Pumping Lemma for Regular Languages
Pumping Lemma for Regular Languages: If $A$ is a regular language, then there is a number $p$ ( the pumping length ) where if $s$ is any string in $A$ of length atleast $p$, then $s$ may be divided into three pieces, $s=xy$, satisfying the following conditions:
- $for \ each \ i>=xy^iz∈A$,
- $|y|>0$, and
When using pumping lemma to prove that a given language is "non regular, we do the following steps:
- Assume $L$ is regular, hence $L$ exhibits pumping lemma for regular langauges
- Since $L$ is regular, it will have $p$ and if we choose a string that is atleast as long as $p$
- Then that string will have decompositions satisfying all the three conditions of the lemma
- This is where we obtain a contradiction by showing that the string will fail to satisfy condition 1 of the lemma and hence our assumption is false and the language is not regular
My question is: Do we need to show that the string fails to satisfy condition 1 for one $i$ or all $i>=0$. I do understand that we need to show that string fails to satisfy for one $i$ but should the proof be able to extrapolate for any value of $i>=0$?
My understanding is that: In order to have a strong valid contradiction, we need to show that our string choice and pumping of the string fails for all values of $i$ according to condition 1
But when we say "for all values of $i>=0$", we know that for $i=1$, the pumped string does exist in the language( because we chose that string ), and sometimes it also exists in the langauge for $i=0$, so are these sort of exceptions?