Answer
See explanation
Work Step by Step
We want to prove, by **structural induction**, that **every integer in \(S\)** is a multiple of 3 (i.e.\ divisible by 3).
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## 1. Definition of \(S\)
\(S\) is defined recursively as follows:
1. **Base**: \(0 \in S\).
2. **Recursion**: If \(s \in S\), then
- \(s + 3 \in S\), and
- \(s - 3 \in S\).
3. **Restriction**: No other integers are in \(S\) besides those obtained by rules (1) and (2).
We want to show that **every** integer \(x\in S\) is divisible by 3.
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## 2. Proof by Structural Induction
### Inductive Hypothesis
Assume that for any integer \(s \in S\), we already know \(3\mid s\). We will check the base case and show that the property “being a multiple of 3” is preserved by the recursion rules.
### Base Case
By rule (1), \(0\in S\). Since \(0\) is clearly a multiple of 3 \(\bigl(0=3\times 0\bigr)\), the base case holds.
### Inductive Step
Suppose \(s\) is in \(S\), and by the inductive hypothesis \(3\mid s\). We must show that each new integer formed by rule (2) also is divisible by 3:
1. **\(s+3\)**:
Since \(3\mid s\), we can write \(s=3k\) for some integer \(k\). Then
\[
s+3 = 3k + 3 = 3(k+1),
\]
which is also a multiple of 3.
2. **\(s-3\)**:
Similarly, \(s-3 = 3k - 3 = 3(k-1)\), which is again a multiple of 3.
Hence both \(s+3\) and \(s-3\) are multiples of 3, so they remain in the set of multiples of 3.
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## 3. Conclusion
By structural induction on the rules defining \(S\), we conclude that **all** integers in \(S\) are multiples of 3. Formally:
\[
\boxed{\text{If }x \in S,\text{ then }3 \mid x.}
\]
