How does Sipser's proof that $0^n1^n$ is not regular work?

Question

In Sipser's Introduction to the Theory of Computation this is how $0^n1^n$ is proved to be not regular

Example 1.73:

Let $B$ be the language $\{0^n1^n|n \ge 0\}$

We use the pumping lemma to prove that $B$ is not regular. The proof is by contradiction. Assume to the contrary that $B$ is regular. Let $p$ be the pumping length given by the pumping lemma. Choose $s$ to be the string $0^p1^p$. Because $s$ is a member of $B$ and $s$ has length more than $p$, the pumping lemma guarantees that $s$ can be split into three pieces, $s = xyz$, where for any $i \ge 0$ the string $xy^iz$ is in $B$. We consider three cases to show that this result is impossible.

• The string $y$ consists only of $0$'s. In this case, the string $xyyz$ has more $0$'s than $1$'s and so is not a member of $B$, violating condition 1 of the pumping lemma. This case is a contradiction.

• The string $y$ consists only of $1$'s. This case also gives a contradiction.

• The string $y$ consists of both $0$'s and $1$'s. In this case, the string $xyyz$ may have the same number of $0$'s and $1$'s, but they will be out of order with some $1$'s before $0$'s. Hence it is not a member of $B$, which is a contradiction.

In the first condition ($y$ consists only of $0$'s), why would the string $xyyz$ have more $0$'s than $1$'s?

Let's say that $|x|=p-2$ and consists only of $0$'s. $|y| = 1$ and $y=0$, $|z| = p$ and consists only of $1$'s. At this point $|y| > 0$ and $|xy| \le p$.

In this case we would have the same number of $0$'s and $1$'s without violating any of the constraints of the pumping lemma. Of course $xyyyz$ would violate it then, but that's not his proof.

My question is, why does he state that the number of $0$'s is larger than the number of $1$'s for $xyyz$?

Answer 1

If $x$ consists of $p-2$ zeros, $y$ is 0 and $z$ consists of $p$ ones, then $xyz$ is $0^{p-1}1^p$, not $0^p 1^p$. Hence, the particular partition you give is impossible. If $|xy| < p$, then $z$ must take the remaining zeros.

Recall that we started with the assumption that the string $0^p 1^p$ was partitioned into three section $x,y,z$. This implies the number of zeros in $x,y,z$ must be $p$, and since $y$ is 0, $xyyz$ will have more $0$'s than $1$'s.

Answer 2

With the way you split your string, $|z|$ is $p +1$ and not $p$. Then, $z$ will be $01^p$ and not $1^p$. And with this, the proof stated by Sipser still holds.