Define a set S as follows: 1. 1 e S 2. If s e S then: 1. Os e S 2. 1s e S We wish to prove using structural induction to prove that every string in S ends in a 1. Predicate: P(s): The last symbol of s is 1. Base Case: P(1) holds since "1" ends with 1. Inductive Hypothesis: Assume that if s e S then P(s) holds. Inductive Step: There are two cases. Case 1) The string is "1s". Since we know S has no empty strings, IH implies P(s) holds, and s ends with a 1, so 1s ends with a 1. Case 2) The string is "Os". Since we know S has no empty strings, IH implies P(s) holds, and s ends with a 1, so Os ends with a 1.
Define a set S as follows: 1. 1 e S 2. If s e S then: 1. Os e S 2. 1s e S We wish to prove using structural induction to prove that every string in S ends in a 1. Predicate: P(s): The last symbol of s is 1. Base Case: P(1) holds since "1" ends with 1. Inductive Hypothesis: Assume that if s e S then P(s) holds. Inductive Step: There are two cases. Case 1) The string is "1s". Since we know S has no empty strings, IH implies P(s) holds, and s ends with a 1, so 1s ends with a 1. Case 2) The string is "Os". Since we know S has no empty strings, IH implies P(s) holds, and s ends with a 1, so Os ends with a 1.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
I need to see solutions. Please dont say mea answer without an answer.I nee do too look at it
Transcribed Image Text:In this exercise, we say that writing two strings next to each other combines them together, e.g. if
v='101' and w='110', then vw is the string '101110'
Define a set S as follows:
1. 1 e S
2. If s e S then:
1. Os e S
2. 1s e S
We wish to prove using structural induction to prove that every string in S ends in a 1.
Predicate: P(s): The last symbol of s is 1.
Base Case: P(1) holds since "1" ends with 1.
Inductive Hypothesis: Assume that if s e S then P(s) holds.
Inductive Step: There are two cases.
Case 1) The string is "1s". Since we know S has no empty strings, IH implies P(s) holds, and s ends
with a 1, so 1s ends with a 1.
SO
Case 2) The string is "Os". Since we know S has no empty strings, IH implies P(s) holds, and s ends
with a 1, so 0s ends with a 1.
Transcribed Image Text:The proof is correct.
The proof is incorrect.
The predicate is not valid.
There is a missing base case.
The Inductive Hypothesis is not valid.
Case 1 is wrong.
Case 2 is wrong.
Expert Solution
Step 1
A set S is defined as follows:
This way of defining a set is called recursive definition of set. Here, the element 1 is a base element (fixed). So, the set S is not empty. It has the element 1.
To Prove: Every string s in set S ends with 1.
Step by step
Solved in 3 steps
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
