Using the Pumping Lemma to prove a language is not regular.

Question

I want to know if my proof is wrong and whether what I am doing works.

$$\sigma = \{0, 1\}$$ $$A = \{0^n1^m \mid n < m\}$$

Claim: A is not regular.

Proof:

Assume A is regular. Let p be the pumping length.

Let $S = 0^p1^{p+1}$

$S = xyz$ where $y \neq \varepsilon $ where $ \varepsilon $ is the empty string and $|xy| \leq p$

$y = 0^k$ where $0<k \leq m$

$x = 0^q$ where $0\leq q < m$

$z = 0^{p-k-q}1^{p+1}$

The pumping lemma says that $xyyz \in A $

$$xyyz = 0^q0^k0^k0^{p-k-q}1^{p+1}$$ $$ = 0^{p+k}1^{p+1} $$

Since $p+k \geq p+1$

$0^{p+k}1^{p+1} $ is not an element of $A$. This is a contradiction.

Therefore, A is not regular.

As written, $L$ is a language with exactly two elements. This is probably not what you want. Do you mean $L = { 0^n1^m \mid n < m }$? (And you refer to it as $A$ afterwards.) — Magdiragdag, Mar 13 '14 at 05:05
Very similar question: http://math.stackexchange.com/q/523994/25554 — MJD, Mar 13 '14 at 12:50

score 1 · Accepted Answer · answered Mar 13 '14 at 19:14

Your argument is correct, but its presentation could be improved. As written, it looks like you choose $x, y, z$ and then apply the pumping lemma to conclude that $xy^2z \in A$. However, the pumping lemma asserts the existence of such $x, y, z$.

So, that part is better phrased as "Let $S = 0^p1^{p+1}$. By the pumping lemma there are $x, y \neq \epsilon, z$ such that $S = xyz$, $|xy| \leq q$, and that $xy^lz \in A$ for all $l$. Then $x = 0^q$ for some $q$, $y = 0^k$ for some $k > 0$, etc.".

Using the Pumping Lemma to prove a language is not regular.

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