I am trying to prove that the follow language is not regular
L = {w ∈ {0, 1}∗ | the number of 1s in w is one more than the number of 0s}
My approach was to prove that it is regular and prove by contraction. So far I have:
p = pumping length
S = 0p1p and I could use 00111 string as an example I supposed. Am I going about this the right way? How could I go forward? I am new to pumping length.
Suppose that $L$ is regular. Then the pumping lemma says that there is a pumping length, $p$, such that if $s\in L$, and $|s|\ge p$, then $s$ can be decomposed as $s=xyz$ in such a way that $|xy|\le p$, $|y|\ge 1$, and $xy^kz\in L$ for each $k\ge 0$. To use this to get a contradiction, you need to start with some $s\in L$ that is at least $p$ in length.
I would start with $s=0^p1^{p+1}$: it has one more $1$ than it has $0$s, so it’s in $L$, and it’s certainly at least $p$ characters long. Moreover, the pumping lemma tells you that it can be decomposed as $xyz$ in such a way that $|xy|\le p$, which means that $xy$ lies entirely within the string of $p$ $0$s. Thus, there are non-negative integers $r$ and $s$ such that $x=0^r$, $y=0^s$, $r+s\le p$, and $s\ge 1$. Now the pumping lemma says that $xy^kz\in L$ for each $k\ge 0$; we’d like to get a contradiction by showing that this simply isn’t true for this particular $s$. So what is $xy^kz$? If $xy=0^{r+s}$, that leaves $p-(r+s)=p-r-s$ $0$s at the beginning of $z$ together with all $p+1$ of the $1$s, so $z=0^{p-r-s}1^{p+1}$, and
$$xy^kz=0^r(0^s)^k0^{p-r-s}1^{p+1}=0^{r+ks+p-r-s}1^{p+1}=0^{p+(k-1)s}1^{p+1}\;.$$
It’s not hard now to find values of $k$ for which $xy^kz\notin L$, and doing that proves that $L$ could not have been regular after all: it doesn’t satisfy the conclusion of the pumping lemma.
