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 € 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.
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 € 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
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Choose all the correct options please. I am quite confused.
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.
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.
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