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So I have a pumping lemma question A{www|w ∈ {a,b}*} I have the correct answer but I'm not fully sure how it works. I'll give the answer just so people know what I'm going with

Assume A is REG let p be the pumping length x ∈ A, x=a^p b, a^p b, a^p b.... |s|=3p+3 where each a^p b is a w

Let s = xyz a split such that 1)sum of i>=0 s'=xy'z ∈ A 2)|x|>0 , 3)|xy| <=p

By (3) y contains only a's and by (2) y contains at least 1 a. Let s'=xyyz, Then s=a^+ ba^p ba^p b,

1)s' ∈ A as it contains contradiction t>p ie. A not an element of REG

Jon
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you want to prove that $\mathcal L=\{www|w\in\{a,b\}^*\}$ is not regular

The word the you chose: $x=a^pba^pba^pb$

$|x|=3p+3>p$ so we can use the pumping lemma

so $x=uvw$ etc...

Note that

$\underbrace{aaaaa...b}_{\text{p+1 latters}}\underbrace{aaaaa...b}_{\text{p+1 latters}}\underbrace{aaaaa...b}_{\text{p+1 latters}}$

So since that $|uv|\leq p$ uv can be here:

$\overbrace{aaa...}^{\text{uv}}b,aaaa....b,aaaa....b$

Or here:

$aaaa....b,aaaa....b,\overbrace{aaa...}^{\text{uv}}b$

Or here:

$aaaa....b,\overbrace{aaa...}^{\text{uv}}baaaa....b$

Or here:

$aaaa...\overbrace{.b,aa}^{\text{uv}}aa...b,aaaa....b$

Or here:

$aaaa....b,aaaa..\overbrace{.b,a}^{\text{uv}}aaa...b$

Now if you choose $i=2$ the word is not in $\mathcal L$