Answer
See explanation
Work Step by Step
Below is a **structural‐induction** proof that **no string** in the recursively defined set \(S\) begins with the digit ‘0’ (i.e.\ there are **no leading zeros**).
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## 1. Definition of the Set \(S\)
\(S\) is defined by:
1. **Base**: The single‐digit strings \(1,2,3,4,5,6,7,8,9\) are in \(S\).
2. **Recursion**:
- If \(s \in S\), then \(s0 \in S\). (Append ‘0’ to the end of \(s\).)
- If \(s \in S\) and \(t \in S\), then \(s\,t \in S\). (Concatenate \(s\) and \(t\).)
3. **Restriction**: No other strings are in \(S\) except those generated by rules (1) and (2).
We must show that **no** string in \(S\) has a leading zero.
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## 2. Statement to Prove
**Claim**: Every string \(x \in S\) begins with one of the digits \(1,2,3,4,5,6,7,8,9\).
Equivalently, **no** string in \(S\) starts with ‘0’, so there are no leading zeros.
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## 3. Proof by Structural Induction
### Inductive Hypothesis
Assume for any string \(s \in S\), the first character of \(s\) is never ‘0’. We verify this for the base elements and then for the recursively formed elements.
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### Base Case
By rule (1), the strings in the base are exactly the one‐digit strings \(1,2,3,4,5,6,7,8,9\).
Clearly, **none** of these starts with ‘0’ (they each start with themselves, a digit from 1 to 9).
Hence the property holds for all base strings.
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### Inductive Step
Suppose \(s \in S\) (and possibly \(t \in S\) also), and by the inductive hypothesis, neither \(s\) nor \(t\) begins with ‘0’. We must show that all new strings formed by the recursion also do not start with ‘0’:
1. **Form \(s0\)**:
By rule (2a), appending a ‘0’ to \(s\) creates \(s0\). The first character of \(s0\) is the same as the first character of \(s\). Since \(s\) did **not** start with ‘0’, \(s0\) also does not start with ‘0’.
2. **Form \(s\,t\)**:
By rule (2b), concatenating \(s\) and \(t\) gives \(s\,t\). The first character of \(s\,t\) is the first character of \(s\). Since \(s\) did not start with ‘0’, the concatenation \(s\,t\) also does not start with ‘0’.
In both cases, the newly formed strings still begin with a digit in \(\{1,2,3,4,5,6,7,8,9\}\), never ‘0’.
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### Conclusion
By structural induction on the formation rules of \(S\), **no** string in \(S\) can have a leading zero. Equivalently, **every** string in \(S\) starts with a digit from \(1\) to \(9\). Thus,
\[
\boxed{\text{There are no leading‐zero strings in }S.}
\]
