Applying the pumping lemma, Theory of Computation

Applying the pumping lemma is not fundamentally di?erent than applying (general) su?x substitution closure or the non-counting property. The pumping lemma is a little more complicated-rather than just the single universal quanti?er ("for all languages L") and single existential quanti?er ("there exists n"), we have a nest of alternating quanti?ers (denoting "for all" as ∀ and "there exists" as ∃):

(∀L)[L regular ⇒

(∃n)[

(∀x)[x ∈ L and |x| ≥ n ⇒

(∃u, v,w)[x = uvw and

|uv| ≤ n and

|v| ≥ 1 and

(∀i ≥ 0)[uviw ∈ L]]]]].

Just as with the lemmas for the local languages, we will approach this as an adversary game. Our proof will consist of a strategy for showing that L fails to satisfy the pumping lemma. Our choices are the "for all"s; the "there exists"s are our adversary's choices. There are just a few more rounds in this game than there were in the lemmas for the local languages. The key things are being clear about which are our choices and which are the adversary's and making sure that our strategy accounts for every legal choice the adversary
might make.

The game starts with our choice of the L we wish to prove to be non regular. Our adversary then chooses some n, we choose a string x ∈ L of length at least n, etc. We win if, at the end of this process, we can choose i such that uviw ∈ L. Of course, our strategy at each step will depend on the choices our adversary has made.

What we end up with is a proof by contradiction. For instance:

To show that Lab = {ajbj| j ≥ 0} is not regular.

Posted Date: 3/21/2013 1:45:12 AM | Location : United States







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