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src/Pure/Pure.thy

author | ballarin |

Tue, 06 Jun 2006 10:05:57 +0200 | |

changeset 19783 | 82f365a14960 |

parent 19121 | d7fd5415a781 |

child 19800 | 5f764272183e |

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

Improved parameter management of locales.

(* Title: Pure/Pure.thy ID: $Id$ The Pure theory. *) header {* The Pure theory *} theory Pure imports ProtoPure begin ML{*set Toplevel.debug*} setup -- {* Common setup of internal components *} 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`]) 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] subsection {* Meta-level conjunction *} locale (open) meta_conjunction_syntax = fixes meta_conjunction :: "prop => prop => prop" (infixr "&&" 2) parse_translation {* [("\<^fixed>meta_conjunction", fn [t, u] => Logic.mk_conjunction (t, u))] *} 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 "PROP A" and "PROP B" show "PROP C" by (rule r) - 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 end