(* ID: $Id: Abstract_NAT.thy,v 1.10 2007/07/11 09:52:28 berghofe Exp $ Author: Makarius *) header {* Abstract Natural Numbers primitive recursion *} theory Abstract_NAT imports Main begin text {* Axiomatic Natural Numbers (Peano) -- a monomorphic theory. *} locale NAT = fixes zero :: 'n and succ :: "'n => 'n" assumes succ_inject [simp]: "(succ m = succ n) = (m = n)" and succ_neq_zero [simp]: "succ m ≠ zero" and induct [case_names zero succ, induct type: 'n]: "P zero ==> (!!n. P n ==> P (succ n)) ==> P n" begin lemma zero_neq_succ [simp]: "zero ≠ succ m" by (rule succ_neq_zero [symmetric]) text {* \medskip Primitive recursion as a (functional) relation -- polymorphic! *} inductive Rec :: "'a => ('n => 'a => 'a) => 'n => 'a => bool" for e :: 'a and r :: "'n => 'a => 'a" where Rec_zero: "Rec e r zero e" | Rec_succ: "Rec e r m n ==> Rec e r (succ m) (r m n)" lemma Rec_functional: fixes x :: 'n shows "∃!y::'a. Rec e r x y" proof - let ?R = "Rec e r" show ?thesis proof (induct x) case zero show "∃!y. ?R zero y" proof show "?R zero e" .. fix y assume "?R zero y" then show "y = e" by cases simp_all qed next case (succ m) from `∃!y. ?R m y` obtain y where y: "?R m y" and yy': "!!y'. ?R m y' ==> y = y'" by blast show "∃!z. ?R (succ m) z" proof from y show "?R (succ m) (r m y)" .. fix z assume "?R (succ m) z" then obtain u where "z = r m u" and "?R m u" by cases simp_all with yy' show "z = r m y" by (simp only:) qed qed qed text {* \medskip The recursion operator -- polymorphic! *} definition rec :: "'a => ('n => 'a => 'a) => 'n => 'a" where "rec e r x = (THE y. Rec e r x y)" lemma rec_eval: assumes Rec: "Rec e r x y" shows "rec e r x = y" unfolding rec_def using Rec_functional and Rec by (rule the1_equality) lemma rec_zero [simp]: "rec e r zero = e" proof (rule rec_eval) show "Rec e r zero e" .. qed lemma rec_succ [simp]: "rec e r (succ m) = r m (rec e r m)" proof (rule rec_eval) let ?R = "Rec e r" have "?R m (rec e r m)" unfolding rec_def using Rec_functional by (rule theI') then show "?R (succ m) (r m (rec e r m))" .. qed text {* \medskip Example: addition (monomorphic) *} definition add :: "'n => 'n => 'n" where "add m n = rec n (λ_ k. succ k) m" lemma add_zero [simp]: "add zero n = n" and add_succ [simp]: "add (succ m) n = succ (add m n)" unfolding add_def by simp_all lemma add_assoc: "add (add k m) n = add k (add m n)" by (induct k) simp_all lemma add_zero_right: "add m zero = m" by (induct m) simp_all lemma add_succ_right: "add m (succ n) = succ (add m n)" by (induct m) simp_all lemma "add (succ (succ (succ zero))) (succ (succ zero)) = succ (succ (succ (succ (succ zero))))" by simp text {* \medskip Example: replication (polymorphic) *} definition repl :: "'n => 'a => 'a list" where "repl n x = rec [] (λ_ xs. x # xs) n" lemma repl_zero [simp]: "repl zero x = []" and repl_succ [simp]: "repl (succ n) x = x # repl n x" unfolding repl_def by simp_all lemma "repl (succ (succ (succ zero))) True = [True, True, True]" by simp end text {* \medskip Just see that our abstract specification makes sense \dots *} interpretation NAT [0 Suc] proof (rule NAT.intro) fix m n show "(Suc m = Suc n) = (m = n)" by simp show "Suc m ≠ 0" by simp fix P assume zero: "P 0" and succ: "!!n. P n ==> P (Suc n)" show "P n" proof (induct n) case 0 show ?case by (rule zero) next case Suc then show ?case by (rule succ) qed qed end
lemma zero_neq_succ:
zero ≠ succ m
lemma Rec_functional:
∃!y. Rec e r x y
lemma rec_eval:
Rec e r x y ==> rec e r x = y
lemma rec_zero:
rec e r zero = e
lemma rec_succ:
rec e r (succ m) = r m (rec e r m)
lemma add_zero:
add zero n = n
and add_succ:
add (succ m) n = succ (add m n)
lemma add_assoc:
add (add k m) n = add k (add m n)
lemma add_zero_right:
add m zero = m
lemma add_succ_right:
add m (succ n) = succ (add m n)
lemma
add (succ (succ (succ zero))) (succ (succ zero)) =
succ (succ (succ (succ (succ zero))))
lemma repl_zero:
repl zero x = []
and repl_succ:
repl (succ n) x = x # repl n x
lemma
repl (succ (succ (succ zero))) True = [True, True, True]