Answer
See explanation
Work Step by Step
We want to show, using structural induction on the set \(S\), that **every string in \(S\) ends with the symbol ‘1’**. Here is a concise proof.
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## Definitions
The set \(S\) is defined recursively as follows:
1. **Base:** The string \(1\) is in \(S\).
2. **Recursion:** If \(s \in S\), then
- \(0s \in S\),
- \(1s \in S\).
3. **Restriction:** Nothing is in \(S\) other than those strings obtained by the above rules (1) and (2).
Our goal is to prove that **every** string \(x \in S\) ends in the symbol ‘1’.
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## Proof by Structural Induction
We proceed by structural (or “rule‐based”) induction on how elements of \(S\) are generated.
### Induction Hypothesis
For any string \(s \in S\), we assume as the inductive hypothesis that \(s\) ends with the symbol ‘1’. We must check this is true for the base case, and then show it holds for all new strings formed by the recursion rules.
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### Base Case
By rule (1), the string \(1\) is in \(S\). Clearly, \(1\) ends with ‘1’. So the statement holds for the base element.
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### Inductive Step
Suppose \(s \in S\) is an arbitrary string **already known** (by induction) to end with ‘1’. We need to check the two ways new strings can be formed from \(s\):
1. **Forming \(0s\):**
By rule (2a), if \(s \in S\), then \(0s \in S\). The string \(0s\) is obtained by prefixing a ‘0’ onto \(s\). Since \(s\) ends with ‘1’, the new string \(0s\) also ends with exactly the same last symbol as \(s\). Hence \(0s\) ends with ‘1’.
2. **Forming \(1s\):**
By rule (2b), if \(s \in S\), then \(1s \in S\). The string \(1s\) is formed by prefixing a ‘1’ onto \(s\). Again, this does not affect the last symbol of \(s\); the new string \(1s\) ends with whatever \(s\) ended with. Since \(s\) ended with ‘1’, \(1s\) also ends with ‘1’.
Thus, **both** newly formed strings \(0s\) and \(1s\) end with ‘1’. This completes the inductive step.
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### Conclusion
By structural induction, every string generated in \(S\)—starting from the base string \(1\) and applying the recursion rules—must end with the symbol ‘1’. Therefore,
\[
\boxed{\text{All strings in }S\text{ end with ‘1’.}}
\]
