author  wenzelm 
Mon, 07 Apr 2008 21:29:46 +0200  
changeset 26572  9178a7f4c4c8 
parent 26570  dbc458262f4c 
child 26958  ed3a58a9eae1 
permissions  rwrr 
15803  1 
(* Title: Pure/Pure.thy 
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ID: $Id$ 

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*) 
15803  4 

26435  5 
section {* Further content for the Pure theory *} 
20627  6 

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subsection {* Metalevel connectives in assumptions *} 
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lemma meta_mp: 

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assumes "PROP P ==> PROP Q" and "PROP P" 
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shows "PROP Q" 
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by (rule `PROP P ==> PROP Q` [OF `PROP P`]) 
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lemmas meta_impE = meta_mp [elim_format] 
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lemma meta_spec: 
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assumes "!!x. PROP P(x)" 
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shows "PROP P(x)" 
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by (rule `!!x. PROP P(x)`) 
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lemmas meta_allE = meta_spec [elim_format] 

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lemma swap_params: 
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"(!!x y. PROP P(x, y)) == (!!y x. PROP P(x, y))" .. 
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21625  27 
subsection {* Embedded terms *} 
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locale (open) meta_term_syntax = 

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fixes meta_term :: "'a => prop" ("TERM _") 

31 

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lemmas [intro?] = termI 

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34 

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subsection {* Metalevel conjunction *} 
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locale (open) meta_conjunction_syntax = 
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fixes meta_conjunction :: "prop => prop => prop" (infixr "&&" 2) 
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lemma all_conjunction: 
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includes meta_conjunction_syntax 
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shows "(!!x. PROP A(x) && PROP B(x)) == ((!!x. PROP A(x)) && (!!x. PROP B(x)))" 
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proof 
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assume conj: "!!x. PROP A(x) && PROP B(x)" 
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show "(!!x. PROP A(x)) && (!!x. PROP B(x))" 
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proof  
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fix x 
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from conj show "PROP A(x)" by (rule conjunctionD1) 
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from conj show "PROP B(x)" by (rule conjunctionD2) 

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qed 
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next 
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assume conj: "(!!x. PROP A(x)) && (!!x. PROP B(x))" 
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fix x 
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show "PROP A(x) && PROP B(x)" 
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proof  

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show "PROP A(x)" by (rule conj [THEN conjunctionD1, rule_format]) 

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show "PROP B(x)" by (rule conj [THEN conjunctionD2, rule_format]) 

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qed 
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qed 
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19121  61 
lemma imp_conjunction: 
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includes meta_conjunction_syntax 
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shows "(PROP A ==> PROP B && PROP C) == (PROP A ==> PROP B) && (PROP A ==> PROP C)" 
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proof 
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assume conj: "PROP A ==> PROP B && PROP C" 
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show "(PROP A ==> PROP B) && (PROP A ==> PROP C)" 
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proof  

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assume "PROP A" 
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from conj [OF `PROP A`] show "PROP B" by (rule conjunctionD1) 
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from conj [OF `PROP A`] show "PROP C" by (rule conjunctionD2) 

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qed 
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next 
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assume conj: "(PROP A ==> PROP B) && (PROP A ==> PROP C)" 
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assume "PROP A" 
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show "PROP B && PROP C" 
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proof  

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from `PROP A` show "PROP B" by (rule conj [THEN conjunctionD1]) 

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from `PROP A` show "PROP C" by (rule conj [THEN conjunctionD2]) 

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qed 
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qed 
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lemma conjunction_imp: 
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includes meta_conjunction_syntax 
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shows "(PROP A && PROP B ==> PROP C) == (PROP A ==> PROP B ==> PROP C)" 
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proof 
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assume r: "PROP A && PROP B ==> PROP C" 
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assume ab: "PROP A" "PROP B" 
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show "PROP C" 

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proof (rule r) 

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from ab show "PROP A && PROP B" . 

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qed 

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next 
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assume r: "PROP A ==> PROP B ==> PROP C" 
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assume conj: "PROP A && PROP B" 
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show "PROP C" 
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proof (rule r) 
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from conj show "PROP A" by (rule conjunctionD1) 
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from conj show "PROP B" by (rule conjunctionD2) 

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qed 
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qed 
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101 