src/Pure/Pure.thy
author paulson
Fri, 05 Oct 2007 09:59:03 +0200
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child 26426 ddac7ef1e991
permissions -rw-r--r--
filtering out some package theorems
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(*  Title:      Pure/Pure.thy
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    ID:         $Id$
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*)
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header {* The Pure theory *}
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theory Pure
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imports ProtoPure
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begin
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setup  -- {* Common setup of internal components *}
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subsection {* Meta-level 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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subsection {* Embedded terms *}
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locale (open) meta_term_syntax =
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  fixes meta_term :: "'a => prop"  ("TERM _")
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lemmas [intro?] = termI
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subsection {* Meta-level 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 "(\<And>x. PROP A(x)) && (\<And>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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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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end