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src/Pure/Pure.thy

author | wenzelm |

Mon, 07 Apr 2008 21:25:21 +0200 | |

changeset 26570 | dbc458262f4c |

parent 26435 | bdce320cd426 |

child 26572 | 9178a7f4c4c8 |

permissions | -rw-r--r-- |

added swap_params;

(* Title: Pure/Pure.thy ID: $Id$ *) section {* Further content for the Pure theory *} subsection {* Meta-level connectives in assumptions *} lemma meta_mp: assumes "PROP P ==> PROP Q" and "PROP P" shows "PROP Q" by (rule `PROP P ==> PROP Q` [OF `PROP P`]) lemmas meta_impE = meta_mp [elim_format] lemma meta_spec: assumes "!!x. PROP P(x)" shows "PROP P(x)" by (rule `!!x. PROP P(x)`) lemmas meta_allE = meta_spec [elim_format] lemma swap_params: "(\<And>x y. PROP P(x, y)) == (\<And>y x. PROP P(x, y))" .. subsection {* Embedded terms *} locale (open) meta_term_syntax = fixes meta_term :: "'a => prop" ("TERM _") lemmas [intro?] = termI subsection {* Meta-level conjunction *} locale (open) meta_conjunction_syntax = fixes meta_conjunction :: "prop => prop => prop" (infixr "&&" 2) lemma all_conjunction: includes meta_conjunction_syntax shows "(!!x. PROP A(x) && PROP B(x)) == ((!!x. PROP A(x)) && (!!x. PROP B(x)))" proof assume conj: "!!x. PROP A(x) && PROP B(x)" show "(\<And>x. PROP A(x)) && (\<And>x. PROP B(x))" proof - fix x from conj show "PROP A(x)" by (rule conjunctionD1) from conj show "PROP B(x)" by (rule conjunctionD2) qed next assume conj: "(!!x. PROP A(x)) && (!!x. PROP B(x))" fix x show "PROP A(x) && PROP B(x)" proof - show "PROP A(x)" by (rule conj [THEN conjunctionD1, rule_format]) show "PROP B(x)" by (rule conj [THEN conjunctionD2, rule_format]) qed qed lemma imp_conjunction: includes meta_conjunction_syntax shows "(PROP A ==> PROP B && PROP C) == (PROP A ==> PROP B) && (PROP A ==> PROP C)" proof assume conj: "PROP A ==> PROP B && PROP C" show "(PROP A ==> PROP B) && (PROP A ==> PROP C)" proof - assume "PROP A" from conj [OF `PROP A`] show "PROP B" by (rule conjunctionD1) from conj [OF `PROP A`] show "PROP C" by (rule conjunctionD2) qed next assume conj: "(PROP A ==> PROP B) && (PROP A ==> PROP C)" assume "PROP A" show "PROP B && PROP C" proof - from `PROP A` show "PROP B" by (rule conj [THEN conjunctionD1]) from `PROP A` show "PROP C" by (rule conj [THEN conjunctionD2]) qed qed lemma conjunction_imp: includes meta_conjunction_syntax shows "(PROP A && PROP B ==> PROP C) == (PROP A ==> PROP B ==> PROP C)" proof assume r: "PROP A && PROP B ==> PROP C" assume ab: "PROP A" "PROP B" show "PROP C" proof (rule r) from ab show "PROP A && PROP B" . qed next assume r: "PROP A ==> PROP B ==> PROP C" assume conj: "PROP A && PROP B" show "PROP C" proof (rule r) from conj show "PROP A" by (rule conjunctionD1) from conj show "PROP B" by (rule conjunctionD2) qed qed