src/HOL/Library/NatPair.thy
changeset 14706 71590b7733b7
parent 14414 3fd75e96145d
child 15131 c69542757a4d
     1.1 --- a/src/HOL/Library/NatPair.thy	Thu May 06 12:43:00 2004 +0200
     1.2 +++ b/src/HOL/Library/NatPair.thy	Thu May 06 14:14:18 2004 +0200
     1.3 @@ -4,89 +4,89 @@
     1.4      Copyright   2003 Technische Universitaet Muenchen
     1.5  *)
     1.6  
     1.7 -header {*
     1.8 -  \title{Pairs of Natural Numbers}
     1.9 -  \author{Stefan Richter}
    1.10 -*}
    1.11 +header {* Pairs of Natural Numbers *}
    1.12  
    1.13  theory NatPair = Main:
    1.14  
    1.15 -text{*An injective function from $\mathbf{N}^2$ to
    1.16 -  $\mathbf{N}$.  Definition and proofs are from 
    1.17 -  Arnold Oberschelp.  Rekursionstheorie.  BI-Wissenschafts-Verlag, 1993
    1.18 -  (page 85).  *}
    1.19 +text{*
    1.20 +  An injective function from @{text "\<nat>\<twosuperior>"} to @{text
    1.21 +  \<nat>}.  Definition and proofs are from \cite[page
    1.22 +  85]{Oberschelp:1993}.
    1.23 +*}
    1.24  
    1.25 -constdefs 
    1.26 +constdefs
    1.27    nat2_to_nat:: "(nat * nat) \<Rightarrow> nat"
    1.28    "nat2_to_nat pair \<equiv> let (n,m) = pair in (n+m) * Suc (n+m) div 2 + n"
    1.29  
    1.30 -lemma dvd2_a_x_suc_a: "2 dvd a * (Suc a)" 
    1.31 -proof (cases "2 dvd a") 
    1.32 +lemma dvd2_a_x_suc_a: "2 dvd a * (Suc a)"
    1.33 +proof (cases "2 dvd a")
    1.34    case True
    1.35    thus ?thesis by (rule dvd_mult2)
    1.36  next
    1.37 -  case False 
    1.38 +  case False
    1.39    hence "Suc (a mod 2) = 2" by (simp add: dvd_eq_mod_eq_0)
    1.40 -  hence "Suc a mod 2 = 0" by (simp add: mod_Suc) 
    1.41 -  hence "2 dvd Suc a" by (simp only:dvd_eq_mod_eq_0) 
    1.42 +  hence "Suc a mod 2 = 0" by (simp add: mod_Suc)
    1.43 +  hence "2 dvd Suc a" by (simp only:dvd_eq_mod_eq_0)
    1.44    thus ?thesis by (rule dvd_mult)
    1.45  qed
    1.46  
    1.47 -lemma assumes eq: "nat2_to_nat (u,v) = nat2_to_nat (x,y)" 
    1.48 +lemma
    1.49 +  assumes eq: "nat2_to_nat (u,v) = nat2_to_nat (x,y)"
    1.50    shows nat2_to_nat_help: "u+v \<le> x+y"
    1.51  proof (rule classical)
    1.52    assume "\<not> ?thesis"
    1.53 -  hence contrapos: "x+y < u+v" 
    1.54 +  hence contrapos: "x+y < u+v"
    1.55      by simp
    1.56 -  have "nat2_to_nat (x,y) < (x+y) * Suc (x+y) div 2 + Suc (x + y)" 
    1.57 +  have "nat2_to_nat (x,y) < (x+y) * Suc (x+y) div 2 + Suc (x + y)"
    1.58      by (unfold nat2_to_nat_def) (simp add: Let_def)
    1.59 -  also have "\<dots> = (x+y)*Suc(x+y) div 2 + 2 * Suc(x+y) div 2" 
    1.60 -    by (simp only: div_mult_self1_is_m) 
    1.61 -  also have "\<dots> = (x+y)*Suc(x+y) div 2 + 2 * Suc(x+y) div 2 
    1.62 -    + ((x+y)*Suc(x+y) mod 2 + 2 * Suc(x+y) mod 2) div 2" 
    1.63 +  also have "\<dots> = (x+y)*Suc(x+y) div 2 + 2 * Suc(x+y) div 2"
    1.64 +    by (simp only: div_mult_self1_is_m)
    1.65 +  also have "\<dots> = (x+y)*Suc(x+y) div 2 + 2 * Suc(x+y) div 2
    1.66 +    + ((x+y)*Suc(x+y) mod 2 + 2 * Suc(x+y) mod 2) div 2"
    1.67    proof -
    1.68 -    have "2 dvd (x+y)*Suc(x+y)" 
    1.69 +    have "2 dvd (x+y)*Suc(x+y)"
    1.70        by (rule dvd2_a_x_suc_a)
    1.71 -    hence "(x+y)*Suc(x+y) mod 2 = 0" 
    1.72 +    hence "(x+y)*Suc(x+y) mod 2 = 0"
    1.73        by (simp only: dvd_eq_mod_eq_0)
    1.74      also
    1.75 -    have "2 * Suc(x+y) mod 2 = 0" 
    1.76 +    have "2 * Suc(x+y) mod 2 = 0"
    1.77        by (rule mod_mult_self1_is_0)
    1.78 -    ultimately have 
    1.79 -      "((x+y)*Suc(x+y) mod 2 + 2 * Suc(x+y) mod 2) div 2 = 0" 
    1.80 -      by simp 
    1.81 -    thus ?thesis 
    1.82 +    ultimately have
    1.83 +      "((x+y)*Suc(x+y) mod 2 + 2 * Suc(x+y) mod 2) div 2 = 0"
    1.84 +      by simp
    1.85 +    thus ?thesis
    1.86        by simp
    1.87 -  qed 
    1.88 -  also have "\<dots> = ((x+y)*Suc(x+y) + 2*Suc(x+y)) div 2" 
    1.89 -    by (rule div_add1_eq[THEN sym])
    1.90 -  also have "\<dots> = ((x+y+2)*Suc(x+y)) div 2" 
    1.91 -    by (simp only: add_mult_distrib[THEN sym])
    1.92 -  also from contrapos have "\<dots> \<le> ((Suc(u+v))*(u+v)) div 2" 
    1.93 -    by (simp only: mult_le_mono div_le_mono) 
    1.94 -  also have "\<dots> \<le> nat2_to_nat (u,v)" 
    1.95 +  qed
    1.96 +  also have "\<dots> = ((x+y)*Suc(x+y) + 2*Suc(x+y)) div 2"
    1.97 +    by (rule div_add1_eq [symmetric])
    1.98 +  also have "\<dots> = ((x+y+2)*Suc(x+y)) div 2"
    1.99 +    by (simp only: add_mult_distrib [symmetric])
   1.100 +  also from contrapos have "\<dots> \<le> ((Suc(u+v))*(u+v)) div 2"
   1.101 +    by (simp only: mult_le_mono div_le_mono)
   1.102 +  also have "\<dots> \<le> nat2_to_nat (u,v)"
   1.103      by (unfold nat2_to_nat_def) (simp add: Let_def)
   1.104 -  finally show ?thesis 
   1.105 +  finally show ?thesis
   1.106      by (simp only: eq)
   1.107  qed
   1.108  
   1.109 -lemma nat2_to_nat_inj: "inj nat2_to_nat"
   1.110 +theorem nat2_to_nat_inj: "inj nat2_to_nat"
   1.111  proof -
   1.112 -  {fix u v x y assume "nat2_to_nat (u,v) = nat2_to_nat (x,y)"
   1.113 +  {
   1.114 +    fix u v x y assume "nat2_to_nat (u,v) = nat2_to_nat (x,y)"
   1.115      hence "u+v \<le> x+y" by (rule nat2_to_nat_help)
   1.116 -    also from prems[THEN sym] have "x+y \<le> u+v" 
   1.117 +    also from prems [symmetric] have "x+y \<le> u+v"
   1.118        by (rule nat2_to_nat_help)
   1.119      finally have eq: "u+v = x+y" .
   1.120 -    with prems have ux: "u=x" 
   1.121 +    with prems have ux: "u=x"
   1.122        by (simp add: nat2_to_nat_def Let_def)
   1.123 -    with eq have vy: "v=y" 
   1.124 +    with eq have vy: "v=y"
   1.125        by simp
   1.126 -    with ux have "(u,v) = (x,y)" 
   1.127 +    with ux have "(u,v) = (x,y)"
   1.128        by simp
   1.129    }
   1.130 -  hence "\<And>x y. nat2_to_nat x = nat2_to_nat y \<Longrightarrow> x=y" 
   1.131 +  hence "\<And>x y. nat2_to_nat x = nat2_to_nat y \<Longrightarrow> x=y"
   1.132      by fast
   1.133 -  thus ?thesis 
   1.134 +  thus ?thesis
   1.135      by (unfold inj_on_def) simp
   1.136  qed
   1.137