Question

Prove that language L on {a, b}, L={ v | v = vR} is not regular

use string ab^nab^na

Answer 1

pumping Lewma - *) If A is a Regular Language, then A has a Pumping Length p such that ocny string 'g' where 18129 may be divided into 3 pasos g=XYZ such that the following conditions must te tove DDYTZEA for ergy izo 214) > 0 3) IXY) SP To prove that a longuage is not regular using Pumping Lemma follow the below steps: cwe prove using contadico op 1. Assime that A is Regular 2. It has to have a pumping Lengon csay D 3 All Stings Longer than p can be Pumped ISIZP 4 Now find a string g' in A such that 18/2P 5-DNide s into xyz 6- show that xyiz&a fer some i 7. Then consides all ways that I can te divided into Xyz

zoata 121=3 L = {v/v = vR3 where L on sov63 lo Assume that Lis Regular 2. Pumping Length According to condition It is pourmdiame Take La fama where m=023, ... I will take ni =) Lzablabla L-СТЬciba 3. Divide Language into 3 Parts Xyz Laababa ca 1x)=1 x y z 4=6 MI=1 string lengthy)=5 case 1: - IXYISh assu case2 -- 14>O lon cases -- L=scyiz izo i will take i-2 l=act? ata (=abbaba of our Assumption is cong, Becase our Language must te paundjeme cv=rRy but sight now v=abloata VR-atlabba mis match xyizal L={v/v = vR3 whese len şar63 is not degulas x pailed L

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