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L = {a^i b^j c^k; i≠j and i≠k and j≠k}.

First approach: I tried two different string to prove it by pumping lemma but non of them is correct. first w = a^m b^m+1 c^m+2 and m is pumping length. for example one case in w = uvxyz is that vxy in is a part. so w = a^m-k a^k b^m+1 c^m+2 for any i >=0 it has to be in the L wi = a^m-k a^ik b^m+1 c^m+2. I cant show that number of a's is equal to number of b's.

Second approach: I converted L into union of 6 different languages {a^ib^jc^k U a^ib^kc^j U a^jb^ic^k U a^jb^kc^i U a^kb^ic^j U a^kb^jc^i ; i

Pedram
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1 Answers1

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I found the answer.

If we pick W= a^m! b^(m+1)! c^(m+2)! then we can prove it.

I am solving for the case that uxy is in a's side. then W= a^m!-k a^k b^(m+1)! c(m+2)! we know that there exist an integer x = m! m / k for 1<= k <= m so we pick i = (1+x) then m! + k x = m! + m! m = m! (m+1) = (m+1)! means the number of a’s = number of b’s which means this string is not in L. This is a contradiction.

Pedram
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