src/HOL/ex/Abstract_NAT.thy
changeset 64920 31044168af84
parent 64919 7e0c8924dfda
child 64921 1cbfe46ad6b1
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64919:7e0c8924dfda 64920:31044168af84
     1 (*  Title:      HOL/ex/Abstract_NAT.thy
       
     2     Author:     Makarius
       
     3 *)
       
     4 
       
     5 section \<open>Abstract Natural Numbers primitive recursion\<close>
       
     6 
       
     7 theory Abstract_NAT
       
     8 imports Main
       
     9 begin
       
    10 
       
    11 text \<open>Axiomatic Natural Numbers (Peano) -- a monomorphic theory.\<close>
       
    12 
       
    13 locale NAT =
       
    14   fixes zero :: 'n
       
    15     and succ :: "'n \<Rightarrow> 'n"
       
    16   assumes succ_inject [simp]: "succ m = succ n \<longleftrightarrow> m = n"
       
    17     and succ_neq_zero [simp]: "succ m \<noteq> zero"
       
    18     and induct [case_names zero succ, induct type: 'n]:
       
    19       "P zero \<Longrightarrow> (\<And>n. P n \<Longrightarrow> P (succ n)) \<Longrightarrow> P n"
       
    20 begin
       
    21 
       
    22 lemma zero_neq_succ [simp]: "zero \<noteq> succ m"
       
    23   by (rule succ_neq_zero [symmetric])
       
    24 
       
    25 
       
    26 text \<open>\<^medskip> Primitive recursion as a (functional) relation -- polymorphic!\<close>
       
    27 
       
    28 inductive Rec :: "'a \<Rightarrow> ('n \<Rightarrow> 'a \<Rightarrow> 'a) \<Rightarrow> 'n \<Rightarrow> 'a \<Rightarrow> bool"
       
    29   for e :: 'a and r :: "'n \<Rightarrow> 'a \<Rightarrow> 'a"
       
    30 where
       
    31   Rec_zero: "Rec e r zero e"
       
    32 | Rec_succ: "Rec e r m n \<Longrightarrow> Rec e r (succ m) (r m n)"
       
    33 
       
    34 lemma Rec_functional:
       
    35   fixes x :: 'n
       
    36   shows "\<exists>!y::'a. Rec e r x y"
       
    37 proof -
       
    38   let ?R = "Rec e r"
       
    39   show ?thesis
       
    40   proof (induct x)
       
    41     case zero
       
    42     show "\<exists>!y. ?R zero y"
       
    43     proof
       
    44       show "?R zero e" ..
       
    45       show "y = e" if "?R zero y" for y
       
    46         using that by cases simp_all
       
    47     qed
       
    48   next
       
    49     case (succ m)
       
    50     from \<open>\<exists>!y. ?R m y\<close>
       
    51     obtain y where y: "?R m y" and yy': "\<And>y'. ?R m y' \<Longrightarrow> y = y'"
       
    52       by blast
       
    53     show "\<exists>!z. ?R (succ m) z"
       
    54     proof
       
    55       from y show "?R (succ m) (r m y)" ..
       
    56     next
       
    57       fix z
       
    58       assume "?R (succ m) z"
       
    59       then obtain u where "z = r m u" and "?R m u"
       
    60         by cases simp_all
       
    61       with yy' show "z = r m y"
       
    62         by (simp only:)
       
    63     qed
       
    64   qed
       
    65 qed
       
    66 
       
    67 
       
    68 text \<open>\<^medskip> The recursion operator -- polymorphic!\<close>
       
    69 
       
    70 definition rec :: "'a \<Rightarrow> ('n \<Rightarrow> 'a \<Rightarrow> 'a) \<Rightarrow> 'n \<Rightarrow> 'a"
       
    71   where "rec e r x = (THE y. Rec e r x y)"
       
    72 
       
    73 lemma rec_eval:
       
    74   assumes Rec: "Rec e r x y"
       
    75   shows "rec e r x = y"
       
    76   unfolding rec_def
       
    77   using Rec_functional and Rec by (rule the1_equality)
       
    78 
       
    79 lemma rec_zero [simp]: "rec e r zero = e"
       
    80 proof (rule rec_eval)
       
    81   show "Rec e r zero e" ..
       
    82 qed
       
    83 
       
    84 lemma rec_succ [simp]: "rec e r (succ m) = r m (rec e r m)"
       
    85 proof (rule rec_eval)
       
    86   let ?R = "Rec e r"
       
    87   have "?R m (rec e r m)"
       
    88     unfolding rec_def using Rec_functional by (rule theI')
       
    89   then show "?R (succ m) (r m (rec e r m))" ..
       
    90 qed
       
    91 
       
    92 
       
    93 text \<open>\<^medskip> Example: addition (monomorphic)\<close>
       
    94 
       
    95 definition add :: "'n \<Rightarrow> 'n \<Rightarrow> 'n"
       
    96   where "add m n = rec n (\<lambda>_ k. succ k) m"
       
    97 
       
    98 lemma add_zero [simp]: "add zero n = n"
       
    99   and add_succ [simp]: "add (succ m) n = succ (add m n)"
       
   100   unfolding add_def by simp_all
       
   101 
       
   102 lemma add_assoc: "add (add k m) n = add k (add m n)"
       
   103   by (induct k) simp_all
       
   104 
       
   105 lemma add_zero_right: "add m zero = m"
       
   106   by (induct m) simp_all
       
   107 
       
   108 lemma add_succ_right: "add m (succ n) = succ (add m n)"
       
   109   by (induct m) simp_all
       
   110 
       
   111 lemma "add (succ (succ (succ zero))) (succ (succ zero)) =
       
   112     succ (succ (succ (succ (succ zero))))"
       
   113   by simp
       
   114 
       
   115 
       
   116 text \<open>\<^medskip> Example: replication (polymorphic)\<close>
       
   117 
       
   118 definition repl :: "'n \<Rightarrow> 'a \<Rightarrow> 'a list"
       
   119   where "repl n x = rec [] (\<lambda>_ xs. x # xs) n"
       
   120 
       
   121 lemma repl_zero [simp]: "repl zero x = []"
       
   122   and repl_succ [simp]: "repl (succ n) x = x # repl n x"
       
   123   unfolding repl_def by simp_all
       
   124 
       
   125 lemma "repl (succ (succ (succ zero))) True = [True, True, True]"
       
   126   by simp
       
   127 
       
   128 end
       
   129 
       
   130 
       
   131 text \<open>\<^medskip> Just see that our abstract specification makes sense \dots\<close>
       
   132 
       
   133 interpretation NAT 0 Suc
       
   134 proof (rule NAT.intro)
       
   135   fix m n
       
   136   show "Suc m = Suc n \<longleftrightarrow> m = n" by simp
       
   137   show "Suc m \<noteq> 0" by simp
       
   138   show "P n"
       
   139     if zero: "P 0"
       
   140     and succ: "\<And>n. P n \<Longrightarrow> P (Suc n)"
       
   141     for P
       
   142   proof (induct n)
       
   143     case 0
       
   144     show ?case by (rule zero)
       
   145   next
       
   146     case Suc
       
   147     then show ?case by (rule succ)
       
   148   qed
       
   149 qed
       
   150 
       
   151 end