Length of mid part of the string in Pumping Lemma This statement of the pumping lemma from Wikipedia.


  Let $L$ be a regular language. Then there exists an integer $p \ge  1$ (depending only on $L$) such that every string $w$ in $L$ of length at least $p$ ($p$ is called the "pumping length") can be written as $w = x y z$ (i.e., $w$ can be divided into three substrings), satisfying the following conditions: 
  
  
  $\lvert y \rvert \ge 1$  
  $\lvert x y \rvert \le p$ and  
  for all $i \ge 0$, $x y^i z \in L$.
  $y$ is the substring that can be pumped (removed or repeated any number of times, and the resulting string is always in $L$).  
  


What confuses me about the definition of pumping lemma are two requirements: $\lvert y \rvert \ge 1$  and $i \ge 0$, $x y^i z$. The way I read it, that we are required to have $y$ length be equal to one or greater, and at the same time, we can completely skip it, since $i \ge 0$, i.e. effectively  $\lvert y \rvert = 0 $.
Intuitively, it makes sense that we should be able to skip $y$ and still have string be in $L$.
Any finite state automaton that accepts an infinite number of words will necessarily have a loop in it. One such loop may go from state $q$ to state $q$ consuming word $y$ – that is, $y$ is the word based on the symbols on the transitions going from $q$ back to $q$. 

The pumping lemma says, more or less, that you can get to such a state $q$ from the initial state by consuming the word $x$ and that you can get to the final state from state $q$ by consuming the word $z$. In the middle you can go through the loop as many times as you like. Thus making the complete word $xy^iz$, where $i$ is the number of times you chose to go through the loop.