Answer
See explanation
Work Step by Step
Below is a standard recursive definition for the set of all binary strings (strings over \(\{0,1\}\)) in which **all the 0’s precede all the 1’s**. Equivalently, these are precisely the strings of the form \(0^m\,1^n\) for nonnegative integers \(m,n\).
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## **Recursive Definition**
Let \(S\) be the smallest set of strings over \(\{0,1\}\) satisfying:
1. **Base Case:**
\[
\epsilon \in S
\]
(where \(\epsilon\) denotes the empty string).
2. **Recursive Steps:**
- **Prefixing by 0:** If \(x \in S\), then \(0x \in S\).
- **Suffixing by 1:** If \(x \in S\), then \(x1 \in S\).
3. **Restriction:**
No other strings are in \(S\) except those obtained by finitely many applications of the above rules, starting from \(\epsilon\).
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## **Why This Definition Works**
1. **All generated strings have all 0’s before 1’s.**
- The empty string \(\epsilon\) trivially has no 0’s or 1’s (so no violation).
- If \(x\) has all its 0’s before its 1’s, then:
- \(0x\) simply adds a 0 at the front; it still has all 0’s (now one more) before all 1’s.
- \(x1\) simply appends a 1 at the end; it still has all 0’s before all 1’s.
2. **Every string of the form \(0^m 1^n\) is generated.**
- Start from \(\epsilon\).
- Apply “prefixing by 0” exactly \(m\) times to get \(0^m\).
- Then apply “suffixing by 1” exactly \(n\) times to get \(0^m 1^n\).
Hence the set \(S\) is *exactly* the set of all strings over \(\{0,1\}\) whose 0’s precede their 1’s.
