Pumping Lemma $A = \{0^n1^n \mid n \geq0\}$. Prove $A$ is not regular

Question

Question : Are my Justifications Correct?

Pumping Lemma
$A = \{0^n1^n \mid n \geq 0\}$. Prove $A$ is not regular :

Suppose $S = 0^p1^p$ and $p = 3$.
Therefore, $S = 000111$. Breaking $S$ into $xyz$

---

Case 1 : $y$ is in the $0's$ part.

Therefore, $xy^iz \notin A $ for $i = 3$, because if $i = 3$, then $S_i = 0 000 111$ Therefore, there is an additional $0$, so $xy^iz \notin A $

---

Case 2 : $y$ is in the $1's$ part.

Therefore, $xy^iz \notin A $ for $i = 3$, because if $i = 3$, then $S_i = 000 111 1$. Therefore, there is an additional $1$, so $xy^iz \notin A $

Also, $|xy| > p$, because as $i$ increases, the amount of $1's$ is not equal to amount of $0's$ ?

---

Case 3 : $y$ contains $0's$ and $1's$

Therefore, $xy^iz \notin A $ for $i\geq1$ ? Because as $i$, the order of the symbols is violated.
Also, $|xy| > p$

---

I am confused at what value of $i$ does $xy^iz \notin A $?
Can someone please clarify?

Answer 1

In your proof, you assume $p=3$. This would only prove that there does not exist a machine [a finite automata] with three states accepting the given language $A = \{0^n 1^n: n \ge 0\}$. To show the language is not regular, you should not assume any fixed value of $p$. Also, you can assume that the length of $xy$ is at most $p$, which means $y$ would contain only 0’s. Here’s a proof:

Suppose there exists a machine with $p$ states accepting the given language $A$. Let $S = 0^p 1^p$. Then, there exist strings $x,y$ and $z$ such that $S = xyz$, $|xy| \le p$, $|y| \ge 1$ and $xy^kz \in A, \forall k \ge 0$. (This can be proved by the pigeonhole principle and is implied by the pumping lemma.) However, the conditions $|xy| \le p$ and $|y| \ge 1$ imply $y$ is a string of one or more zeros, whence $xyyz \in A$, a contradiction because $xyyz$ contains more $0$’s than $1$’s.