src/FOL/ex/Nat.thy
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Sun, 27 Oct 2019 21:51:14 -0400
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(*  Title:      FOL/ex/Nat.thy
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    Author:     Lawrence C Paulson, Cambridge University Computer Laboratory
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    Copyright   1992  University of Cambridge
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*)
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section \<open>Theory of the natural numbers: Peano's axioms, primitive recursion\<close>
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theory Nat
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  imports FOL
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begin
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typedecl nat
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instance nat :: \<open>term\<close> ..
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axiomatization
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  Zero :: \<open>nat\<close>  (\<open>0\<close>) and
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  Suc :: \<open>nat \<Rightarrow> nat\<close> and
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  rec :: \<open>[nat, 'a, [nat, 'a] \<Rightarrow> 'a] \<Rightarrow> 'a\<close>
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where
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  induct: \<open>\<lbrakk>P(0); \<And>x. P(x) \<Longrightarrow> P(Suc(x))\<rbrakk> \<Longrightarrow> P(n)\<close> and
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  Suc_inject: \<open>Suc(m)=Suc(n) \<Longrightarrow> m=n\<close> and
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  Suc_neq_0: \<open>Suc(m)=0 \<Longrightarrow> R\<close> and
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  rec_0: \<open>rec(0,a,f) = a\<close> and
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  rec_Suc: \<open>rec(Suc(m), a, f) = f(m, rec(m,a,f))\<close>
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definition add :: \<open>[nat, nat] \<Rightarrow> nat\<close>  (infixl \<open>+\<close> 60)
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  where \<open>m + n \<equiv> rec(m, n, \<lambda>x y. Suc(y))\<close>
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subsection \<open>Proofs about the natural numbers\<close>
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lemma Suc_n_not_n: \<open>Suc(k) \<noteq> k\<close>
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apply (rule_tac n = \<open>k\<close> in induct)
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apply (rule notI)
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apply (erule Suc_neq_0)
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apply (rule notI)
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apply (erule notE)
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apply (erule Suc_inject)
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done
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lemma \<open>(k+m)+n = k+(m+n)\<close>
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apply (rule induct)
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back
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back
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back
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back
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back
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back
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oops
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lemma add_0 [simp]: \<open>0+n = n\<close>
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apply (unfold add_def)
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apply (rule rec_0)
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done
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lemma add_Suc [simp]: \<open>Suc(m)+n = Suc(m+n)\<close>
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apply (unfold add_def)
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apply (rule rec_Suc)
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done
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lemma add_assoc: \<open>(k+m)+n = k+(m+n)\<close>
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apply (rule_tac n = \<open>k\<close> in induct)
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apply simp
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apply simp
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done
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lemma add_0_right: \<open>m+0 = m\<close>
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apply (rule_tac n = \<open>m\<close> in induct)
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apply simp
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apply simp
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done
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lemma add_Suc_right: \<open>m+Suc(n) = Suc(m+n)\<close>
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apply (rule_tac n = \<open>m\<close> in induct)
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apply simp_all
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done
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lemma
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  assumes prem: \<open>\<And>n. f(Suc(n)) = Suc(f(n))\<close>
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  shows \<open>f(i+j) = i+f(j)\<close>
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apply (rule_tac n = \<open>i\<close> in induct)
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apply simp
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apply (simp add: prem)
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done
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end