author wenzelm Wed, 08 Mar 2017 10:50:59 +0100 changeset 65151 a7394aa4d21c parent 63465 d7610beb98bc child 68553 333998becebe permissions -rw-r--r--
tuned proofs;
```
Author:      Tim Makarios <tjm1983 at gmail.com>, 2012

Originally from the AFP entry Tarskis_Geometry
*)

imports Complex_Main
begin

definition discrim :: "real \<Rightarrow> real \<Rightarrow> real \<Rightarrow> real"
where "discrim a b c \<equiv> b\<^sup>2 - 4 * a * c"

lemma complete_square:
fixes a b c x :: "real"
assumes "a \<noteq> 0"
shows "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow> (2 * a * x + b)\<^sup>2 = discrim a b c"
proof -
have "4 * a\<^sup>2 * x\<^sup>2 + 4 * a * b * x + 4 * a * c = 4 * a * (a * x\<^sup>2 + b * x + c)"
with \<open>a \<noteq> 0\<close>
have "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow> 4 * a\<^sup>2 * x\<^sup>2 + 4 * a * b * x + 4 * a * c = 0"
by simp
then show "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow> (2 * a * x + b)\<^sup>2 = discrim a b c"
by (simp add: discrim_def power2_eq_square algebra_simps)
qed

lemma discriminant_negative:
fixes a b c x :: real
assumes "a \<noteq> 0"
and "discrim a b c < 0"
shows "a * x\<^sup>2 + b * x + c \<noteq> 0"
proof -
have "(2 * a * x + b)\<^sup>2 \<ge> 0"
by simp
with \<open>discrim a b c < 0\<close> have "(2 * a * x + b)\<^sup>2 \<noteq> discrim a b c"
by arith
with complete_square and \<open>a \<noteq> 0\<close> show "a * x\<^sup>2 + b * x + c \<noteq> 0"
by simp
qed

lemma plus_or_minus_sqrt:
fixes x y :: real
assumes "y \<ge> 0"
shows "x\<^sup>2 = y \<longleftrightarrow> x = sqrt y \<or> x = - sqrt y"
proof
assume "x\<^sup>2 = y"
then have "sqrt (x\<^sup>2) = sqrt y"
by simp
then have "sqrt y = \<bar>x\<bar>"
by simp
then show "x = sqrt y \<or> x = - sqrt y"
by auto
next
assume "x = sqrt y \<or> x = - sqrt y"
then have "x\<^sup>2 = (sqrt y)\<^sup>2 \<or> x\<^sup>2 = (- sqrt y)\<^sup>2"
by auto
with \<open>y \<ge> 0\<close> show "x\<^sup>2 = y"
by simp
qed

lemma divide_non_zero:
fixes x y z :: real
assumes "x \<noteq> 0"
shows "x * y = z \<longleftrightarrow> y = z / x"
proof
show "y = z / x" if "x * y = z"
using \<open>x \<noteq> 0\<close> that by (simp add: field_simps)
show "x * y = z" if "y = z / x"
using \<open>x \<noteq> 0\<close> that by simp
qed

lemma discriminant_nonneg:
fixes a b c x :: real
assumes "a \<noteq> 0"
and "discrim a b c \<ge> 0"
shows "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow>
x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a)"
proof -
from complete_square and plus_or_minus_sqrt and assms
have "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow>
(2 * a) * x + b = sqrt (discrim a b c) \<or>
(2 * a) * x + b = - sqrt (discrim a b c)"
by simp
also have "\<dots> \<longleftrightarrow> (2 * a) * x = (-b + sqrt (discrim a b c)) \<or>
(2 * a) * x = (-b - sqrt (discrim a b c))"
by auto
also from \<open>a \<noteq> 0\<close> and divide_non_zero [of "2 * a" x]
have "\<dots> \<longleftrightarrow> x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a)"
by simp
finally show "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow>
x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a)" .
qed

lemma discriminant_zero:
fixes a b c x :: real
assumes "a \<noteq> 0"
and "discrim a b c = 0"
shows "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow> x = -b / (2 * a)"

theorem discriminant_iff:
fixes a b c x :: real
assumes "a \<noteq> 0"
shows "a * x\<^sup>2 + b * x + c = 0 \<longleftrightarrow>
discrim a b c \<ge> 0 \<and>
(x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a))"
proof
assume "a * x\<^sup>2 + b * x + c = 0"
with discriminant_negative and \<open>a \<noteq> 0\<close> have "\<not>(discrim a b c < 0)"
by auto
then have "discrim a b c \<ge> 0"
by simp
with discriminant_nonneg and \<open>a * x\<^sup>2 + b * x + c = 0\<close> and \<open>a \<noteq> 0\<close>
have "x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a)"
by simp
with \<open>discrim a b c \<ge> 0\<close>
show "discrim a b c \<ge> 0 \<and>
(x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a))" ..
next
assume "discrim a b c \<ge> 0 \<and>
(x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a))"
then have "discrim a b c \<ge> 0" and
"x = (-b + sqrt (discrim a b c)) / (2 * a) \<or>
x = (-b - sqrt (discrim a b c)) / (2 * a)"
by simp_all
with discriminant_nonneg and \<open>a \<noteq> 0\<close> show "a * x\<^sup>2 + b * x + c = 0"
by simp
qed

lemma discriminant_nonneg_ex:
fixes a b c :: real
assumes "a \<noteq> 0"
and "discrim a b c \<ge> 0"
shows "\<exists> x. a * x\<^sup>2 + b * x + c = 0"
by (auto simp: discriminant_nonneg assms)

lemma discriminant_pos_ex:
fixes a b c :: real
assumes "a \<noteq> 0"
and "discrim a b c > 0"
shows "\<exists>x y. x \<noteq> y \<and> a * x\<^sup>2 + b * x + c = 0 \<and> a * y\<^sup>2 + b * y + c = 0"
proof -
let ?x = "(-b + sqrt (discrim a b c)) / (2 * a)"
let ?y = "(-b - sqrt (discrim a b c)) / (2 * a)"
from \<open>discrim a b c > 0\<close> have "sqrt (discrim a b c) \<noteq> 0"
by simp
then have "sqrt (discrim a b c) \<noteq> - sqrt (discrim a b c)"
by arith
with \<open>a \<noteq> 0\<close> have "?x \<noteq> ?y"
by simp
moreover from assms have "a * ?x\<^sup>2 + b * ?x + c = 0" and "a * ?y\<^sup>2 + b * ?y + c = 0"
using discriminant_nonneg [of a b c ?x]
and discriminant_nonneg [of a b c ?y]
by simp_all
ultimately show ?thesis
by blast
qed

lemma discriminant_pos_distinct:
fixes a b c x :: real
assumes "a \<noteq> 0"
and "discrim a b c > 0"
shows "\<exists> y. x \<noteq> y \<and> a * y\<^sup>2 + b * y + c = 0"
proof -
from discriminant_pos_ex and \<open>a \<noteq> 0\<close> and \<open>discrim a b c > 0\<close>
obtain w and z where "w \<noteq> z"
and "a * w\<^sup>2 + b * w + c = 0" and "a * z\<^sup>2 + b * z + c = 0"
by blast
show "\<exists>y. x \<noteq> y \<and> a * y\<^sup>2 + b * y + c = 0"
proof (cases "x = w")
case True
with \<open>w \<noteq> z\<close> have "x \<noteq> z"
by simp
with \<open>a * z\<^sup>2 + b * z + c = 0\<close> show ?thesis
by auto
next
case False
with \<open>a * w\<^sup>2 + b * w + c = 0\<close> show ?thesis
by auto
qed
qed

end
```