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I was given the following question:

Use the Pumping Lemma with length to prove that the following language is non-regular:
$L = \{b^na^{100}b^{2n}, \text{where n} = 1, 2, 3,...\}$

Use the prompts below to complete the proof
Assume

Then there exists

We choose any word w =

Thus we can write w =

The according to the pumping lemma with length,

There is/are __ possible choice(s) for y

If y is pumped in each of the above case(s),

Now, my answer, actually how far I came, is as follow:

Assume L is a regular language.
Then there exists an FA with, say k states that accepts L.
We choose any word w = $b^na^{100}b^2n$ in L.
Thus we can write w = $b^ka^{100}b^{2k}$
The according to the pumping lemma with length, w may be written as w = xyz such that |x| + |y| <= k and |y| > 0

Then I don't know further. I do know that y is 1 b or more. y is also at most k b's.

I would appreciate it if you will correct my mistakes and also finish the answer. Thank you

Clint Theron
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1 Answers1

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You’ve filled in the first two blanks correctly, but after that you’ve gone a bit astray, I think because you’re not seeing how the contradiction is going to arise.

It’s true that you can start with any word in $L$, so $w$ could be $b^na^{100}b^{2n}$ for any $n\in\Bbb Z^+$, and the pumping lemma would say something about other members of $L$, but if we want to make matters easy for ourselves, we’ll take $w=b^ka^{100}b^{2k}$. Then the pumping lemma says that we can write $w=xyz$, where $|y|\ge 1$, $|xy|\le k$, and $xy^nz\in L$ for each $n\ge 0$. So what can $y$ be? We know that $xy$ is the first $|xy|$ characters of $w$, and $|xy|\le k$, so $xy$ is contained in the first $k$ characters of $w$. And $w$ begins with $k$ $b$s, so $xy$ is a string of $b$s, say $xy=b^\ell$. What are the possible choices for $\ell$? It has to be at least $1$: $x$ can be the empty string, but we know that $|y|\ge 1$. And it can be at most $k$, since we know that $|xy|\le k$. Thus, $1\le\ell\le k$, and there are therefore $k$ possible choices for $y$, from $b^1$ through $b^k$.

Suppose that we pump $y$ in any of these cases: if we pump $y$ up to $y^2$, we change $y=b^\ell$ to $y^2=b^{2\ell}$, adding $\ell$ $b$s, and we get the word $b^{k+\ell}a^{100}b^{2k}$. And this is our contradiction: on the one hand the pumping lemma says that this must be in $L$, but on the other hand it clearly is not in $L$, since $2k\ne 2(k+\ell)$ (because $\ell\ne 0$). Thus, the initial assumption that $L$ is regular must have been false.

It is possible to use other initial words to get a contradiction, but this is the simplest.

Brian M. Scott
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  • I think your hypotheses is correct; I'm not seeing how the contradiction is going to arise simply because I don't understand how we can pump y. The language clearly states what the length of y can be so how can we pump the length of y above its limits. Do you understand my confusion? There are k possible choices for y but we are pumping y to twice that, then, yes, a contradiction will arise. It should arise within the definition of the language then it would make sense to me. Will you explain what I'm missing here. – Clint Theron May 28 '20 at 21:53
  • @ClintTheron: You’re ignoring what the pumping lemma actually says: if $L$ is regular, $w\in L$, and $|w|\ge k$, then $w$ can always be decomposed into three substrings, $w=xyz$, with the following three properties: $|y|\ge 1$; $|xy|\le k$; and $xy^nz\in L$ for every $n\ge 0$. To understand why this is so, you have to understand the proof of the pumping lemma, but for now just accept that it is true. This means that if you have some language $L$ and find a word $w\in L$ such that ... – Brian M. Scott May 28 '20 at 22:01
  • ... $|w|\ge k$, but any decomposition of $w$ as $xyz$ that satisfies the first two conditions ($|y|\ge 1$ and $|xy|\le k$) fails to satisfy the third, because there is at least one $n\ge 0$ such that $xy^nz$ is not in $L$, then your language could not have been regular in the first place. That’s what I did: I showed that if we start with $w=b^ka^{100}b^{2k}$, which is certainly in $L$, then any decomposition $w=xyz$ that satisfies the length conditions fails to satisfy the pumping condition, because $xy^2z$ provably is not in $L$. (In fact the only $n\ge 0$ for which ... – Brian M. Scott May 28 '20 at 22:05
  • ... $xy^nz\in L$ is $n=1$.) – Brian M. Scott May 28 '20 at 22:05
  • I don't get why you went back to n when you say, "Then the pumping lemma says that we can write...., and $xy^nz\in L$ for each $n\ge 0$." Why not k? – Clint Theron Jun 02 '20 at 22:50
  • I think you also made a typo here, "And w begins with k b's, so xy is a string of a's". I think it should be a string of b's – Clint Theron Jun 02 '20 at 22:57
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    @Clint: I don’t use $k$ as a variable there because we’re already using $k$ for a specific number, namely, the pumping length (the number of states in the finite automaton that accepts $L$). But yes, you did indeed catch a typo; I’ll fix it as soon as I post this comment. – Brian M. Scott Jun 03 '20 at 02:31
  • I get now why you used n as variable. It's because you stating the properties of the pumping lemma. Thank you for the clarification. I appreciate it :) – Clint Theron Jun 03 '20 at 19:34
  • @Clint: You’re welcome. – Brian M. Scott Jun 03 '20 at 19:39