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is the language $$\{a^j b^j c^k\; : \;(j = 2 \ OR\ j=k+i)\ AND\; i, k \gt 0\}$$ context free? if $j=2$ only I would say yes but when I add the other condition ($j=k+i$) if I use pumping lemma I get stuck. On the case $j=k+i$ I can rewrite the string in the form $a^k a^i b^k b^i c^k=a^k a^i b^i b^k c^k$ then I choose for the pumping lemma the string $s=a^p a^p b^p b^p c^p$ . but for vwx i can choose a substring that contains equal numbers of a and b inside the $a^p b^p$ and if I pump the i of the string $uv^i wx^i y$ I still obtain a string of that language so I think the pumping lemma holds for this language but I still think that in this case is not context free. Do i make mistakes or wrong assumptions? Is the language context free or not? how can I demonstrate it?

vin2098
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  • Welcome to MSE. To format math expression properly, you need to put $ signs around them, so $j=2$ is typeset as $j=2$. – saulspatz Jan 11 '20 at 14:49
  • Have you tried the pumping lemma? – saulspatz Jan 11 '20 at 14:56
  • yes but i got stuck. if I use pumping lemma on the case $j=k+i$ I can rewrite the string in the form $ a^k a^i b^k b^i c^k = a^k a^i b^i b^k c^k$ then i choose for the pumping lemma the string $s=a^p a^p b^p b^p c^p$ . but for vwx i can choose a substring that contains equal numbers of a and b inside the $a^p b^p$ and if i pump the i of the string $u v^i w x^i y$ i still obtain a string of that language so I think the pumping lemma holds for this language. Do i make mistakes or wrong assumptions? – vin2098 Jan 11 '20 at 16:32
  • I don't have time to look at it right now. I'll try later. Meanwhile, you'll find you'll get a lot better response if you copy your comment to the body of the question. Many people won't answer if you don't show your efforts, and most won't bother hunting through the comments. – saulspatz Jan 11 '20 at 16:51
  • yes, thank you. – vin2098 Jan 11 '20 at 17:01

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