prove a^nb^nc^n is not regular

There are languages which are NOT context free ( LECTURE 11) Just as for regular languages we employ the pumping lemma in a Consider the language L = {anbncn n ≥ 0} Opponent picks p L = {ww w ∈ Σ∗} is NOT a CFL ( Prove it using pumping lemma) and na(w) = nb(w) = nc(w)} a CFL? L ︸︷︷︸

and z are strings and y ̸= ϵ, xy ≤ n and • x(yk)z ∈ L for all k ≥ 0 Proof: Because L is regular, there is a finite automaton M such that L is the language 

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,

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 

L4= {w ∈ Σ∗ na(w) + nb(w) = nc(w)},Σ= {a, b, c} (a) L1 ∩ L2 = Use the Pumping Lemma to prove the language L below is not regular L={w ∈ Σ∗ na( w) 

(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 

Show that the following languages are not context-free (a) L = {an2 : n ≥ 0} What is wrong with the following "proof" that anb2nan is context free? (i) From the fact that G is context free, it follows that there is no regular expression for L(G)

23 oct 2005 · (b) Show that B = {anbncii ≤ n} is not context free Proof: Let z = akbkck, and suppose z = uvwxy such that vx = ϵ and vwx ≤ k There are 

Note: We can construct an NFA N for each case and find a DFA M equivalent to N 2 9 2 more non-regular languages proved by Pumping lemma 1 Note that L = {w ∈ {a, b, c}∗ na(w) + nb(w) > nc(w) or na(w) + nb(w) < nc(w)} 15