src/HOL/Real/HahnBanach/Linearform.thy
author ballarin
Tue Jul 15 16:50:09 2008 +0200 (2008-07-15)
changeset 27611 2c01c0bdb385
parent 25762 c03e9d04b3e4
child 27612 d3eb431db035
permissions -rw-r--r--
Removed uses of context element includes.
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(*  Title:      HOL/Real/HahnBanach/Linearform.thy
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    ID:         $Id$
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    Author:     Gertrud Bauer, TU Munich
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*)
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header {* Linearforms *}
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theory Linearform imports VectorSpace begin
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text {*
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  A \emph{linear form} is a function on a vector space into the reals
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  that is additive and multiplicative.
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*}
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locale linearform = var V + var f +
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  constrains V :: "'a\<Colon>{minus, plus, zero, uminus} set"
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  assumes add [iff]: "x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> f (x + y) = f x + f y"
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    and mult [iff]: "x \<in> V \<Longrightarrow> f (a \<cdot> x) = a * f x"
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declare linearform.intro [intro?]
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lemma (in linearform) neg [iff]:
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  assumes "vectorspace V"
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  shows "x \<in> V \<Longrightarrow> f (- x) = - f x"
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proof -
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  interpret vectorspace [V] by fact
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  assume x: "x \<in> V"
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  hence "f (- x) = f ((- 1) \<cdot> x)" by (simp add: negate_eq1)
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  also from x have "... = (- 1) * (f x)" by (rule mult)
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  also from x have "... = - (f x)" by simp
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  finally show ?thesis .
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qed
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lemma (in linearform) diff [iff]:
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  assumes "vectorspace V"
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  shows "x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> f (x - y) = f x - f y"
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proof -
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  interpret vectorspace [V] by fact
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  assume x: "x \<in> V" and y: "y \<in> V"
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  hence "x - y = x + - y" by (rule diff_eq1)
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  also have "f ... = f x + f (- y)" by (rule add) (simp_all add: x y)
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  also have "f (- y) = - f y" using `vectorspace V` y by (rule neg)
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  finally show ?thesis by simp
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qed
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text {* Every linear form yields @{text 0} for the @{text 0} vector. *}
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lemma (in linearform) zero [iff]:
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  assumes "vectorspace V"
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  shows "f 0 = 0"
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proof -
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  interpret vectorspace [V] by fact
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  have "f 0 = f (0 - 0)" by simp
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  also have "\<dots> = f 0 - f 0" using `vectorspace V` by (rule diff) simp_all
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  also have "\<dots> = 0" by simp
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  finally show ?thesis .
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qed
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end