I need help with question 2 please.

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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I need help with question 2 please. 

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 = xyz so that for any
i > 0, s' = xy' z E B. Let's consider three cases for the partition y:
ry'z
1. The string (partition) y consists only of 0's. Then, s' = xyyz =
has more 0'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' = xyyz = xy²z
may have the same number of 0's and l's, but some of the l's will
come before some of the 0'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"#0³n | n > 0}. Hint: consider s = 0P#0²r#0³P.
C = {w| the number of 1s equals the number of 2s and the number of 3s equals the number of 4s
and E = {1,2, 3,4}. Hint: consider s = 1P3P2P4P.
Transcribed Image Text: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 = xyz so that for any i > 0, s' = xy' z E B. Let's consider three cases for the partition y: ry'z 1. The string (partition) y consists only of 0's. Then, s' = xyyz = has more 0'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' = xyyz = xy²z may have the same number of 0's and l's, but some of the l's will come before some of the 0'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"#0³n | n > 0}. Hint: consider s = 0P#0²r#0³P. C = {w| the number of 1s equals the number of 2s and the number of 3s equals the number of 4s and E = {1,2, 3,4}. Hint: consider s = 1P3P2P4P.
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