somebody give me this answer but it is not correct and he did not explain it to

Question

somebody give me this answer but it is not correct and he did not explain it to me so I gave him a bad evaluation

so please make sure to give the correct and explained the answer

those are the wrong answer I think

Let s=bp-3a4, such that length of s is greater than p. Here i=0 , j=p-3 and m=4

and p-3>4.

Then y must be “ba” and xy2z=babaa4, which is not in L since b comes after a.

5. Use the e pumping lemma for regular languages to prove that the anguage L given below is not regular. The alphabet of L is = { a, b, c} You may write your proof in your preferred format, but your proof must precisely specify the information required for each of the five blanks in the following template. Note that the third blank requires an integer as an exponent. Use p as the pumping length. Define all variables that you use in your proof. contradicts the pumping lemma because for all x, y, z e2 -is not Proof. The sentence s". such that xyz-s, iflxyl sp and lyl >0, then y must be-- --, and xy-z. in L because QED.

Explanation / Answer

pumping lemma is applied to show any language is not regular.

if L is regular then it satisfies pumping lemma, if L does not satsify pumping lemma then it is non regular

So now to prove a language L is not regular then we have to consider the following steps:

1. we have to assume that the L is regular.

2. so now the pumping lemma will hold the language L.

3. select w such that |w| > c

select y such that |y| > 1

select x such that |xy| < c

4. assign the remaning string to z.

5. select k such that the resulting string is not in L.

let u prove that L is non regular in this language L={ci bj an | j>0, i>0 and i+j>m }

1. first consider L is regular and n is the no of states

2. let w=cn bn an then |w|=3n> n

3. by pumping lemma let w=xyz where xy< p

4. let x=apy=bqand z= ar where p+q+r=n, p q r are not equal to zero.

5. let k=1, then xy2z = apa2qarbn

6. number of as (p+2q+r)=(p+q+r)+r=n+q

7. hence  xy2z = an+q bn since q is not equal to zero  xy2z is not in the form of cn bn an

8. thus xy2z is not in L. hence L is not regular.

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