src/HOL/Real/Rational.thy
 author nipkow Mon, 16 Aug 2004 14:22:27 +0200 changeset 15131 c69542757a4d parent 15013 34264f5e4691 child 15140 322485b816ac permissions -rw-r--r--
```
(*  Title: HOL/Library/Rational.thy
ID:    \$Id\$
Author: Markus Wenzel, TU Muenchen
*)

theory Rational
import Quotient
files ("rat_arith.ML")
begin

subsection {* Fractions *}

subsubsection {* The type of fractions *}

typedef fraction = "{(a, b) :: int \<times> int | a b. b \<noteq> 0}"
proof
show "(0, 1) \<in> ?fraction" by simp
qed

constdefs
fract :: "int => int => fraction"
"fract a b == Abs_fraction (a, b)"
num :: "fraction => int"
"num Q == fst (Rep_fraction Q)"
den :: "fraction => int"
"den Q == snd (Rep_fraction Q)"

lemma fract_num [simp]: "b \<noteq> 0 ==> num (fract a b) = a"
by (simp add: fract_def num_def fraction_def Abs_fraction_inverse)

lemma fract_den [simp]: "b \<noteq> 0 ==> den (fract a b) = b"
by (simp add: fract_def den_def fraction_def Abs_fraction_inverse)

lemma fraction_cases [case_names fract, cases type: fraction]:
"(!!a b. Q = fract a b ==> b \<noteq> 0 ==> C) ==> C"
proof -
assume r: "!!a b. Q = fract a b ==> b \<noteq> 0 ==> C"
obtain a b where "Q = fract a b" and "b \<noteq> 0"
by (cases Q) (auto simp add: fract_def fraction_def)
thus C by (rule r)
qed

lemma fraction_induct [case_names fract, induct type: fraction]:
"(!!a b. b \<noteq> 0 ==> P (fract a b)) ==> P Q"
by (cases Q) simp

subsubsection {* Equivalence of fractions *}

instance fraction :: eqv ..

equiv_fraction_def: "Q \<sim> R == num Q * den R = num R * den Q"

lemma equiv_fraction_iff [iff]:
"b \<noteq> 0 ==> b' \<noteq> 0 ==> (fract a b \<sim> fract a' b') = (a * b' = a' * b)"

instance fraction :: equiv
proof
fix Q R S :: fraction
{
show "Q \<sim> Q"
proof (induct Q)
fix a b :: int
assume "b \<noteq> 0" and "b \<noteq> 0"
with refl show "fract a b \<sim> fract a b" ..
qed
next
assume "Q \<sim> R" and "R \<sim> S"
show "Q \<sim> S"
proof (insert prems, induct Q, induct R, induct S)
fix a b a' b' a'' b'' :: int
assume b: "b \<noteq> 0" and b': "b' \<noteq> 0" and b'': "b'' \<noteq> 0"
assume "fract a b \<sim> fract a' b'" hence eq1: "a * b' = a' * b" ..
assume "fract a' b' \<sim> fract a'' b''" hence eq2: "a' * b'' = a'' * b'" ..
have "a * b'' = a'' * b"
proof cases
assume "a' = 0"
with b' eq1 eq2 have "a = 0 \<and> a'' = 0" by auto
thus ?thesis by simp
next
assume a': "a' \<noteq> 0"
from eq1 eq2 have "(a * b') * (a' * b'') = (a' * b) * (a'' * b')" by simp
hence "(a * b'') * (a' * b') = (a'' * b) * (a' * b')" by (simp only: mult_ac)
with a' b' show ?thesis by simp
qed
thus "fract a b \<sim> fract a'' b''" ..
qed
next
show "Q \<sim> R ==> R \<sim> Q"
proof (induct Q, induct R)
fix a b a' b' :: int
assume b: "b \<noteq> 0" and b': "b' \<noteq> 0"
assume "fract a b \<sim> fract a' b'"
hence "a * b' = a' * b" ..
hence "a' * b = a * b'" ..
thus "fract a' b' \<sim> fract a b" ..
qed
}
qed

lemma eq_fraction_iff [iff]:
"b \<noteq> 0 ==> b' \<noteq> 0 ==> (\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor>) = (a * b' = a' * b)"

subsubsection {* Operations on fractions *}

text {*
We define the basic arithmetic operations on fractions and
demonstrate their ``well-definedness'', i.e.\ congruence with respect
to equivalence of fractions.
*}

instance fraction :: "{zero, one, plus, minus, times, inverse, ord}" ..

zero_fraction_def: "0 == fract 0 1"
one_fraction_def: "1 == fract 1 1"
fract (num Q * den R + num R * den Q) (den Q * den R)"
minus_fraction_def: "-Q == fract (-(num Q)) (den Q)"
mult_fraction_def: "Q * R == fract (num Q * num R) (den Q * den R)"
inverse_fraction_def: "inverse Q == fract (den Q) (num Q)"
le_fraction_def: "Q \<le> R ==
(num Q * den R) * (den Q * den R) \<le> (num R * den Q) * (den Q * den R)"

lemma is_zero_fraction_iff: "b \<noteq> 0 ==> (\<lfloor>fract a b\<rfloor> = \<lfloor>0\<rfloor>) = (a = 0)"

"\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor> ==> \<lfloor>fract c d\<rfloor> = \<lfloor>fract c' d'\<rfloor>
==> b \<noteq> 0 ==> b' \<noteq> 0 ==> d \<noteq> 0 ==> d' \<noteq> 0
==> \<lfloor>fract a b + fract c d\<rfloor> = \<lfloor>fract a' b' + fract c' d'\<rfloor>"
proof -
assume neq: "b \<noteq> 0"  "b' \<noteq> 0"  "d \<noteq> 0"  "d' \<noteq> 0"
assume "\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor>" hence eq1: "a * b' = a' * b" ..
assume "\<lfloor>fract c d\<rfloor> = \<lfloor>fract c' d'\<rfloor>" hence eq2: "c * d' = c' * d" ..
have "\<lfloor>fract (a * d + c * b) (b * d)\<rfloor> = \<lfloor>fract (a' * d' + c' * b') (b' * d')\<rfloor>"
proof
show "(a * d + c * b) * (b' * d') = (a' * d' + c' * b') * (b * d)"
(is "?lhs = ?rhs")
proof -
have "?lhs = (a * b') * (d * d') + (c * d') * (b * b')"
also have "... = (a' * b) * (d * d') + (c' * d) * (b * b')"
by (simp only: eq1 eq2)
also have "... = ?rhs"
finally show "?lhs = ?rhs" .
qed
from neq show "b * d \<noteq> 0" by simp
from neq show "b' * d' \<noteq> 0" by simp
qed
qed

theorem minus_fraction_cong:
"\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor> ==> b \<noteq> 0 ==> b' \<noteq> 0
==> \<lfloor>-(fract a b)\<rfloor> = \<lfloor>-(fract a' b')\<rfloor>"
proof -
assume neq: "b \<noteq> 0"  "b' \<noteq> 0"
assume "\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor>"
hence "a * b' = a' * b" ..
hence "-a * b' = -a' * b" by simp
hence "\<lfloor>fract (-a) b\<rfloor> = \<lfloor>fract (-a') b'\<rfloor>" ..
with neq show ?thesis by (simp add: minus_fraction_def)
qed

theorem mult_fraction_cong:
"\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor> ==> \<lfloor>fract c d\<rfloor> = \<lfloor>fract c' d'\<rfloor>
==> b \<noteq> 0 ==> b' \<noteq> 0 ==> d \<noteq> 0 ==> d' \<noteq> 0
==> \<lfloor>fract a b * fract c d\<rfloor> = \<lfloor>fract a' b' * fract c' d'\<rfloor>"
proof -
assume neq: "b \<noteq> 0"  "b' \<noteq> 0"  "d \<noteq> 0"  "d' \<noteq> 0"
assume "\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor>" hence eq1: "a * b' = a' * b" ..
assume "\<lfloor>fract c d\<rfloor> = \<lfloor>fract c' d'\<rfloor>" hence eq2: "c * d' = c' * d" ..
have "\<lfloor>fract (a * c) (b * d)\<rfloor> = \<lfloor>fract (a' * c') (b' * d')\<rfloor>"
proof
from eq1 eq2 have "(a * b') * (c * d') = (a' * b) * (c' * d)" by simp
thus "(a * c) * (b' * d') = (a' * c') * (b * d)" by (simp add: mult_ac)
from neq show "b * d \<noteq> 0" by simp
from neq show "b' * d' \<noteq> 0" by simp
qed
with neq show "\<lfloor>fract a b * fract c d\<rfloor> = \<lfloor>fract a' b' * fract c' d'\<rfloor>"
qed

theorem inverse_fraction_cong:
"\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor> ==> \<lfloor>fract a b\<rfloor> \<noteq> \<lfloor>0\<rfloor> ==> \<lfloor>fract a' b'\<rfloor> \<noteq> \<lfloor>0\<rfloor>
==> b \<noteq> 0 ==> b' \<noteq> 0
==> \<lfloor>inverse (fract a b)\<rfloor> = \<lfloor>inverse (fract a' b')\<rfloor>"
proof -
assume neq: "b \<noteq> 0"  "b' \<noteq> 0"
assume "\<lfloor>fract a b\<rfloor> \<noteq> \<lfloor>0\<rfloor>" and "\<lfloor>fract a' b'\<rfloor> \<noteq> \<lfloor>0\<rfloor>"
with neq obtain "a \<noteq> 0" and "a' \<noteq> 0" by (simp add: is_zero_fraction_iff)
assume "\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor>"
hence "a * b' = a' * b" ..
hence "b * a' = b' * a" by (simp only: mult_ac)
hence "\<lfloor>fract b a\<rfloor> = \<lfloor>fract b' a'\<rfloor>" ..
with neq show ?thesis by (simp add: inverse_fraction_def)
qed

theorem le_fraction_cong:
"\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor> ==> \<lfloor>fract c d\<rfloor> = \<lfloor>fract c' d'\<rfloor>
==> b \<noteq> 0 ==> b' \<noteq> 0 ==> d \<noteq> 0 ==> d' \<noteq> 0
==> (fract a b \<le> fract c d) = (fract a' b' \<le> fract c' d')"
proof -
assume neq: "b \<noteq> 0"  "b' \<noteq> 0"  "d \<noteq> 0"  "d' \<noteq> 0"
assume "\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor>" hence eq1: "a * b' = a' * b" ..
assume "\<lfloor>fract c d\<rfloor> = \<lfloor>fract c' d'\<rfloor>" hence eq2: "c * d' = c' * d" ..

let ?le = "\<lambda>a b c d. ((a * d) * (b * d) \<le> (c * b) * (b * d))"
{
fix a b c d x :: int assume x: "x \<noteq> 0"
have "?le a b c d = ?le (a * x) (b * x) c d"
proof -
from x have "0 < x * x" by (auto simp add: zero_less_mult_iff)
hence "?le a b c d =
((a * d) * (b * d) * (x * x) \<le> (c * b) * (b * d) * (x * x))"
also have "... = ?le (a * x) (b * x) c d"
finally show ?thesis .
qed
} note le_factor = this

let ?D = "b * d" and ?D' = "b' * d'"
from neq have D: "?D \<noteq> 0" by simp
from neq have "?D' \<noteq> 0" by simp
hence "?le a b c d = ?le (a * ?D') (b * ?D') c d"
by (rule le_factor)
also have "... = ((a * b') * ?D * ?D' * d * d' \<le> (c * d') * ?D * ?D' * b * b')"
also have "... = ((a' * b) * ?D * ?D' * d * d' \<le> (c' * d) * ?D * ?D' * b * b')"
by (simp only: eq1 eq2)
also have "... = ?le (a' * ?D) (b' * ?D) c' d'"
also from D have "... = ?le a' b' c' d'"
by (rule le_factor [symmetric])
finally have "?le a b c d = ?le a' b' c' d'" .
with neq show ?thesis by (simp add: le_fraction_def)
qed

subsection {* Rational numbers *}

subsubsection {* The type of rational numbers *}

typedef (Rat)
rat = "UNIV :: fraction quot set" ..

lemma RatI [intro, simp]: "Q \<in> Rat"

constdefs
fraction_of :: "rat => fraction"
"fraction_of q == pick (Rep_Rat q)"
rat_of :: "fraction => rat"
"rat_of Q == Abs_Rat \<lfloor>Q\<rfloor>"

theorem rat_of_equality [iff?]: "(rat_of Q = rat_of Q') = (\<lfloor>Q\<rfloor> = \<lfloor>Q'\<rfloor>)"

lemma rat_of: "\<lfloor>Q\<rfloor> = \<lfloor>Q'\<rfloor> ==> rat_of Q = rat_of Q'" ..

constdefs
Fract :: "int => int => rat"
"Fract a b == rat_of (fract a b)"

theorem Fract_inverse: "\<lfloor>fraction_of (Fract a b)\<rfloor> = \<lfloor>fract a b\<rfloor>"
by (simp add: fraction_of_def rat_of_def Fract_def Abs_Rat_inverse pick_inverse)

theorem Fract_equality [iff?]:
"(Fract a b = Fract c d) = (\<lfloor>fract a b\<rfloor> = \<lfloor>fract c d\<rfloor>)"

theorem eq_rat:
"b \<noteq> 0 ==> d \<noteq> 0 ==> (Fract a b = Fract c d) = (a * d = c * b)"

theorem Rat_cases [case_names Fract, cases type: rat]:
"(!!a b. q = Fract a b ==> b \<noteq> 0 ==> C) ==> C"
proof -
assume r: "!!a b. q = Fract a b ==> b \<noteq> 0 ==> C"
obtain x where "q = Abs_Rat x" by (cases q)
moreover obtain Q where "x = \<lfloor>Q\<rfloor>" by (cases x)
moreover obtain a b where "Q = fract a b" and "b \<noteq> 0" by (cases Q)
ultimately have "q = Fract a b" by (simp only: Fract_def rat_of_def)
thus ?thesis by (rule r)
qed

theorem Rat_induct [case_names Fract, induct type: rat]:
"(!!a b. b \<noteq> 0 ==> P (Fract a b)) ==> P q"
by (cases q) simp

subsubsection {* Canonical function definitions *}

text {*
Note that the unconditional version below is much easier to read.
*}

theorem rat_cond_function:
"(!!q r. P \<lfloor>fraction_of q\<rfloor> \<lfloor>fraction_of r\<rfloor> ==>
f q r == g (fraction_of q) (fraction_of r)) ==>
(!!a b a' b' c d c' d'.
\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor> ==> \<lfloor>fract c d\<rfloor> = \<lfloor>fract c' d'\<rfloor> ==>
P \<lfloor>fract a b\<rfloor> \<lfloor>fract c d\<rfloor> ==> P \<lfloor>fract a' b'\<rfloor> \<lfloor>fract c' d'\<rfloor> ==>
b \<noteq> 0 ==> b' \<noteq> 0 ==> d \<noteq> 0 ==> d' \<noteq> 0 ==>
g (fract a b) (fract c d) = g (fract a' b') (fract c' d')) ==>
P \<lfloor>fract a b\<rfloor> \<lfloor>fract c d\<rfloor> ==>
f (Fract a b) (Fract c d) = g (fract a b) (fract c d)"
(is "PROP ?eq ==> PROP ?cong ==> ?P ==> _")
proof -
assume eq: "PROP ?eq" and cong: "PROP ?cong" and P: ?P
have "f (Abs_Rat \<lfloor>fract a b\<rfloor>) (Abs_Rat \<lfloor>fract c d\<rfloor>) = g (fract a b) (fract c d)"
proof (rule quot_cond_function)
fix X Y assume "P X Y"
with eq show "f (Abs_Rat X) (Abs_Rat Y) == g (pick X) (pick Y)"
by (simp add: fraction_of_def pick_inverse Abs_Rat_inverse)
next
fix Q Q' R R' :: fraction
show "\<lfloor>Q\<rfloor> = \<lfloor>Q'\<rfloor> ==> \<lfloor>R\<rfloor> = \<lfloor>R'\<rfloor> ==>
P \<lfloor>Q\<rfloor> \<lfloor>R\<rfloor> ==> P \<lfloor>Q'\<rfloor> \<lfloor>R'\<rfloor> ==> g Q R = g Q' R'"
by (induct Q, induct Q', induct R, induct R') (rule cong)
qed
thus ?thesis by (unfold Fract_def rat_of_def)
qed

theorem rat_function:
"(!!q r. f q r == g (fraction_of q) (fraction_of r)) ==>
(!!a b a' b' c d c' d'.
\<lfloor>fract a b\<rfloor> = \<lfloor>fract a' b'\<rfloor> ==> \<lfloor>fract c d\<rfloor> = \<lfloor>fract c' d'\<rfloor> ==>
b \<noteq> 0 ==> b' \<noteq> 0 ==> d \<noteq> 0 ==> d' \<noteq> 0 ==>
g (fract a b) (fract c d) = g (fract a' b') (fract c' d')) ==>
f (Fract a b) (Fract c d) = g (fract a b) (fract c d)"
proof -
case rule_context from this TrueI
show ?thesis by (rule rat_cond_function)
qed

subsubsection {* Standard operations on rational numbers *}

instance rat :: "{zero, one, plus, minus, times, inverse, ord}" ..

zero_rat_def: "0 == rat_of 0"
one_rat_def: "1 == rat_of 1"
add_rat_def: "q + r == rat_of (fraction_of q + fraction_of r)"
minus_rat_def: "-q == rat_of (-(fraction_of q))"
diff_rat_def: "q - r == q + (-(r::rat))"
mult_rat_def: "q * r == rat_of (fraction_of q * fraction_of r)"
inverse_rat_def: "inverse q ==
if q=0 then 0 else rat_of (inverse (fraction_of q))"
divide_rat_def: "q / r == q * inverse (r::rat)"
le_rat_def: "q \<le> r == fraction_of q \<le> fraction_of r"
less_rat_def: "q < r == q \<le> r \<and> q \<noteq> (r::rat)"
abs_rat_def: "\<bar>q\<bar> == if q < 0 then -q else (q::rat)"

theorem zero_rat: "0 = Fract 0 1"
by (simp add: zero_rat_def zero_fraction_def rat_of_def Fract_def)

theorem one_rat: "1 = Fract 1 1"
by (simp add: one_rat_def one_fraction_def rat_of_def Fract_def)

theorem add_rat: "b \<noteq> 0 ==> d \<noteq> 0 ==>
Fract a b + Fract c d = Fract (a * d + c * b) (b * d)"
proof -
have "Fract a b + Fract c d = rat_of (fract a b + fract c d)"
also
assume "b \<noteq> 0"  "d \<noteq> 0"
hence "fract a b + fract c d = fract (a * d + c * b) (b * d)"
finally show ?thesis by (unfold Fract_def)
qed

theorem minus_rat: "b \<noteq> 0 ==> -(Fract a b) = Fract (-a) b"
proof -
have "-(Fract a b) = rat_of (-(fract a b))"
by (rule rat_function, rule minus_rat_def, rule rat_of, rule minus_fraction_cong)
also assume "b \<noteq> 0" hence "-(fract a b) = fract (-a) b"
finally show ?thesis by (unfold Fract_def)
qed

theorem diff_rat: "b \<noteq> 0 ==> d \<noteq> 0 ==>
Fract a b - Fract c d = Fract (a * d - c * b) (b * d)"

theorem mult_rat: "b \<noteq> 0 ==> d \<noteq> 0 ==>
Fract a b * Fract c d = Fract (a * c) (b * d)"
proof -
have "Fract a b * Fract c d = rat_of (fract a b * fract c d)"
by (rule rat_function, rule mult_rat_def, rule rat_of, rule mult_fraction_cong)
also
assume "b \<noteq> 0"  "d \<noteq> 0"
hence "fract a b * fract c d = fract (a * c) (b * d)"
finally show ?thesis by (unfold Fract_def)
qed

theorem inverse_rat: "Fract a b \<noteq> 0 ==> b \<noteq> 0 ==>
inverse (Fract a b) = Fract b a"
proof -
assume neq: "b \<noteq> 0" and nonzero: "Fract a b \<noteq> 0"
hence "\<lfloor>fract a b\<rfloor> \<noteq> \<lfloor>0\<rfloor>"
by (simp add: zero_rat eq_rat is_zero_fraction_iff)
with _ inverse_fraction_cong [THEN rat_of]
have "inverse (Fract a b) = rat_of (inverse (fract a b))"
proof (rule rat_cond_function)
fix q assume cond: "\<lfloor>fraction_of q\<rfloor> \<noteq> \<lfloor>0\<rfloor>"
have "q \<noteq> 0"
proof (cases q)
fix a b assume "b \<noteq> 0" and "q = Fract a b"
from this cond show ?thesis
by (simp add: Fract_inverse is_zero_fraction_iff zero_rat eq_rat)
qed
thus "inverse q == rat_of (inverse (fraction_of q))"
qed
also from neq nonzero have "inverse (fract a b) = fract b a"
finally show ?thesis by (unfold Fract_def)
qed

theorem divide_rat: "Fract c d \<noteq> 0 ==> b \<noteq> 0 ==> d \<noteq> 0 ==>
Fract a b / Fract c d = Fract (a * d) (b * c)"
proof -
assume neq: "b \<noteq> 0"  "d \<noteq> 0" and nonzero: "Fract c d \<noteq> 0"
hence "c \<noteq> 0" by (simp add: zero_rat eq_rat)
with neq nonzero show ?thesis
by (simp add: divide_rat_def inverse_rat mult_rat)
qed

theorem le_rat: "b \<noteq> 0 ==> d \<noteq> 0 ==>
(Fract a b \<le> Fract c d) = ((a * d) * (b * d) \<le> (c * b) * (b * d))"
proof -
have "(Fract a b \<le> Fract c d) = (fract a b \<le> fract c d)"
by (rule rat_function, rule le_rat_def, rule le_fraction_cong)
also
assume "b \<noteq> 0"  "d \<noteq> 0"
hence "(fract a b \<le> fract c d) = ((a * d) * (b * d) \<le> (c * b) * (b * d))"
finally show ?thesis .
qed

theorem less_rat: "b \<noteq> 0 ==> d \<noteq> 0 ==>
(Fract a b < Fract c d) = ((a * d) * (b * d) < (c * b) * (b * d))"
by (simp add: less_rat_def le_rat eq_rat order_less_le)

theorem abs_rat: "b \<noteq> 0 ==> \<bar>Fract a b\<bar> = Fract \<bar>a\<bar> \<bar>b\<bar>"
by (simp add: abs_rat_def minus_rat zero_rat less_rat eq_rat)
(auto simp add: mult_less_0_iff zero_less_mult_iff order_le_less
split: abs_split)

subsubsection {* The ordered field of rational numbers *}

lemma rat_add_assoc: "(q + r) + s = q + (r + (s::rat))"
by (induct q, induct r, induct s)

lemma rat_add_0: "0 + q = (q::rat)"

lemma rat_left_minus: "(-q) + q = (0::rat)"

instance rat :: field
proof
fix q r s :: rat
show "(q + r) + s = q + (r + s)"
show "q + r = r + q"
show "0 + q = q"
show "(-q) + q = 0"
by (rule rat_left_minus)
show "q - r = q + (-r)"
show "(q * r) * s = q * (r * s)"
by (induct q, induct r, induct s) (simp add: mult_rat mult_ac)
show "q * r = r * q"
by (induct q, induct r) (simp add: mult_rat mult_ac)
show "1 * q = q"
by (induct q) (simp add: one_rat mult_rat)
show "(q + r) * s = q * s + r * s"
by (induct q, induct r, induct s)
show "q \<noteq> 0 ==> inverse q * q = 1"
by (induct q) (simp add: inverse_rat mult_rat one_rat zero_rat eq_rat)
show "q / r = q * inverse r"
show "0 \<noteq> (1::rat)"
by (simp add: zero_rat one_rat eq_rat)
qed

instance rat :: linorder
proof
fix q r s :: rat
{
assume "q \<le> r" and "r \<le> s"
show "q \<le> s"
proof (insert prems, induct q, induct r, induct s)
fix a b c d e f :: int
assume neq: "b \<noteq> 0"  "d \<noteq> 0"  "f \<noteq> 0"
assume 1: "Fract a b \<le> Fract c d" and 2: "Fract c d \<le> Fract e f"
show "Fract a b \<le> Fract e f"
proof -
from neq obtain bb: "0 < b * b" and dd: "0 < d * d" and ff: "0 < f * f"
by (auto simp add: zero_less_mult_iff linorder_neq_iff)
have "(a * d) * (b * d) * (f * f) \<le> (c * b) * (b * d) * (f * f)"
proof -
from neq 1 have "(a * d) * (b * d) \<le> (c * b) * (b * d)"
with ff show ?thesis by (simp add: mult_le_cancel_right)
qed
also have "... = (c * f) * (d * f) * (b * b)"
by (simp only: mult_ac)
also have "... \<le> (e * d) * (d * f) * (b * b)"
proof -
from neq 2 have "(c * f) * (d * f) \<le> (e * d) * (d * f)"
with bb show ?thesis by (simp add: mult_le_cancel_right)
qed
finally have "(a * f) * (b * f) * (d * d) \<le> e * b * (b * f) * (d * d)"
by (simp only: mult_ac)
with dd have "(a * f) * (b * f) \<le> (e * b) * (b * f)"
with neq show ?thesis by (simp add: le_rat)
qed
qed
next
assume "q \<le> r" and "r \<le> q"
show "q = r"
proof (insert prems, induct q, induct r)
fix a b c d :: int
assume neq: "b \<noteq> 0"  "d \<noteq> 0"
assume 1: "Fract a b \<le> Fract c d" and 2: "Fract c d \<le> Fract a b"
show "Fract a b = Fract c d"
proof -
from neq 1 have "(a * d) * (b * d) \<le> (c * b) * (b * d)"
also have "... \<le> (a * d) * (b * d)"
proof -
from neq 2 have "(c * b) * (d * b) \<le> (a * d) * (d * b)"
thus ?thesis by (simp only: mult_ac)
qed
finally have "(a * d) * (b * d) = (c * b) * (b * d)" .
moreover from neq have "b * d \<noteq> 0" by simp
ultimately have "a * d = c * b" by simp
with neq show ?thesis by (simp add: eq_rat)
qed
qed
next
show "q \<le> q"
by (induct q) (simp add: le_rat)
show "(q < r) = (q \<le> r \<and> q \<noteq> r)"
by (simp only: less_rat_def)
show "q \<le> r \<or> r \<le> q"
by (induct q, induct r) (simp add: le_rat mult_ac, arith)
}
qed

instance rat :: ordered_field
proof
fix q r s :: rat
show "q \<le> r ==> s + q \<le> s + r"
proof (induct q, induct r, induct s)
fix a b c d e f :: int
assume neq: "b \<noteq> 0"  "d \<noteq> 0"  "f \<noteq> 0"
assume le: "Fract a b \<le> Fract c d"
show "Fract e f + Fract a b \<le> Fract e f + Fract c d"
proof -
let ?F = "f * f" from neq have F: "0 < ?F"
from neq le have "(a * d) * (b * d) \<le> (c * b) * (b * d)"
with F have "(a * d) * (b * d) * ?F * ?F \<le> (c * b) * (b * d) * ?F * ?F"
qed
qed
show "q < r ==> 0 < s ==> s * q < s * r"
proof (induct q, induct r, induct s)
fix a b c d e f :: int
assume neq: "b \<noteq> 0"  "d \<noteq> 0"  "f \<noteq> 0"
assume le: "Fract a b < Fract c d"
assume gt: "0 < Fract e f"
show "Fract e f * Fract a b < Fract e f * Fract c d"
proof -
let ?E = "e * f" and ?F = "f * f"
from neq gt have "0 < ?E"
by (auto simp add: zero_rat less_rat le_rat order_less_le eq_rat)
moreover from neq have "0 < ?F"
moreover from neq le have "(a * d) * (b * d) < (c * b) * (b * d)"
ultimately have "(a * d) * (b * d) * ?E * ?F < (c * b) * (b * d) * ?E * ?F"
with neq show ?thesis
by (simp add: less_rat mult_rat mult_ac)
qed
qed
show "\<bar>q\<bar> = (if q < 0 then -q else q)"
by (simp only: abs_rat_def)
qed

instance rat :: division_by_zero
proof
show "inverse 0 = (0::rat)"  by (simp add: inverse_rat_def)
qed

subsection {* Various Other Results *}

lemma minus_rat_cancel [simp]: "b \<noteq> 0 ==> Fract (-a) (-b) = Fract a b"

theorem Rat_induct_pos [case_names Fract, induct type: rat]:
assumes step: "!!a b. 0 < b ==> P (Fract a b)"
shows "P q"
proof (cases q)
have step': "!!a b. b < 0 ==> P (Fract a b)"
proof -
fix a::int and b::int
assume b: "b < 0"
hence "0 < -b" by simp
hence "P (Fract (-a) (-b))" by (rule step)
thus "P (Fract a b)" by (simp add: order_less_imp_not_eq [OF b])
qed
case (Fract a b)
thus "P q" by (force simp add: linorder_neq_iff step step')
qed

lemma zero_less_Fract_iff:
"0 < b ==> (0 < Fract a b) = (0 < a)"
by (simp add: zero_rat less_rat order_less_imp_not_eq2 zero_less_mult_iff)

lemma Fract_add_one: "n \<noteq> 0 ==> Fract (m + n) n = Fract m n + 1"
apply (insert add_rat [of concl: m n 1 1])
done

lemma Fract_of_nat_eq: "Fract (of_nat k) 1 = of_nat k"
apply (induct k)
done

lemma Fract_of_int_eq: "Fract k 1 = of_int k"
proof (cases k rule: int_cases)
case (nonneg n)
thus ?thesis by (simp add: int_eq_of_nat Fract_of_nat_eq)
next
case (neg n)
hence "Fract k 1 = - (Fract (of_nat (Suc n)) 1)"
by (simp only: minus_rat int_eq_of_nat)
also have "... =  - (of_nat (Suc n))"
by (simp only: Fract_of_nat_eq)
finally show ?thesis
by (simp add: only: prems int_eq_of_nat of_int_minus of_int_of_nat_eq)
qed

subsection {* Numerals and Arithmetic *}

instance rat :: number ..

rat_number_of_def: "(number_of w :: rat) == of_int (Rep_Bin w)"
--{*the type constraint is essential!*}

instance rat :: number_ring