抽水引理（没有常规语言）可以证明这一点吗？

Question

我需要证明给定的语言不是正常语言，这行得通吗？

语言是M={a^ma^lcb^(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.

是或不是？ 如果您能告诉我我的错误，我将非常感谢

Answer 1

你的想法是对的。 只是一些细节：

“固定”而不是“确定”（德语翻译？）

您需要将选择的n和常数与抽运引理（未选择）区分开。

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.