Irregularity of $a^ib^jc^k∣\text{ if }i=1\text{ then }j=k$ using pumping lemma

Question

This question has been asked here several times. However none of the examples I saw actually tried to prove it's irregularity using the standard pumping lemma.

I know how to prove irregularity using the specific case of $a^1b^pc^p$ however I'm struggling to build the pumping lemma for the whole language $a^ib^jc^k$ I know this won't prove the language to be regular, but I'm more interested in how to deal with lemmas with multiple variables on the exponents. Let's suppose I've divided the string into the following:

$ x = \epsilon$
$ y = b^\beta$
$ z = b^{p-\beta}c^m$

Let $n = 2$. I get
$xy^2z= b^{p-\beta}c^n$

But this doesn't seem like I can find a pumping length for this language as it has too many variables. Is there a way to make this pumping lemma work? (I know it can't be used to prove the language regular, but I'm trying to find at least one pumping length which works for this language)

Answer 1

If I understand correctly, you want to explicitly show that language $\{a^i b^j c^k | i, j, k \in \mathbb{N}\}$ satisfies pumping lemma (it should be true, as this language is clearly regular).

Notation of pumping lemma: there exist pumping length $p$ s.t. any word $w$ with $|w| \geq p$ can be written as $w = xyz$ where $|y| > 0$, $|xy| \leq p$ and for all $n$, $x y^n z$ is in our language.

For our language, we can take $p = 1$, and represent word $w$ as $\epsilon . y . z$ ($.$ denotes concatenation), where $|y| = 1$.

As $w$ is in our language, $w = a^i b^j c^k$, as $|w| > 0$, at least one of $i, j, k$ is non-zero.

If $i > 0$, they $y = a$ and we have $a^{i - 1 + n} b^j c^k$ is in our language - it's true by definition.

Cases $i = 0, j > 0$ and $i = j = 0, k > 0$ are similar.

Answer 2

The pumping lemma states that if $L$ is regular, there is a constant $N$ so that for all $\sigma \in L$ with $\lvert \sigma \rvert \ge N$ there is a division $\sigma = \alpha \beta \gamma$ with $\lvert \alpha \beta \rvert \le N$ and $\beta \ne \varepsilon$ for all $k$ it is $\alpha \beta^k \gamma \in L$.

Sadly, in this case if you pick e.g. $\sigma = a b^N c^N$, you can always pick $\alpha = \varepsilon$, $\beta = a$ and $\gamma = b^N c^N$, and everything works out beautifully. No contradiction to be had.

If you check out the proof of the pumping lemma, it is clear that you can prove variants that for example state $\lvert \beta \gamma \rvert \le N$, and $\beta$ can be pumped. Use that one to get a contradiction.

(Nice question, by the way.)