src/HOL/Isar_examples/Fibonacci.thy
author haftmann
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(*  Title:      HOL/Isar_examples/Fibonacci.thy
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    ID:         $Id$
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    Author:     Gertrud Bauer
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    Copyright   1999 Technische Universitaet Muenchen
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The Fibonacci function.  Demonstrates the use of recdef.  Original
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tactic script by Lawrence C Paulson.
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Fibonacci numbers: proofs of laws taken from
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  R. L. Graham, D. E. Knuth, O. Patashnik.
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  Concrete Mathematics.
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  (Addison-Wesley, 1989)
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*)
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header {* Fib and Gcd commute *}
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theory Fibonacci
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imports Primes
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begin
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text_raw {*
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 \footnote{Isar version by Gertrud Bauer.  Original tactic script by
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 Larry Paulson.  A few proofs of laws taken from
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 \cite{Concrete-Math}.}
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*}
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subsection {* Fibonacci numbers *}
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fun fib :: "nat \<Rightarrow> nat" where
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  "fib 0 = 0"
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  | "fib (Suc 0) = 1"
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  | "fib (Suc (Suc x)) = fib x + fib (Suc x)"
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lemma [simp]: "0 < fib (Suc n)"
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  by (induct n rule: fib.induct) simp_all
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text {* Alternative induction rule. *}
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theorem fib_induct:
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    "P 0 ==> P 1 ==> (!!n. P (n + 1) ==> P n ==> P (n + 2)) ==> P (n::nat)"
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  by (induct rule: fib.induct) simp_all
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subsection {* Fib and gcd commute *}
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text {* A few laws taken from \cite{Concrete-Math}. *}
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lemma fib_add:
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  "fib (n + k + 1) = fib (k + 1) * fib (n + 1) + fib k * fib n"
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  (is "?P n")
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  -- {* see \cite[page 280]{Concrete-Math} *}
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proof (induct n rule: fib_induct)
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  show "?P 0" by simp
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  show "?P 1" by simp
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  fix n
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  have "fib (n + 2 + k + 1)
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    = fib (n + k + 1) + fib (n + 1 + k + 1)" by simp
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  also assume "fib (n + k + 1)
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    = fib (k + 1) * fib (n + 1) + fib k * fib n"
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      (is " _ = ?R1")
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  also assume "fib (n + 1 + k + 1)
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    = fib (k + 1) * fib (n + 1 + 1) + fib k * fib (n + 1)"
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      (is " _ = ?R2")
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  also have "?R1 + ?R2
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    = fib (k + 1) * fib (n + 2 + 1) + fib k * fib (n + 2)"
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    by (simp add: add_mult_distrib2)
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  finally show "?P (n + 2)" .
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qed
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lemma gcd_fib_Suc_eq_1: "gcd (fib n, fib (n + 1)) = 1" (is "?P n")
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proof (induct n rule: fib_induct)
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  show "?P 0" by simp
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  show "?P 1" by simp
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  fix n
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  have "fib (n + 2 + 1) = fib (n + 1) + fib (n + 2)"
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    by simp
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  also have "gcd (fib (n + 2), ...) = gcd (fib (n + 2), fib (n + 1))"
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    by (simp only: gcd_add2')
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  also have "... = gcd (fib (n + 1), fib (n + 1 + 1))"
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    by (simp add: gcd_commute)
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  also assume "... = 1"
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  finally show "?P (n + 2)" .
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qed
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lemma gcd_mult_add: "0 < n ==> gcd (n * k + m, n) = gcd (m, n)"
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proof -
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  assume "0 < n"
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  then have "gcd (n * k + m, n) = gcd (n, m mod n)"
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    by (simp add: gcd_non_0 add_commute)
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  also from `0 < n` have "... = gcd (m, n)" by (simp add: gcd_non_0)
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  finally show ?thesis .
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qed
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lemma gcd_fib_add: "gcd (fib m, fib (n + m)) = gcd (fib m, fib n)"
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proof (cases m)
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  case 0
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  then show ?thesis by simp
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next
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  case (Suc k)
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  then have "gcd (fib m, fib (n + m)) = gcd (fib (n + k + 1), fib (k + 1))"
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    by (simp add: gcd_commute)
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  also have "fib (n + k + 1)
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    = fib (k + 1) * fib (n + 1) + fib k * fib n"
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    by (rule fib_add)
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  also have "gcd (..., fib (k + 1)) = gcd (fib k * fib n, fib (k + 1))"
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    by (simp add: gcd_mult_add)
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  also have "... = gcd (fib n, fib (k + 1))"
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    by (simp only: gcd_fib_Suc_eq_1 gcd_mult_cancel)
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  also have "... = gcd (fib m, fib n)"
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    using Suc by (simp add: gcd_commute)
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  finally show ?thesis .
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qed
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lemma gcd_fib_diff:
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  assumes "m <= n"
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  shows "gcd (fib m, fib (n - m)) = gcd (fib m, fib n)"
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proof -
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  have "gcd (fib m, fib (n - m)) = gcd (fib m, fib (n - m + m))"
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    by (simp add: gcd_fib_add)
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  also from `m <= n` have "n - m + m = n" by simp
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  finally show ?thesis .
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qed
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lemma gcd_fib_mod:
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  assumes "0 < m"
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  shows "gcd (fib m, fib (n mod m)) = gcd (fib m, fib n)"
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proof (induct n rule: nat_less_induct)
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  case (1 n) note hyp = this
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  show ?case
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  proof -
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    have "n mod m = (if n < m then n else (n - m) mod m)"
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      by (rule mod_if)
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    also have "gcd (fib m, fib ...) = gcd (fib m, fib n)"
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    proof (cases "n < m")
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      case True then show ?thesis by simp
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    next
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      case False then have "m <= n" by simp
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      from `0 < m` and False have "n - m < n" by simp
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      with hyp have "gcd (fib m, fib ((n - m) mod m))
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        = gcd (fib m, fib (n - m))" by simp
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      also have "... = gcd (fib m, fib n)"
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        using `m <= n` by (rule gcd_fib_diff)
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      finally have "gcd (fib m, fib ((n - m) mod m)) =
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        gcd (fib m, fib n)" .
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      with False show ?thesis by simp
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    qed
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    finally show ?thesis .
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  qed
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qed
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theorem fib_gcd: "fib (gcd (m, n)) = gcd (fib m, fib n)" (is "?P m n")
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proof (induct m n rule: gcd_induct)
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  fix m show "fib (gcd (m, 0)) = gcd (fib m, fib 0)" by simp
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  fix n :: nat assume n: "0 < n"
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  then have "gcd (m, n) = gcd (n, m mod n)" by (rule gcd_non_0)
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  also assume hyp: "fib ... = gcd (fib n, fib (m mod n))"
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  also from n have "... = gcd (fib n, fib m)" by (rule gcd_fib_mod)
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  also have "... = gcd (fib m, fib n)" by (rule gcd_commute)
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  finally show "fib (gcd (m, n)) = gcd (fib m, fib n)" .
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qed
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end