Answer
See explanation
Work Step by Step
We have the set \(P\) of well‐formed parentheses defined by:
1. **Base:** \(\texttt{()}\in P.\)
2. **Recursive “wrap” rule:** If \(E\in P,\) then \(\texttt{(}E\texttt{)}\in P.\)
3. **Recursive “concatenate” rule:** If \(E\in P\) and \(F\in P,\) then the concatenation \(E\,F\in P.\)
We want to show derivations for the following strings:
- (a) \(\texttt{()(())}\)
- (b) \(\texttt{(())(())}\)
Below are step‐by‐step proofs (derivations) that each is in \(P\).
---
## (a) Derivation of \(\texttt{()(())}\)
1. By the **base rule**, \(\texttt{()}\in P.\)
2. Again, by the base rule, we have another copy of \(\texttt{()}\in P.\)
3. Apply the **“wrap” rule** to \(\texttt{()}\). If \(E=\texttt{()}\in P,\) then \(\texttt{(}E\texttt{)}=\texttt{(())}\in P.\)
4. Now apply the **concatenate** rule to \(\texttt{()}\) and \(\texttt{(())}\). Since both are in \(P,\) their concatenation
\[
\texttt{()}\;\texttt{(())} \;=\; \texttt{()(())}
\]
is also in \(P.\)
Hence \(\texttt{()(())}\in P.\)
---
## (b) Derivation of \(\texttt{(())(())}\)
1. By the **base rule**, \(\texttt{()}\in P.\)
2. Apply the **“wrap” rule** to \(\texttt{()}\). Then \(\texttt{(())}\in P.\)
3. We repeat the same step to get another copy of \(\texttt{(())}\in P\).
(Concretely, again take \(\texttt{()}\) in \(P\) and wrap it to get a second \(\texttt{(())}\).)
4. Now apply the **concatenate** rule to \(\texttt{(())}\) and \(\texttt{(())}\). Their concatenation
\[
\texttt{(())}\;\texttt{(())} \;=\; \texttt{(())(())}
\]
is in \(P\).
Hence \(\texttt{(())(())}\in P.\)
