Question

1. Let L = {ambm cn | m <n}. Use the pumping lemma to show that L is not a CFL.

Answer 1

Pumping Lemmacfor CFL) is used to prove that a Language is not context face * If A is a context free Language, then, A has a Pumping Length's such that any string & Whose ISIP may be divided into 5 Pieces 9=UNXYZ Such that the following conditions must te tove Duvixyiz is in a for every izo 2) Ivylo 3) IVXYLP * To prove that a Lango is not context free using Pumping Lemma (foro CFL) {we Prole using contradictions 1) Assume that A is context free ii) It has to have a puning Lengon (Say D 717) All strings Longer than p can be pumped 1512P PV Now find a string & in A such that 15127 V Divide s into UV X Y Z VI) Show that uvixyiz & A for some i vii) Then consider the ways that I can be divided into Uvxyz the Vill show that some of these can satisfy all 3 Pumping conditions at the same time it I can't te Pumped Contadichon

L = {ammen, men 1. Assume that Ľis context free 2. Pumping Length According to condition let m=2 than 923 L={2}}? 33=çaabbccc} 3. Divide Latrage into spots UVxyz l=saabbccc} uzoa lur I V X Y Z v=66 IV -2 x =c IxI=1 soning Length (n) =7 4 = c 140-1 z = ( IZISI Cose 1:-1vxylch Low casca-- Irgs>O 30 case 3 - Lauvixyiz izo i will take i-2 - L=aacto3cco²c i aabbbbbc c CCC & forled of) out Assumporon is asong Becake sightx equou no of a's and t's not these and msnis fased B65C40's (Soyled UvixqZ&L fg gore i - Sammen meng is not contest be
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