Don't set at all. Let for some choice of . Now, suppose there is some decomposition with , . Hence, must consist only of 's, and it must have at least one. Now pump up .
Hi
I am trying to prove if language is not regular. my language is {a^i b^j c^l | where i=j}.
I assume this language is regular and that there exist constant p
I am trying to pick my string w but i don't know what are constrains on l.
So if i pick my string on at least p length it will be w = a^p b^p c^?
Do I have to consider 3 cases such as :
w = a^p b^p c^(l<p)
w = a^p b^p c^(l>p)
w = a^p b^p c^(l=p)
Thanks
hi thanks so much. It looks like i can do few things:
I can say w0 = xy^0z = xz (this actually drops y out) and therefore we have at least 1 less 'a' than b
I can say w2 = xy^2z = which adds at least 1 'a' and makes it more 'a' than 'b'
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