抽水引理(没有常规语言)可以证明这一点吗?

时间:2019-01-17 23:05:09

标签: regular-language proof pumping-lemma

我需要证明给定语言不是常规语言,这行得通吗?

语言为M={a^m a^l c b^(m+l)|m,l in N},其中 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.

是还是不? 如果您能告诉我我的错误,我将非常感谢

1 个答案:

答案 0 :(得分:0)

您的想法是正确的。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.