Let's break down the proof step by step and then provide code to check palindromes using the recursive definition. --- ### **Step 1: Restate The Problem** We have two definitions of a palindrome: 1. **Definition 1 (D1):** A string that reads the same forward and backward. 2. **Definition 2 (D2):** The recursive rules (epsilon, single character, and wrapping). We need to **prove by induction that these definitions are equivalent**: - Every string that satisfies D1 also satisfies D2. - Every string that satisfies D2 also satisfies D1. --- ## **Step 2: Prove D2 ⇒ D1 (All Recursively-Defined Palindromes are Reversible Palindromes)** **Proof by induction on the length of the string:** **Base case:** - Length 0: `ε` is a palindrome. Clearly, reading `ε` forward or backward is the same. - Length 1: Any symbol `a` is a palindrome. A single character reads the same forward and backward. **Inductive step:** Assume any string of length ≤ n that satisfies rules 1-3 is a palindrome (i.e., reads the same forward and backward). Consider a string of length n+2 constructed by rule 3: `axa`, where `x` is a palindrome of length n, and `a` is any symbol. - The reverse of `axa` is `a x^R a`, where `x^R` is the reverse of `x`. - By the inductive hypothesis, `x` is a palindrome, so `x^R = x`. - Therefore, the reverse is `a x a`, which is the original string. **Conclusion:** Any string generated by rules 1–3 reads the same forward and backward. **D2 ⇒ D1 is proven.** --- ## **Step 3: Prove D1 ⇒ D2 (All Reversible Palindromes Can Be Built Recursively)** **Proof by induction on the length of the string:** **Base case:** - Length 0: Only `ε`, which matches rule 1. - Length 1: Only one symbol, which matches rule 2. **Inductive step:** Suppose for length ≤ n, any palindrome can be constructed by rules 1–3. Consider a palindrome `s` of length n+2. - Since `s` reads the same forward and backward, the first and last symbols must be the same. Let `s = a x a`, where `a` is a symbol and `x` is the substring in the middle. - The substring `x` must also be a palindrome (since if you remove the first and last character, the remainder must still read the same forward and backward). - By inductive hypothesis, `x` can be constructed using the rules. - By rule 3, `a x a` is a palindrome. **Conclusion:** Any palindrome by D1 can be built using rules 1–3. **D1 ⇒ D2 is proven.** --- ## **Step 4: Final Conclusion** **The two definitions are equivalent.** --- ## **Step 5: Code Implementation** ### **Python Code: Recursive Palindrome Check (Rule-based)** ```python def is_palindrome_recursive(s): # Rule 1: epsilon (empty string) is a palindrome if len(s) == 0: return True # Rule 2: any single character is a palindrome if len(s) == 1: return True # Rule 3: wrap with same character if s[0] == s[-1]: return is_palindrome_recursive(s[1:-1]) # Rule 4: otherwise, not a palindrome return False # Test cases print(is_palindrome_recursive("")) # True print(is_palindrome_recursive("a")) # True print(is_palindrome_recursive("aba")) # True print(is_palindrome_recursive("abba")) # True print(is_palindrome_recursive("abc")) # False ``` ### **Python Code: Standard Palindrome Check (By Reversal)** ```python def is_palindrome_standard(s): return s == s[::-1] # Test cases print(is_palindrome_standard("")) # True print(is_palindrome_standard("a")) # True print(is_palindrome_standard("aba")) # True print(is_palindrome_standard("abba")) # True print(is_palindrome_standard("abc")) # False ``` --- ## **Summary** - **Proof:** Both definitions are equivalent, as shown by induction. - **Code:** Both recursive and standard checks confirm the equivalence. If you need code in another language or a more formal mathematical write-up, let me know!