Need help with question one at the bottom by following the example above.Thank-you Example ProofProve that the language B = {0*1*|n > 0} is not a regular language.Proof. Assume that B is regular and let p be the pumping length for thelanguage. Choose the string s = 0P1P so that |s| > p. By the pumpinglemma (PL) for regular languages, we can partition s = ryz so that for anyi > 0, s' = ry'z e B. Let's consider three cases for the partition y:1. The string (partition) y consists only of O's. Then, s' = ryyz = ry²zhas more O's than l's. Clearly s' 4 B and so condition number 1 ofthe PL is violated. This is a contradiction for our assumption that Bis regular.2. The string (partition) y consists only of l’s. Using the same argumentfrom the previous case, we obtain another contradiction.3. The string (partition) y consists of 0's and l's. Then, s' = ryyz = ry²zmay have the same number of O's and l's, but some of the l's willcome before some of the 0O's and violate membership in the languageB. Hence, we have another contradiction with our assumption of reg-ularity.Therefore, we cannot avoid a contradiction with any possible y partition andconclude that the language B cannot be a regular language.InstructionsUse the Pumping Lemma for CFLS to prove that each of the following lan-guages is not context-free.1A = {0"#0²"#0*n | n >0}. Hint: consider s = 0"#0#0*.%3D

Pumping lemma for CFL is used to prove the language is not cfl using some valid string s. String s…

Answered: Use the Pumping Lemma for CFLS to prove…

Use the Pumping Lemma for CFLS to prove that each of the following lan- guages is not context-free. 1 A = {0*#0²m#0®» | n>0}. Hint: consider s = 0P#0#0®.

Computer Networking: A Top-Down Approach (7th Edition)

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Question

Need help with question one at the bottom by following the example above.

Thank-you

Example Proof
Prove that the language B = {0*1*|n > 0} is not a regular language.
Proof. Assume that B is regular and let p be the pumping length for the
language. Choose the string s = 0P1P so that |s| > p. By the pumping
lemma (PL) for regular languages, we can partition s = ryz so that for any
i > 0, s' = ry'z e B. Let's consider three cases for the partition y:
1. The string (partition) y consists only of O's. Then, s' = ryyz = ry²z
has more O's than l's. Clearly s' 4 B and so condition number 1 of
the PL is violated. This is a contradiction for our assumption that B
is regular.
2. The string (partition) y consists only of l’s. Using the same argument
from the previous case, we obtain another contradiction.
3. The string (partition) y consists of 0's and l's. Then, s' = ryyz = ry²z
may have the same number of O's and l's, but some of the l's will
come before some of the 0O's and violate membership in the language
B. Hence, we have another contradiction with our assumption of reg-
ularity.
Therefore, we cannot avoid a contradiction with any possible y partition and
conclude that the language B cannot be a regular language.
Instructions
Use the Pumping Lemma for CFLS to prove that each of the following lan-
guages is not context-free.
1
A = {0"#0²"#0*n | n >0}. Hint: consider s = 0"#0#0*.
%3D

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