar rules. It would require some kind of memory, so at least pushdown automaton. Are you sure it isn't context-free language? – cdm May 11 '18 at 08:21

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Consider the regular language R = a*b*cd. The intersection of two regular languages must be a regular language. The intersection of L and R is a^n b^n cd. However, this is easily shown not to be regular using the pumping lemma or Myhill-Nerode theorem. This is a contradiction, so L must not be regular.

Patrick87
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