Answer
See explanation
Work Step by Step
We want to show by **structural induction** that **every integer in \(S\) is a multiple of 5**. Concretely, if \(x\in S\), then \(5\mid x\).
---
## 1. Definition of \(S\)
\(S\) is defined recursively as follows:
1. **Base:** The integers \(0\) and \(5\) are in \(S\).
2. **Recursion:** If \(s\in S\) and \(t\in S\), then
- \(s + t \in S\), and
- \(s - t \in S\).
3. **Restriction:** Nothing else is in \(S\) apart from what is generated by (1) and (2).
We must prove that every integer formed in this manner is divisible by 5.
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## 2. Proof by Structural Induction
### Inductive Hypothesis
Assume for any integers \(s,t\in S\), we already know \(5\mid s\) and \(5\mid t\). We will verify this for the base elements and show it is preserved under the recursive steps.
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### Base Case
By rule (1), \(0\in S\) and \(5\in S\).
- Clearly \(5\mid 0\) since \(0 = 5\times 0\).
- Also \(5\mid 5\) since \(5 = 5\times 1\).
So the base elements are indeed multiples of 5.
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### Inductive Step
Suppose \(s\) and \(t\) are already in \(S\), and by the inductive hypothesis \(5\mid s\) and \(5\mid t\). We must show that the new integers generated also lie in the set of multiples of 5:
1. **\(s + t\):**
Since \(5\mid s\) and \(5\mid t\), we can write \(s=5k\) and \(t=5m\) for some integers \(k,m\). Then
\[
s+t = 5k + 5m = 5(k+m),
\]
which is a multiple of 5. Hence \(5\mid (s+t)\).
2. **\(s - t\):**
Similarly, \(s-t = 5k - 5m = 5(k-m)\). So \(5\mid (s-t)\).
Therefore any new integer formed from \(s\) and \(t\) by addition or subtraction remains divisible by 5.
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## 3. Conclusion
By structural induction on the rules defining \(S\), **every** integer in \(S\) is a multiple of 5. Formally:
\[
\boxed{\text{If }x \in S,\text{ then }5 \mid x.}
\]
