我需要证明给定的语言不规律，这可行吗？
语言是M={a^m a^l c b^(m+l)|m,l in N}，字母= {a,b,c}。
证明：
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.
是啊还是不？如果你能告诉我我的错误，我将非常感激
你的想法是对的。只是一些细节：
“固定”而非“坚定”（德语翻译？）
你需要区分你选择的n和常数与泵浦引理（你没有选择）。
所以：
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.