Pumping lemma :
There are 3 mistakes in this proof could you identify and show the corrections .please!
Please comment if you have any doubt regarding this question.I
will respond as soon as possible.Thank you.
In the first case of decomposition if a is not in the list
u = e is fine but v will not be of length n+1 as |uv|<=n and other thing is uv will not always have n length.
So v = . k>=0. w is also incorrecct as we have taken v incorrect.
w = . k+l = n,l>=1.
As we have taken the decomposition wrong for the first case in proof of pumping it down to i = 0
uw = where l>=1.
In the second case where we pumped up to i = 3.
we got and in the proof it is mentioned that as l>=1 we have n+2l not equal to n
But there is a mistake in this as we can write as
and this belongs to the language as there is a mirror line after the b block.
So it is better if we consider a different i value.
we have to choose i such that 2n+(i-1)l will be odd as if it is even then we can divide into two parts.
This is a tough task instead we need to choose our initial string carefully to make our later proof easy.
For example if we take a string .
then if we pump it up then we will pump a's and it is easy to prove.
Pumping lemma : There are 3 mistakes in this proof could you identify and show the...
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