Question

Pumping lemma :

There are 3 mistakes in this proof could you identify and show the corrections .please!

(b) Proof attempt: L = {w(w)R WEL*}l is not regular We choose the word x = abba. Then 2 = 2n+2 > n and I EL (mirror axis between the two b-blocks). We consider all decompositions 2 = uvw where v > 1, Uv Sn: 1) (the a is not in u:) u = €, v = ab", w=b"a 2) (the a is in u:) u = abk, v = bl, w = bn-k-lbra where k > 0,1 > 1, k +6+1 <n For each decomposition there exists an index i such that uv'w & L: 1) i = 0, then wvw = ba & L, because the word contains only one a, i.e. an odd number of as (i.e. there is some a in the word that does not have a mirror image in the other half of the word) 2) i = 3, then uvw = abb3lyn klima = abk+32+n k lyra = abn+24b"a & L, since because of (> 1 we have n +21 7n According to the Pumping Lemma L is therefore not regular.

Answer 1

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 = $ab^{k}$ . k>=0. w is also incorrecct as we have taken v incorrect.

w = . k+l = n,l>=1.

FIL

As we have taken the decomposition wrong for the first case in proof of pumping it down to i = 0

uw = where l>=1.

FIL

In the second case where we pumped up to i = 3.

we got $ab^{n+2l}b^{n}a$ 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 $ab^{n+2l}b^{n}a$ as $ab^{n+l}b^{n+l}a$

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.

uvlw= abklin-k-16" abk+il+n-k-l+na ab2n+(i-1)

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 $a^{n}bba^{n}$ .

then if we pump it up then we will pump a's and it is easy to prove.