I want to come up with a language that satisfies the pumping lemma while not being a regular expression.

I thought of $\{0^i1^j: i > j > 0\} $. The pumping seems to work just fine, and this is not a regular expression, as it would not be possible to create an automata that recognizes it. This last part is where I'm having a bit of trouble, as I think I have the right idea, but can't formalize it. This language is just a regular expression $\big(\{0^i : i \in \mathbb{N}\}\big)$ concatenated with something that is not a regular expression $\big(\{0^i1^i: i \in \mathbb{N}\}\big)$.

Is my thought process correct? Can this be generalized? If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE? I believe so, because we know that if $L$ is a RE then so is $L^r$. And so we end up with $(L')^rL^r$, and we can use the contrapositive of the pumping lemma to prove this language is not regular. Any insight would be helpful!


2 Answers 2


Yes, $L_>=\{0^i1^j: i > j > 0\} $ is a non-regular language.

However, it does not satisfy the pumping lemma for regular languages. Assume for the sake of contradiction that $p$ is a pumping length for $L_>$. Consider $w=0^{p+1}1^p$. Let $w=xyz$ with $|y|>1$ and $|xy|<p$. Then $y$ contains 0 only. There is no more 0's than 1's in $xy^0z=xz$. That contradicts that $p$ is a pumping length.

The most we can say is that all words in $L_>$ can be pumped up.

If $L$ is a RE and $L'$ is not, then $LL'$ is also not a RE?

No. A simple counterexample is $L=\Sigma^*$, the language of all words while $L'$ is any non-regular language that contains the empty word such as $\{a^{n^2}\mid n\in \Bbb N_0\}$. $LL'$ is still the language of all words, which is regular.

Exercise 1. Find two languages $A$ and $B$ such that

  • $A$ is regular
  • $B$ is not regular
  • $A\subsetneq B$
  • $AB$ is regular.

Exercise 2. Find a non-regular language that satisfies the pumping lemma for regular languages.

  • $\begingroup$ Thank you for your answer, apreciate it a lot! In regards to Exercise 1, if you choose A = $\varnothing$, then I believe it works :) $\endgroup$ Jun 29, 2019 at 18:29
  • $\begingroup$ @mathmathmath nice answer. To make it more difficult, I should have added $A$ is not empty. $\endgroup$
    – John L.
    Jun 29, 2019 at 19:30

For a language that isn't regular, but satisfies the pumping lemma, take:

$$ L = \{a b^r c^r \colon r \ge 1\} \cup \{a^r b^s c^t \colon r \ne 1 \wedge s, t \ge 1\} $$

Given $\sigma = a b^r c^r$, you can pump the starting $a$, the result is always in $L$. Given a string that starts with no $a$, you can pump some starting $b$s; if it starts with 2 or more $a$, you can repeat the first two $a$s at will.

Suppose $L$ is regular. Then $L \cap \mathcal{L}(a b^* c^*) = \{a b^r c^r \colon r \ge 1\}$ is regular, but an easy use of the pumping lemma shows this not to be the case.


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