a^nb^n is not regular

In general, for a finite automaton with n states, the same argument can be done for any string of length greater than or equal to n Theorem 1 1 (2 4 1) Let L be a 

(j) If L1 and L2 are nonregular languages, then L1 ∪ L2 is also not regular 4 Show that the language L = {anbm : n ≠ m} is not regular 5 Prove or disprove the 

If L is a regular language, then there is an integer n > 0 with the property that: (*) for any string x ∈ L where x ≥ n, there are strings u, v, w such that (i) x = uvw,

Suppose that language A is recognized by an NFA N, and language B is the collection of strings not accepted by some DFA M Prove that A ◦ B is a regular 

us to prove that a certain language is not regular 4 1 The pumping lemma Theorem 4 1 Given a regular language L, then there is a number n ∈ N such

To show A1 = {anbn n ≥ 0} is not regular Let k be the constant of the pumping lemma (PL) Consider x = ǫ, y = ak and z = bk This meets the requirements of 

{ 0n1n n ≥ 0 } is non-regular – Assume L 1 is regular – Then there is a DFA M = (Q, Σ, δ, q Then n + k b = n + n b = n (1+b), which is not prime (since b ≥ 1)

28 oct 2010 · Prove that the following languages are not regular using the pumping lemma a L = {0n1m0n m, n ≥ 0} Answer To prove that L is not a 

is not regular Ex { 0 1 n 0} n n Limits of FA NO Can FA recognize all '' computable'' languages? Pumping Lemma > Pumping Lemma If A is a regular 

a regular expression is a nonregular language or an irregular language 2 strings w ∈ L, where w ≥ n, there exists a factoring of w = xyz, such that: • y = Λ