Thread: Pumping Lemma proof...

Hello,
Could you please let me know what is wrong with the following proof:
proof that L = a*b* is not regular:Suppose L is regular, so the Pumping Lemma applies to L. Let p be the pumping constant for L.Consider w = a b^p ∈ L.
| w | ≥ p, so the pumping lemma applies to w.
Thus there exist x, y, z ∈ { a, b }* such that w = xyz and 1 ≤ | y | ≤ p.
Take y = abn (n ≤ p), so x = ε and z = bp-n.
Now w2 = x y^2 z = a b^n a b^p ∉ L.
This contradicts the Pumping Lemma and so L is not regular.

L = a*b* is not regular

Before reading the proof: how is it not regular if you define L by a regular expression? Did you mean ?

well a*b* is considered for regular so it gets to contradiction.

Take y = abn (n ≤ p), so x = ε and z = bp-n.

This is not clear. You don't have control over x, y, and z: they are determined by the (proof of the) Pumping Lemma. One you fixed w, you are given x, y and z that satisfy the properties stated in the lemma.

well a*b* is considered for regular so it gets to contradiction.

What is the definition of L?