(* Title: HOL/Real/HahnBanach/VectorSpace.thy
ID: $Id$
Author: Gertrud Bauer, TU Munich
*)
header {* Vector spaces *}
theory VectorSpace = Aux:
subsection {* Signature *}
text {*
For the definition of real vector spaces a type @{typ 'a} of the
sort @{text "{plus, minus, zero}"} is considered, on which a real
scalar multiplication @{text \<cdot>} is declared.
*}
consts
prod :: "real \<Rightarrow> 'a::{plus, minus, zero} \<Rightarrow> 'a" (infixr "'(*')" 70)
syntax (xsymbols)
prod :: "real \<Rightarrow> 'a \<Rightarrow> 'a" (infixr "\<cdot>" 70)
syntax (HTML output)
prod :: "real \<Rightarrow> 'a \<Rightarrow> 'a" (infixr "\<cdot>" 70)
subsection {* Vector space laws *}
text {*
A \emph{vector space} is a non-empty set @{text V} of elements from
@{typ 'a} with the following vector space laws: The set @{text V} is
closed under addition and scalar multiplication, addition is
associative and commutative; @{text "- x"} is the inverse of @{text
x} w.~r.~t.~addition and @{text 0} is the neutral element of
addition. Addition and multiplication are distributive; scalar
multiplication is associative and the real number @{text "1"} is
the neutral element of scalar multiplication.
*}
locale vectorspace = var V +
assumes non_empty [iff, intro?]: "V \<noteq> {}"
and add_closed [iff]: "x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> x + y \<in> V"
and mult_closed [iff]: "x \<in> V \<Longrightarrow> a \<cdot> x \<in> V"
and add_assoc: "x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> z \<in> V \<Longrightarrow> (x + y) + z = x + (y + z)"
and add_commute: "x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> x + y = y + x"
and diff_self [simp]: "x \<in> V \<Longrightarrow> x - x = 0"
and add_zero_left [simp]: "x \<in> V \<Longrightarrow> 0 + x = x"
and add_mult_distrib1: "x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> a \<cdot> (x + y) = a \<cdot> x + a \<cdot> y"
and add_mult_distrib2: "x \<in> V \<Longrightarrow> (a + b) \<cdot> x = a \<cdot> x + b \<cdot> x"
and mult_assoc: "x \<in> V \<Longrightarrow> (a * b) \<cdot> x = a \<cdot> (b \<cdot> x)"
and mult_1 [simp]: "x \<in> V \<Longrightarrow> 1 \<cdot> x = x"
and negate_eq1: "x \<in> V \<Longrightarrow> - x = (- 1) \<cdot> x"
and diff_eq1: "x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> x - y = x + - y"
lemma (in vectorspace) negate_eq2: "x \<in> V \<Longrightarrow> (- 1) \<cdot> x = - x"
by (rule negate_eq1 [symmetric])
lemma (in vectorspace) negate_eq2a: "x \<in> V \<Longrightarrow> -1 \<cdot> x = - x"
by (simp add: negate_eq1)
lemma (in vectorspace) diff_eq2: "x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> x + - y = x - y"
by (rule diff_eq1 [symmetric])
lemma (in vectorspace) diff_closed [iff]: "x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> x - y \<in> V"
by (simp add: diff_eq1 negate_eq1)
lemma (in vectorspace) neg_closed [iff]: "x \<in> V \<Longrightarrow> - x \<in> V"
by (simp add: negate_eq1)
lemma (in vectorspace) add_left_commute:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> z \<in> V \<Longrightarrow> x + (y + z) = y + (x + z)"
proof -
assume xyz: "x \<in> V" "y \<in> V" "z \<in> V"
hence "x + (y + z) = (x + y) + z"
by (simp only: add_assoc)
also from xyz have "... = (y + x) + z" by (simp only: add_commute)
also from xyz have "... = y + (x + z)" by (simp only: add_assoc)
finally show ?thesis .
qed
theorems (in vectorspace) add_ac =
add_assoc add_commute add_left_commute
text {* The existence of the zero element of a vector space
follows from the non-emptiness of carrier set. *}
lemma (in vectorspace) zero [iff]: "0 \<in> V"
proof -
from non_empty obtain x where x: "x \<in> V" by blast
then have "0 = x - x" by (rule diff_self [symmetric])
also from x have "... \<in> V" by (rule diff_closed)
finally show ?thesis .
qed
lemma (in vectorspace) add_zero_right [simp]:
"x \<in> V \<Longrightarrow> x + 0 = x"
proof -
assume x: "x \<in> V"
from this and zero have "x + 0 = 0 + x" by (rule add_commute)
also from x have "... = x" by (rule add_zero_left)
finally show ?thesis .
qed
lemma (in vectorspace) mult_assoc2:
"x \<in> V \<Longrightarrow> a \<cdot> b \<cdot> x = (a * b) \<cdot> x"
by (simp only: mult_assoc)
lemma (in vectorspace) diff_mult_distrib1:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> a \<cdot> (x - y) = a \<cdot> x - a \<cdot> y"
by (simp add: diff_eq1 negate_eq1 add_mult_distrib1 mult_assoc2)
lemma (in vectorspace) diff_mult_distrib2:
"x \<in> V \<Longrightarrow> (a - b) \<cdot> x = a \<cdot> x - (b \<cdot> x)"
proof -
assume x: "x \<in> V"
have " (a - b) \<cdot> x = (a + - b) \<cdot> x"
by (simp add: real_diff_def)
also have "... = a \<cdot> x + (- b) \<cdot> x"
by (rule add_mult_distrib2)
also from x have "... = a \<cdot> x + - (b \<cdot> x)"
by (simp add: negate_eq1 mult_assoc2)
also from x have "... = a \<cdot> x - (b \<cdot> x)"
by (simp add: diff_eq1)
finally show ?thesis .
qed
lemmas (in vectorspace) distrib =
add_mult_distrib1 add_mult_distrib2
diff_mult_distrib1 diff_mult_distrib2
text {* \medskip Further derived laws: *}
lemma (in vectorspace) mult_zero_left [simp]:
"x \<in> V \<Longrightarrow> 0 \<cdot> x = 0"
proof -
assume x: "x \<in> V"
have "0 \<cdot> x = (1 - 1) \<cdot> x" by simp
also have "... = (1 + - 1) \<cdot> x" by simp
also have "... = 1 \<cdot> x + (- 1) \<cdot> x"
by (rule add_mult_distrib2)
also from x have "... = x + (- 1) \<cdot> x" by simp
also from x have "... = x + - x" by (simp add: negate_eq2a)
also from x have "... = x - x" by (simp add: diff_eq2)
also from x have "... = 0" by simp
finally show ?thesis .
qed
lemma (in vectorspace) mult_zero_right [simp]:
"a \<cdot> 0 = (0::'a)"
proof -
have "a \<cdot> 0 = a \<cdot> (0 - (0::'a))" by simp
also have "... = a \<cdot> 0 - a \<cdot> 0"
by (rule diff_mult_distrib1) simp_all
also have "... = 0" by simp
finally show ?thesis .
qed
lemma (in vectorspace) minus_mult_cancel [simp]:
"x \<in> V \<Longrightarrow> (- a) \<cdot> - x = a \<cdot> x"
by (simp add: negate_eq1 mult_assoc2)
lemma (in vectorspace) add_minus_left_eq_diff:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> - x + y = y - x"
proof -
assume xy: "x \<in> V" "y \<in> V"
hence "- x + y = y + - x" by (simp add: add_commute)
also from xy have "... = y - x" by (simp add: diff_eq1)
finally show ?thesis .
qed
lemma (in vectorspace) add_minus [simp]:
"x \<in> V \<Longrightarrow> x + - x = 0"
by (simp add: diff_eq2)
lemma (in vectorspace) add_minus_left [simp]:
"x \<in> V \<Longrightarrow> - x + x = 0"
by (simp add: diff_eq2 add_commute)
lemma (in vectorspace) minus_minus [simp]:
"x \<in> V \<Longrightarrow> - (- x) = x"
by (simp add: negate_eq1 mult_assoc2)
lemma (in vectorspace) minus_zero [simp]:
"- (0::'a) = 0"
by (simp add: negate_eq1)
lemma (in vectorspace) minus_zero_iff [simp]:
"x \<in> V \<Longrightarrow> (- x = 0) = (x = 0)"
proof
assume x: "x \<in> V"
{
from x have "x = - (- x)" by (simp add: minus_minus)
also assume "- x = 0"
also have "- ... = 0" by (rule minus_zero)
finally show "x = 0" .
next
assume "x = 0"
then show "- x = 0" by simp
}
qed
lemma (in vectorspace) add_minus_cancel [simp]:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> x + (- x + y) = y"
by (simp add: add_assoc [symmetric] del: add_commute)
lemma (in vectorspace) minus_add_cancel [simp]:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> - x + (x + y) = y"
by (simp add: add_assoc [symmetric] del: add_commute)
lemma (in vectorspace) minus_add_distrib [simp]:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> - (x + y) = - x + - y"
by (simp add: negate_eq1 add_mult_distrib1)
lemma (in vectorspace) diff_zero [simp]:
"x \<in> V \<Longrightarrow> x - 0 = x"
by (simp add: diff_eq1)
lemma (in vectorspace) diff_zero_right [simp]:
"x \<in> V \<Longrightarrow> 0 - x = - x"
by (simp add: diff_eq1)
lemma (in vectorspace) add_left_cancel:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> z \<in> V \<Longrightarrow> (x + y = x + z) = (y = z)"
proof
assume x: "x \<in> V" and y: "y \<in> V" and z: "z \<in> V"
{
from y have "y = 0 + y" by simp
also from x y have "... = (- x + x) + y" by simp
also from x y have "... = - x + (x + y)"
by (simp add: add_assoc neg_closed)
also assume "x + y = x + z"
also from x z have "- x + (x + z) = - x + x + z"
by (simp add: add_assoc [symmetric] neg_closed)
also from x z have "... = z" by simp
finally show "y = z" .
next
assume "y = z"
then show "x + y = x + z" by (simp only:)
}
qed
lemma (in vectorspace) add_right_cancel:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> z \<in> V \<Longrightarrow> (y + x = z + x) = (y = z)"
by (simp only: add_commute add_left_cancel)
lemma (in vectorspace) add_assoc_cong:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> x' \<in> V \<Longrightarrow> y' \<in> V \<Longrightarrow> z \<in> V
\<Longrightarrow> x + y = x' + y' \<Longrightarrow> x + (y + z) = x' + (y' + z)"
by (simp only: add_assoc [symmetric])
lemma (in vectorspace) mult_left_commute:
"x \<in> V \<Longrightarrow> a \<cdot> b \<cdot> x = b \<cdot> a \<cdot> x"
by (simp add: real_mult_commute mult_assoc2)
lemma (in vectorspace) mult_zero_uniq:
"x \<in> V \<Longrightarrow> x \<noteq> 0 \<Longrightarrow> a \<cdot> x = 0 \<Longrightarrow> a = 0"
proof (rule classical)
assume a: "a \<noteq> 0"
assume x: "x \<in> V" "x \<noteq> 0" and ax: "a \<cdot> x = 0"
from x a have "x = (inverse a * a) \<cdot> x" by simp
also have "... = inverse a \<cdot> (a \<cdot> x)" by (rule mult_assoc)
also from ax have "... = inverse a \<cdot> 0" by simp
also have "... = 0" by simp
finally have "x = 0" .
thus "a = 0" by contradiction
qed
lemma (in vectorspace) mult_left_cancel:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> a \<noteq> 0 \<Longrightarrow> (a \<cdot> x = a \<cdot> y) = (x = y)"
proof
assume x: "x \<in> V" and y: "y \<in> V" and a: "a \<noteq> 0"
from x have "x = 1 \<cdot> x" by simp
also from a have "... = (inverse a * a) \<cdot> x" by simp
also from x have "... = inverse a \<cdot> (a \<cdot> x)"
by (simp only: mult_assoc)
also assume "a \<cdot> x = a \<cdot> y"
also from a y have "inverse a \<cdot> ... = y"
by (simp add: mult_assoc2)
finally show "x = y" .
next
assume "x = y"
then show "a \<cdot> x = a \<cdot> y" by (simp only:)
qed
lemma (in vectorspace) mult_right_cancel:
"x \<in> V \<Longrightarrow> x \<noteq> 0 \<Longrightarrow> (a \<cdot> x = b \<cdot> x) = (a = b)"
proof
assume x: "x \<in> V" and neq: "x \<noteq> 0"
{
from x have "(a - b) \<cdot> x = a \<cdot> x - b \<cdot> x"
by (simp add: diff_mult_distrib2)
also assume "a \<cdot> x = b \<cdot> x"
with x have "a \<cdot> x - b \<cdot> x = 0" by simp
finally have "(a - b) \<cdot> x = 0" .
with x neq have "a - b = 0" by (rule mult_zero_uniq)
thus "a = b" by simp
next
assume "a = b"
then show "a \<cdot> x = b \<cdot> x" by (simp only:)
}
qed
lemma (in vectorspace) eq_diff_eq:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> z \<in> V \<Longrightarrow> (x = z - y) = (x + y = z)"
proof
assume x: "x \<in> V" and y: "y \<in> V" and z: "z \<in> V"
{
assume "x = z - y"
hence "x + y = z - y + y" by simp
also from y z have "... = z + - y + y"
by (simp add: diff_eq1)
also have "... = z + (- y + y)"
by (rule add_assoc) (simp_all add: y z)
also from y z have "... = z + 0"
by (simp only: add_minus_left)
also from z have "... = z"
by (simp only: add_zero_right)
finally show "x + y = z" .
next
assume "x + y = z"
hence "z - y = (x + y) - y" by simp
also from x y have "... = x + y + - y"
by (simp add: diff_eq1)
also have "... = x + (y + - y)"
by (rule add_assoc) (simp_all add: x y)
also from x y have "... = x" by simp
finally show "x = z - y" ..
}
qed
lemma (in vectorspace) add_minus_eq_minus:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> x + y = 0 \<Longrightarrow> x = - y"
proof -
assume x: "x \<in> V" and y: "y \<in> V"
from x y have "x = (- y + y) + x" by simp
also from x y have "... = - y + (x + y)" by (simp add: add_ac)
also assume "x + y = 0"
also from y have "- y + 0 = - y" by simp
finally show "x = - y" .
qed
lemma (in vectorspace) add_minus_eq:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> x - y = 0 \<Longrightarrow> x = y"
proof -
assume x: "x \<in> V" and y: "y \<in> V"
assume "x - y = 0"
with x y have eq: "x + - y = 0" by (simp add: diff_eq1)
with _ _ have "x = - (- y)"
by (rule add_minus_eq_minus) (simp_all add: x y)
with x y show "x = y" by simp
qed
lemma (in vectorspace) add_diff_swap:
"a \<in> V \<Longrightarrow> b \<in> V \<Longrightarrow> c \<in> V \<Longrightarrow> d \<in> V \<Longrightarrow> a + b = c + d
\<Longrightarrow> a - c = d - b"
proof -
assume vs: "a \<in> V" "b \<in> V" "c \<in> V" "d \<in> V"
and eq: "a + b = c + d"
then have "- c + (a + b) = - c + (c + d)"
by (simp add: add_left_cancel)
also have "... = d" by (rule minus_add_cancel)
finally have eq: "- c + (a + b) = d" .
from vs have "a - c = (- c + (a + b)) + - b"
by (simp add: add_ac diff_eq1)
also from vs eq have "... = d + - b"
by (simp add: add_right_cancel)
also from vs have "... = d - b" by (simp add: diff_eq2)
finally show "a - c = d - b" .
qed
lemma (in vectorspace) vs_add_cancel_21:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> z \<in> V \<Longrightarrow> u \<in> V
\<Longrightarrow> (x + (y + z) = y + u) = (x + z = u)"
proof
assume vs: "x \<in> V" "y \<in> V" "z \<in> V" "u \<in> V"
{
from vs have "x + z = - y + y + (x + z)" by simp
also have "... = - y + (y + (x + z))"
by (rule add_assoc) (simp_all add: vs)
also from vs have "y + (x + z) = x + (y + z)"
by (simp add: add_ac)
also assume "x + (y + z) = y + u"
also from vs have "- y + (y + u) = u" by simp
finally show "x + z = u" .
next
assume "x + z = u"
with vs show "x + (y + z) = y + u"
by (simp only: add_left_commute [of x])
}
qed
lemma (in vectorspace) add_cancel_end:
"x \<in> V \<Longrightarrow> y \<in> V \<Longrightarrow> z \<in> V \<Longrightarrow> (x + (y + z) = y) = (x = - z)"
proof
assume vs: "x \<in> V" "y \<in> V" "z \<in> V"
{
assume "x + (y + z) = y"
with vs have "(x + z) + y = 0 + y"
by (simp add: add_ac)
with vs have "x + z = 0"
by (simp only: add_right_cancel add_closed zero)
with vs show "x = - z" by (simp add: add_minus_eq_minus)
next
assume eq: "x = - z"
hence "x + (y + z) = - z + (y + z)" by simp
also have "... = y + (- z + z)"
by (rule add_left_commute) (simp_all add: vs)
also from vs have "... = y" by simp
finally show "x + (y + z) = y" .
}
qed
end