(* Title: HOL/Isar_Examples/Fibonacci.thy
Author: Gertrud Bauer
Copyright 1999 Technische Universitaet Muenchen
The Fibonacci function. Original
tactic script by Lawrence C Paulson.
Fibonacci numbers: proofs of laws taken from
R. L. Graham, D. E. Knuth, O. Patashnik.
Concrete Mathematics.
(Addison-Wesley, 1989)
*)
section \<open>Fib and Gcd commute\<close>
theory Fibonacci
imports "../Number_Theory/Primes"
begin
text_raw \<open>\<^footnote>\<open>Isar version by Gertrud Bauer. Original tactic script by Larry
Paulson. A few proofs of laws taken from @{cite "Concrete-Math"}.\<close>\<close>
declare One_nat_def [simp]
subsection \<open>Fibonacci numbers\<close>
fun fib :: "nat \<Rightarrow> nat" where
"fib 0 = 0"
| "fib (Suc 0) = 1"
| "fib (Suc (Suc x)) = fib x + fib (Suc x)"
lemma [simp]: "fib (Suc n) > 0"
by (induct n rule: fib.induct) simp_all
text \<open>Alternative induction rule.\<close>
theorem fib_induct:
fixes n :: nat
shows "P 0 \<Longrightarrow> P 1 \<Longrightarrow> (\<And>n. P (n + 1) \<Longrightarrow> P n \<Longrightarrow> P (n + 2)) \<Longrightarrow> P n"
by (induct rule: fib.induct) simp_all
subsection \<open>Fib and gcd commute\<close>
text \<open>A few laws taken from @{cite "Concrete-Math"}.\<close>
lemma fib_add:
"fib (n + k + 1) = fib (k + 1) * fib (n + 1) + fib k * fib n"
(is "?P n")
\<comment> \<open>see @{cite \<open>page 280\<close> "Concrete-Math"}\<close>
proof (induct n rule: fib_induct)
show "?P 0" by simp
show "?P 1" by simp
fix n
have "fib (n + 2 + k + 1)
= fib (n + k + 1) + fib (n + 1 + k + 1)" by simp
also assume "fib (n + k + 1)
= fib (k + 1) * fib (n + 1) + fib k * fib n"
(is " _ = ?R1")
also assume "fib (n + 1 + k + 1)
= fib (k + 1) * fib (n + 1 + 1) + fib k * fib (n + 1)"
(is " _ = ?R2")
also have "?R1 + ?R2
= fib (k + 1) * fib (n + 2 + 1) + fib k * fib (n + 2)"
by (simp add: add_mult_distrib2)
finally show "?P (n + 2)" .
qed
lemma gcd_fib_Suc_eq_1: "gcd (fib n) (fib (n + 1)) = 1" (is "?P n")
proof (induct n rule: fib_induct)
show "?P 0" by simp
show "?P 1" by simp
fix n
have "fib (n + 2 + 1) = fib (n + 1) + fib (n + 2)"
by simp
also have "\<dots> = fib (n + 2) + fib (n + 1)"
by simp
also have "gcd (fib (n + 2)) \<dots> = gcd (fib (n + 2)) (fib (n + 1))"
by (rule gcd_add2_nat)
also have "\<dots> = gcd (fib (n + 1)) (fib (n + 1 + 1))"
by (simp add: gcd.commute)
also assume "\<dots> = 1"
finally show "?P (n + 2)" .
qed
lemma gcd_mult_add: "(0::nat) < n \<Longrightarrow> gcd (n * k + m) n = gcd m n"
proof -
assume "0 < n"
then have "gcd (n * k + m) n = gcd n (m mod n)"
by (simp add: gcd_non_0_nat add.commute)
also from \<open>0 < n\<close> have "\<dots> = gcd m n"
by (simp add: gcd_non_0_nat)
finally show ?thesis .
qed
lemma gcd_fib_add: "gcd (fib m) (fib (n + m)) = gcd (fib m) (fib n)"
proof (cases m)
case 0
then show ?thesis by simp
next
case (Suc k)
then have "gcd (fib m) (fib (n + m)) = gcd (fib (n + k + 1)) (fib (k + 1))"
by (simp add: gcd.commute)
also have "fib (n + k + 1)
= fib (k + 1) * fib (n + 1) + fib k * fib n"
by (rule fib_add)
also have "gcd \<dots> (fib (k + 1)) = gcd (fib k * fib n) (fib (k + 1))"
by (simp add: gcd_mult_add)
also have "\<dots> = gcd (fib n) (fib (k + 1))"
by (simp only: gcd_fib_Suc_eq_1 gcd_mult_cancel)
also have "\<dots> = gcd (fib m) (fib n)"
using Suc by (simp add: gcd.commute)
finally show ?thesis .
qed
lemma gcd_fib_diff:
assumes "m \<le> n"
shows "gcd (fib m) (fib (n - m)) = gcd (fib m) (fib n)"
proof -
have "gcd (fib m) (fib (n - m)) = gcd (fib m) (fib (n - m + m))"
by (simp add: gcd_fib_add)
also from \<open>m \<le> n\<close> have "n - m + m = n"
by simp
finally show ?thesis .
qed
lemma gcd_fib_mod:
assumes "0 < m"
shows "gcd (fib m) (fib (n mod m)) = gcd (fib m) (fib n)"
proof (induct n rule: nat_less_induct)
case (1 n) note hyp = this
show ?case
proof -
have "n mod m = (if n < m then n else (n - m) mod m)"
by (rule mod_if)
also have "gcd (fib m) (fib \<dots>) = gcd (fib m) (fib n)"
proof (cases "n < m")
case True
then show ?thesis by simp
next
case False
then have "m \<le> n" by simp
from \<open>0 < m\<close> and False have "n - m < n"
by simp
with hyp have "gcd (fib m) (fib ((n - m) mod m))
= gcd (fib m) (fib (n - m))" by simp
also have "\<dots> = gcd (fib m) (fib n)"
using \<open>m \<le> n\<close> by (rule gcd_fib_diff)
finally have "gcd (fib m) (fib ((n - m) mod m)) =
gcd (fib m) (fib n)" .
with False show ?thesis by simp
qed
finally show ?thesis .
qed
qed
theorem fib_gcd: "fib (gcd m n) = gcd (fib m) (fib n)" (is "?P m n")
proof (induct m n rule: gcd_nat_induct)
fix m
show "fib (gcd m 0) = gcd (fib m) (fib 0)"
by simp
fix n :: nat
assume n: "0 < n"
then have "gcd m n = gcd n (m mod n)"
by (simp add: gcd_non_0_nat)
also assume hyp: "fib \<dots> = gcd (fib n) (fib (m mod n))"
also from n have "\<dots> = gcd (fib n) (fib m)"
by (rule gcd_fib_mod)
also have "\<dots> = gcd (fib m) (fib n)"
by (rule gcd.commute)
finally show "fib (gcd m n) = gcd (fib m) (fib n)" .
qed
end