Answer
See explanation
Work Step by Step
We want to show, by **structural induction**, that **every string in \(S\) begins with the symbol ‘\(a\)’**.
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## Definitions
The set \(S\) is defined recursively as follows:
1. **Base:** The single‐letter string \(a\) is in \(S\).
2. **Recursion:** If \(s\in S\), then:
- \(s\,a\in S\) (append ‘\(a\)’ to the end of \(s\)),
- \(s\,b\in S\) (append ‘\(b\)’ to the end of \(s\)).
3. **Restriction:** Nothing is in \(S\) except those strings obtained by the above rules (1) and (2).
We must prove that **every** string \(x \in S\) **starts** with the symbol ‘\(a\)’.
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## Proof by Structural Induction
### Induction Hypothesis
Assume for an arbitrary string \(s\in S\), \(s\) begins with ‘\(a\)’. We need to check the base case and the recursive construction steps.
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### Base Case
By the base rule, the string \(a\) is in \(S\). Clearly, \(a\) starts with the letter ‘\(a\)’.
Hence the property holds for the base string.
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### Inductive Step
Suppose \(s\) is in \(S\) and, by the inductive hypothesis, \(s\) **already** starts with ‘\(a\)’. We must show that **any** new string formed from \(s\) by the recursion rules also starts with ‘\(a\)’:
1. **Form \(s\,a\):**
If we append ‘\(a\)’ to the end of \(s\), the resulting string is \(s\,a\). Since appending a character at the end does **not** affect the first character of \(s\), \(s\,a\) will begin with the same first character as \(s\). By hypothesis, \(s\) starts with ‘\(a\)’, so \(s\,a\) also starts with ‘\(a\)’.
2. **Form \(s\,b\):**
Similarly, if we append ‘\(b\)’ to the end of \(s\), the new string is \(s\,b\). Again, this does not change the first character. Since \(s\) started with ‘\(a\)’, \(s\,b\) also starts with ‘\(a\)’.
In both cases, the newly formed strings still begin with ‘\(a\)’. This completes the inductive step.
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## Conclusion
By structural induction on the recursive definition of \(S\), **every** string in \(S\) must begin with ‘\(a\)’. Formally,
\[
\boxed{\text{All strings in }S\text{ start with the symbol ‘\(a\)’.}}
\]
