Where is my mistake in proving a language is not regular? Trying to prove that $\big\{0^n1^m | n < m\big\}$ is not a regular language. If I take $w=0^n1^{n+x}=xyz$ with obvious $|w| > n$ then $|xy|\leq n $. Furthemore if I let $z=1^{n+x}$ then $x$ and $y$ contain $0$'s and given that $y\neq\epsilon$ then y must contain at least one $0$
Pumping lemma says that for all k $k \in \mathbb{N}$ $xy^kz \in L$ but if I take $k=x+10$ then suppose $w=0^{n-1}1y^{x+10}1^{n+x}$ then $ w \notin L$ because $n + x + 9 > n + x$.
Have I made a mistake anywhere in the following proof?
 A: I hope it can help you
we proof by contradiction. So, we first must assume something, then contradict it. Here we will assume the L
 is regular, then get a contradiction.


*

*assume L is regular 

*By the pumping lemma there exists a pumping length $p\ge1$ 

*take string $w=0^p1^{p+1}\in L $ such that $|w|\ge p$

*breaks $w$ into $xyz$  such that $|y|\ge1 \,,\,|xy|\le p  $ then $\,\,xy^iz\in L $ $\,\,\,\,\forall \,i\ge 0$
$$w=\overbrace{\underbrace{0000....0000}_{xy}\underbrace{...111....1111}_z}^{2p+1}$$
case 1:
$$w=\overbrace{\underbrace{0000....0000}_{|xy|=q=p}\underbrace{111....1111}_z}^{2p+1}$$
case 2:
$$w=\overbrace{\underbrace{0000....000}_{|xy|=q\lt p}\underbrace{\underbrace{00...00}_{p-q}111....1111}_z}^{2p+1}$$
To show that $L$ is not regular language, you need to consider all the decompositions of $w=xyz$ $.^{[1]}$


*

*$ |xy| = q\,\,\,,$ $1\le q \le p$

*$y=0^r\,\,\,\,\,\,\,$ , $\,1\le r\le q$

*from $[4] $ we have $\forall i\ge0$  then by the pumping lemma  $xy^iz\in L $ so  $q+ir +(p-q)=p+ir\lt p+1$ , ($n_0(w)+ir\lt n_1(w)$).
for $i=1$ : $xy^1z$ then $p+r\lt p+1$   so $xy^1z\notin L$ is a contradiction
we find integer $i=1$ : and we have ${p+ir \ge p+1}$ ,($n_0(w)+ir\ge n_1(w)$) for $i=1$,this is contradiction to the definition of our language $L$ Therefore the language $L$
is not regular. 

[1]  $xy$ contain $0$'s ($n_0(w)= p$) but you can't just let $z=1^{p+1}$ because you need to consider all the decompositions.  $z=0^t1^{p+1}$ such that $0\le t \le p-1$ 
