src/HOL/Library/Quotient.thy
 changeset 10278 ea1bf4b6255c parent 10250 ca93fe25a84b child 10285 6949e17f314a
1.1 --- a/src/HOL/Library/Quotient.thy	Thu Oct 19 21:22:44 2000 +0200
1.2 +++ b/src/HOL/Library/Quotient.thy	Thu Oct 19 21:23:15 2000 +0200
1.3 @@ -162,51 +162,6 @@
1.4  lemma quotE [elim]: "R \<in> quot ==> (!!a. R = {x. a \<sim> x} ==> C) ==> C"
1.5    by (unfold quot_def) blast
1.7 -
1.8 -text {*
1.9 - \medskip Standard properties of type-definitions.\footnote{(FIXME)
1.10 - Better incorporate these into the typedef package?}
1.11 -*}
1.12 -
1.13 -theorem Rep_quot_inject: "(Rep_quot x = Rep_quot y) = (x = y)"
1.14 -proof
1.15 -  assume "Rep_quot x = Rep_quot y"
1.16 -  hence "Abs_quot (Rep_quot x) = Abs_quot (Rep_quot y)" by (simp only:)
1.17 -  thus "x = y" by (simp only: Rep_quot_inverse)
1.18 -next
1.19 -  assume "x = y"
1.20 -  thus "Rep_quot x = Rep_quot y" by simp
1.21 -qed
1.22 -
1.23 -theorem Abs_quot_inject:
1.24 -  "x \<in> quot ==> y \<in> quot ==> (Abs_quot x = Abs_quot y) = (x = y)"
1.25 -proof
1.26 -  assume "Abs_quot x = Abs_quot y"
1.27 -  hence "Rep_quot (Abs_quot x) = Rep_quot (Abs_quot y)" by simp
1.28 -  also assume "x \<in> quot" hence "Rep_quot (Abs_quot x) = x" by (rule Abs_quot_inverse)
1.29 -  also assume "y \<in> quot" hence "Rep_quot (Abs_quot y) = y" by (rule Abs_quot_inverse)
1.30 -  finally show "x = y" .
1.31 -next
1.32 -  assume "x = y"
1.33 -  thus "Abs_quot x = Abs_quot y" by simp
1.34 -qed
1.35 -
1.36 -theorem Rep_quot_induct: "y \<in> quot ==> (!!x. P (Rep_quot x)) ==> P y"
1.37 -proof -
1.38 -  assume "!!x. P (Rep_quot x)" hence "P (Rep_quot (Abs_quot y))" .
1.39 -  also assume "y \<in> quot" hence "Rep_quot (Abs_quot y) = y" by (rule Abs_quot_inverse)
1.40 -  finally show "P y" .
1.41 -qed
1.42 -
1.43 -theorem Abs_quot_induct: "(!!y. y \<in> quot ==> P (Abs_quot y)) ==> P x"
1.44 -proof -
1.45 -  assume r: "!!y. y \<in> quot ==> P (Abs_quot y)"
1.46 -  have "Rep_quot x \<in> quot" by (rule Rep_quot)
1.47 -  hence "P (Abs_quot (Rep_quot x))" by (rule r)
1.48 -  also have "Abs_quot (Rep_quot x) = x" by (rule Rep_quot_inverse)
1.49 -  finally show "P x" .
1.50 -qed
1.51 -
1.52  text {*
1.53   \medskip Abstracted equivalence classes are the canonical
1.54   representation of elements of a quotient type.
1.55 @@ -217,13 +172,11 @@
1.56    "\<lfloor>a\<rfloor> == Abs_quot {x. a \<sim> x}"
1.58  theorem quot_rep: "\<exists>a. A = \<lfloor>a\<rfloor>"
1.59 -proof (unfold eqv_class_def)
1.60 -  show "\<exists>a. A = Abs_quot {x. a \<sim> x}"
1.61 -  proof (induct A rule: Abs_quot_induct)
1.62 -    fix R assume "R \<in> quot"
1.63 -    hence "\<exists>a. R = {x. a \<sim> x}" by blast
1.64 -    thus "\<exists>a. Abs_quot R = Abs_quot {x. a \<sim> x}" by blast
1.65 -  qed
1.66 +proof (cases A)
1.67 +  fix R assume R: "A = Abs_quot R"
1.68 +  assume "R \<in> quot" hence "\<exists>a. R = {x. a \<sim> x}" by blast
1.69 +  with R have "\<exists>a. A = Abs_quot {x. a \<sim> x}" by blast
1.70 +  thus ?thesis by (unfold eqv_class_def)
1.71  qed
1.73  lemma quot_cases [case_names rep, cases type: quot]:
1.74 @@ -302,10 +255,10 @@
1.75    show "a \<in> domain" ..
1.76  qed
1.78 -theorem pick_inverse: "\<lfloor>pick A\<rfloor> = (A::'a::equiv quot)"   (* FIXME tune proof *)
1.79 +theorem pick_inverse: "\<lfloor>pick A\<rfloor> = (A::'a::equiv quot)"
1.80  proof (cases A)
1.81    fix a assume a: "A = \<lfloor>a\<rfloor>"
1.82 -  hence "pick A \<sim> a" by (simp only: pick_eqv)
1.83 +  hence "pick A \<sim> a" by simp
1.84    hence "\<lfloor>pick A\<rfloor> = \<lfloor>a\<rfloor>" by simp
1.85    with a show ?thesis by simp
1.86  qed