正则语言泵引理：能否将首个条件修改为“对每个i>0”？

Great question(s)! Let's unpack this thoroughly—both of your questions are asking the same core thing: can we replace the first condition of the pumping lemma for regular languages from "for every i ≥ 0, xyⁱz ∈ L" to "for every i > 0, xyⁱz ∈ L"?

First, let's recap the standard pumping lemma to set context:

For any regular language L, there exists a pumping length p such that every string s ∈ L with |s| ≥ p can be split into s = xyz where:

1. For every i ≥ 0, xyⁱz ∈ L;
2. |y| ≥ 1;
3. |xy| ≤ p.

Now, let's break down the two key angles of your question:

1. Do all regular languages satisfy the modified condition (i > 0)? #

Absolutely. If a regular language satisfies the original condition (i ≥ 0), it automatically satisfies the modified one—since i > 0 is just a subset of i ≥ 0. Every regular language will still pass the check if we only require pumping for i ≥ 1.

2. Is modifying the condition a good idea for the pumping lemma's purpose? #

No, and here's why: the pumping lemma is used as a necessary condition to prove languages are not regular. The original condition (including i=0) gives us a stronger check that can catch non-regular languages that the modified condition would miss.

Consider this example of a non-regular language:
L = { aⁿbᵐ | n ≥ m ≥ 1 } ∪ { aᵏ | k ≥ 0 }

This language is non-regular, but it satisfies the modified pumping condition:

• For any long string s ∈ L that's just a's: split x=ε, y=a, z=a^(k-1). For any i>0, xyⁱz = a^(i + k-1) which is in L.
• For any long string s = aⁿbᵐ (n ≥ m ≥1): split x=a^(n-m), y=a, z=a^(m-1)b^m. For any i>0, xyⁱz = a^(n+i-1)b^m, where n+i-1 ≥ m (since n ≥ m and i ≥1), so this string is in L.

But the original pumping lemma catches that L is non-regular: take s = aᵐbᵐ (long enough for the pumping length p). Split into xyz, and when i=0, xz = a^(n-m)a^(m-1)b^m = a^(n-1)b^m. If n=m, this becomes a^(m-1)b^m, which is not in L (since m-1 < m). This contradicts the original pumping lemma, proving L is non-regular.

With the modified condition (i>0), we couldn't use this contradiction—L would pass the modified check, even though it's non-regular. So the modified condition weakens the pumping lemma's ability to prove non-regularity.

To sum up:

• Regular languages do satisfy the i>0 condition, so if you're only asking whether the condition holds for regular languages, the answer is yes.
• But you shouldn't modify the pumping lemma's first condition to i>0 if you want to use it to prove languages are non-regular—it loses critical power to distinguish non-regular languages that "fail" only when i=0.

This applies equally to the general pumping lemma and the version in Sipser's Introduction to the Theory of Computation—the core logic holds for both.

内容的提问来源于stack exchange，提问作者Anonemous