src/FOL/ex/Nat.thy

tuned;

(*  Title:      FOL/ex/Nat.thy
    Author:     Lawrence C Paulson, Cambridge University Computer Laboratory
    Copyright   1992  University of Cambridge
*)

section \<open>Theory of the natural numbers: Peano's axioms, primitive recursion\<close>

theory Nat
imports FOL
begin

typedecl nat
instance nat :: "term" ..

axiomatization
  Zero :: nat  ("0") and
  Suc :: "nat => nat" and
  rec :: "[nat, 'a, [nat, 'a] => 'a] => 'a"
where
  induct: "[| P(0);  !!x. P(x) ==> P(Suc(x)) |]  ==> P(n)" and
  Suc_inject: "Suc(m)=Suc(n) ==> m=n" and
  Suc_neq_0: "Suc(m)=0 ==> R" and
  rec_0: "rec(0,a,f) = a" and
  rec_Suc: "rec(Suc(m), a, f) = f(m, rec(m,a,f))"

definition add :: "[nat, nat] => nat"  (infixl "+" 60)
  where "m + n == rec(m, n, %x y. Suc(y))"


subsection \<open>Proofs about the natural numbers\<close>

lemma Suc_n_not_n: "Suc(k) \<noteq> k"
apply (rule_tac n = k in induct)
apply (rule notI)
apply (erule Suc_neq_0)
apply (rule notI)
apply (erule notE)
apply (erule Suc_inject)
done

lemma "(k+m)+n = k+(m+n)"
apply (rule induct)
back
back
back
back
back
back
oops

lemma add_0 [simp]: "0+n = n"
apply (unfold add_def)
apply (rule rec_0)
done

lemma add_Suc [simp]: "Suc(m)+n = Suc(m+n)"
apply (unfold add_def)
apply (rule rec_Suc)
done

lemma add_assoc: "(k+m)+n = k+(m+n)"
apply (rule_tac n = k in induct)
apply simp
apply simp
done

lemma add_0_right: "m+0 = m"
apply (rule_tac n = m in induct)
apply simp
apply simp
done

lemma add_Suc_right: "m+Suc(n) = Suc(m+n)"
apply (rule_tac n = m in induct)
apply simp_all
done

lemma
  assumes prem: "!!n. f(Suc(n)) = Suc(f(n))"
  shows "f(i+j) = i+f(j)"
apply (rule_tac n = i in induct)
apply simp
apply (simp add: prem)
done

end