1. ## pumping lemma

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

2. ## Re: pumping lemma

Don't set $l$ at all. Let $w = a^pb^pc^l$ for some choice of $l$. Now, suppose there is some decomposition $w = xyz$ with $|xy|\le p$, $|y|\ge 1$. Hence, $y$ must consist only of $a$'s, and it must have at least one. Now pump up $y$.

3. ## Re: pumping lemma

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'
....
....

exactly

5. ## Re: pumping lemma

thank you for input