Fibonacci's Rabbits

GEK1517 Tutorial 6 / Question 1.
Starting with the letter b, consider the rewriting rules
b → a; a → ab.
For example, the first few 'words' formed by the above rewriting rules are
b, a, ab, aba, abaab, abaababa, ...
Let C(n) be the number of letters in the n-th word (n is a positive integer). For instances [sic],
C(1) = 1, C(2) = 1, C(3) = 2, C(4) = 3, C(5) = 5, C(6) = 8, ...
Prove that
C(n+1) = C(n) + C(n-1)
for all integers n ≥ 2. (Caution: Math. induction may not work well here.)

Spoiler: Lecturer's solution is more elegant, so please bear with me here.

Let number of 'b's in n-th word: B_n.
Let number of 'a's in n-th word: A_n.

Because of the rewriting rules b → a; a → ab.
'b's in a word is generated only from 'a's in the previous word
∴ B_n = A_n-1
'a's in a word is generated from both 'a's and 'b's in the previous word.
∴ A_n = A_n-1 + B_n-1

Now total number of letters in n-th word is C(n)
Total number of letters in n-th word is sum of 'a's and 'b's.
∴ C(n) = A_n + B_n
now
C(n-1) = A_n-1 + B_n-1 = A_n
C(n+1) = A_n+1 + B_n+1

To show that C(n+1) = C(n) + C(n-1), express all terms in A_n and B_n.
C(n+1) = C(n) + C(n-1)
⇒ A_n+1 + B_n+1 (LHS)
= A_n + B_n + A_n = 2A_n + B_n (RHS)
∵ B_n = A_n-1,
B_n+1 = A_n.
∵ A_n = A_n-1 + B_n-1,
A_n+1 = A_n + B_n.
Sub back into LHS:
LHS = A_n + A_n + B_n
= 2A_n + B_n = RHS.

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