Answer
See explanation
Work Step by Step
We have the set \(S\) of all strings over the alphabet \(\{a,b\}\) defined by:
1. **Base**: The null (empty) string \(\varepsilon\) is in \(S\).
2. **Recursion**: If \(s \in S\), then:
- \(s\,a \in S\) (concatenate an \(a\) on the right),
- \(s\,b \in S\) (concatenate a \(b\) on the right).
3. **Restriction**: No other strings are in \(S\) except those obtained by rules (1) and (2).
We want to derive (show) that the strings **\(aab\)** and **\(bb\)** belong to \(S\).
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## (a) Derivation for \(aab\)
1. **Base**: \(\varepsilon \in S\).
2. From \(\varepsilon \in S\), by rule (2a) we get \(\varepsilon a = a \in S\).
3. From \(a \in S\), again by rule (2a) we get \(a a = aa \in S\).
4. From \(aa \in S\), by rule (2b) we get \(aa b = aab \in S\).
Hence \(aab \in S\) by these four steps.
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## (b) Derivation for \(bb\)
1. **Base**: \(\varepsilon \in S\).
2. From \(\varepsilon \in S\), by rule (2b) we get \(\varepsilon b = b \in S\).
3. From \(b \in S\), by rule (2b) we get \(b b = bb \in S\).
Thus \(bb \in S\).
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### Conclusion
Using the base case \(\varepsilon \in S\) and the recursion rules (adjoining \(a\) or \(b\) to any string in \(S\)), we have shown explicit derivations proving **\(aab\in S\)** and **\(bb\in S\)**.
