author  wenzelm 
Sun, 26 Nov 2006 23:43:53 +0100  
changeset 21539  c5cf9243ad62 
parent 16417  9bc16273c2d4 
child 25989  3267d0694d93 
permissions  rwrr 
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(* Title: FOL/ex/Natural_Numbers.thy 
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ID: $Id$ 
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Author: Markus Wenzel, TU Munich 
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*) 
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header {* Natural numbers *} 
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theory Natural_Numbers imports FOL begin 
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12371  10 
text {* 
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Theory of the natural numbers: Peano's axioms, primitive recursion. 

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(Modernized version of Larry Paulson's theory "Nat".) \medskip 

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*} 

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typedecl nat 
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arities nat :: "term" 
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consts 
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Zero :: nat ("0") 
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Suc :: "nat => nat" 
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rec :: "[nat, 'a, [nat, 'a] => 'a] => 'a" 
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axioms 
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induct [case_names 0 Suc, induct type: nat]: 
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"P(0) ==> (!!x. P(x) ==> P(Suc(x))) ==> P(n)" 
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Suc_inject: "Suc(m) = Suc(n) ==> m = n" 
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Suc_neq_0: "Suc(m) = 0 ==> R" 
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rec_0: "rec(0, a, f) = a" 
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rec_Suc: "rec(Suc(m), a, f) = f(m, rec(m, a, f))" 
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lemma Suc_n_not_n: "Suc(k) \<noteq> k" 
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proof (induct k) 
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show "Suc(0) \<noteq> 0" 
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proof 
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assume "Suc(0) = 0" 
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thus False by (rule Suc_neq_0) 
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qed 
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fix n assume hyp: "Suc(n) \<noteq> n" 
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show "Suc(Suc(n)) \<noteq> Suc(n)" 
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proof 
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assume "Suc(Suc(n)) = Suc(n)" 
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hence "Suc(n) = n" by (rule Suc_inject) 
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with hyp show False by contradiction 
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qed 
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qed 
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constdefs 
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add :: "[nat, nat] => nat" (infixl "+" 60) 
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"m + n == rec(m, n, \<lambda>x y. Suc(y))" 
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lemma add_0 [simp]: "0 + n = n" 
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by (unfold add_def) (rule rec_0) 
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lemma add_Suc [simp]: "Suc(m) + n = Suc(m + n)" 
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by (unfold add_def) (rule rec_Suc) 
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lemma add_assoc: "(k + m) + n = k + (m + n)" 
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by (induct k) simp_all 
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lemma add_0_right: "m + 0 = m" 
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by (induct m) simp_all 
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lemma add_Suc_right: "m + Suc(n) = Suc(m + n)" 
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by (induct m) simp_all 
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lemma "(!!n. f(Suc(n)) = Suc(f(n))) ==> f(i + j) = i + f(j)" 
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proof  
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assume "!!n. f(Suc(n)) = Suc(f(n))" 
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thus ?thesis by (induct i) simp_all 

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qed 
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end 