Is this proof with the pumping lemma (no regular language) ok?

Question

I need to proof that a given language is not regular, could this work?

The language is M={a^ma^lcb^(m+l)|m,l in N} with the alphabet = {a,b,c} .

Proof:

Be n in N arbitrary but firm. We choose the word w=a^(2n)cb^(2n) with w in M and |w|>=n.
Be w=xyz a arbitrary decomposition with y!=lambda and |xy|<=n.
Then we have x=a^(2i), y=a^(2j) and z= a^(2n-2i-2j)cb^(2n) for j!=0 and 2(i+j)<=2n.
Now we choose k=0. The we have xy^0z=a^(2n-2i)cb^(2n).
=> xy^0z is not in M because 2n-2i!=2n for j!=0.
=> M is no regular language.

Yeah or nah? If you could tell me my mistakes i would be very thankful

Answer 1

Your idea is right. Just a few details:

"fixed" not "firm" (translation from German?)

You need to distinguish the n you choose and the constant from the pumping lemma (which you do not choose).

So:

Let K be the constant for M from the pumping lemma and let n be a natural number such that n>K.
We choose the word w=a^(2n)cb^(2n) with w in M  and |w|>=K.
Be w=xyz a arbitrary decomposition with y!=lambda and |xy|<=n.
Then we have x=a^(2i), y=a^(2j) and z= a^(2n-2i-2j)cb^(2n) for j!=0 and 2(i+j)<=2n.
Now we choose k=0. The resulting word is xy^0z=a^(2n-2i)cb^(2n).
xy^0z is not in M because 2n-2i!=2n for j!=0.
=> M is no regular language.