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L = {a^n b c^n| i is greater than 1 and less than 100 , n is greater than 1}

I think i misunderstood the pumping lemma for cfl. why cant i pick a word z = a^n c b^n then break it apart to u= a^s v = a^n-s w=epsilon x=b ,y= b^n then pump it with i=0 then get a contradiction since 0 b's doesnt satisfies the language ? I'm probably missing something here.

gil
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  • What formulation of the pumping lemma are you using? Sipser's statement of it in Introduction to the Theory of Computation requires that all strings _s_ in _L_ at least as long as the pumping length can be split into *five* pieces, _s = uvxyz_. You seem to be splitting into four pieces. (NB: Your _z_ is Sipser's _s_, and you're missing his _z_.) – Sage Mitchell Jul 17 '16 at 17:04
  • What is the i in the definition of the language? In your factorization the c does not appear. Probably you mean x=c. – Peter Leupold Jul 18 '16 at 10:26

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The lemma says that there is a factorization that can be pumped. Not that every possible factorization can be pumped.

Your factorization would indeed lead out of the language, but there are others that do not.

Peter Leupold
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