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John L.
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How can I show Show the pumping lemma is not a universal method for proving CFLnot context-free

I know that the pumping lemma is not powerful enough to prove a language is not context-free, but I don't understand how to show it. I now

I have the same question as this one Show that the Pumping Lemma for CFLs is not powerful enough to prove that the language L = {aibjck |i ≠j ≠ k ≠ i } is not context freeShow that the Pumping Lemma for CFLs is not powerful enough to prove that the language L = {aibjck |i ≠j ≠ k ≠ i } is not context free, but I couldn't understand the answer in this.

Please explain to me in detail, how can I show $L = \{ a^i b^j c^k | i ≠ j ≠ k ≠ i \}$ satisfy the pumping lemma?

How can I show the pumping lemma is not a universal method for proving CFL

I know that the pumping lemma is not powerful enough to prove a language is not context-free, but I don't understand how to show it. I now have the same question as this one Show that the Pumping Lemma for CFLs is not powerful enough to prove that the language L = {aibjck |i ≠j ≠ k ≠ i } is not context free but I couldn't understand the answer in this

Please explain to me in detail, how can I show $L = \{ a^i b^j c^k | i ≠ j ≠ k ≠ i \}$ satisfy the pumping lemma?

Show the pumping lemma is not a universal method for proving not context-free

I know that the pumping lemma is not powerful enough to prove a language is not context-free, but I don't understand how to show it.

I have the same question as this one Show that the Pumping Lemma for CFLs is not powerful enough to prove that the language L = {aibjck |i ≠j ≠ k ≠ i } is not context free, but I couldn't understand the answer in this.

Please explain to me in detail, how can I show $L = \{ a^i b^j c^k | i ≠ j ≠ k ≠ i \}$ satisfy the pumping lemma?

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YX L
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How can I show the pumping lemma is not a universal method for proving CFL

I know that the pumping lemma is not powerful enough to prove a language is not context-free, but I don't understand how to show it. I now have the same question as this one Show that the Pumping Lemma for CFLs is not powerful enough to prove that the language L = {aibjck |i ≠j ≠ k ≠ i } is not context free but I couldn't understand the answer in this

Please explain to me in detail, how can I show $L = \{ a^i b^j c^k | i ≠ j ≠ k ≠ i \}$ satisfy the pumping lemma?