--- a/src/HOL/Library/Quotient.thy Thu Oct 19 21:22:44 2000 +0200
+++ b/src/HOL/Library/Quotient.thy Thu Oct 19 21:23:15 2000 +0200
@@ -162,51 +162,6 @@
lemma quotE [elim]: "R \<in> quot ==> (!!a. R = {x. a \<sim> x} ==> C) ==> C"
by (unfold quot_def) blast
-
-text {*
- \medskip Standard properties of type-definitions.\footnote{(FIXME)
- Better incorporate these into the typedef package?}
-*}
-
-theorem Rep_quot_inject: "(Rep_quot x = Rep_quot y) = (x = y)"
-proof
- assume "Rep_quot x = Rep_quot y"
- hence "Abs_quot (Rep_quot x) = Abs_quot (Rep_quot y)" by (simp only:)
- thus "x = y" by (simp only: Rep_quot_inverse)
-next
- assume "x = y"
- thus "Rep_quot x = Rep_quot y" by simp
-qed
-
-theorem Abs_quot_inject:
- "x \<in> quot ==> y \<in> quot ==> (Abs_quot x = Abs_quot y) = (x = y)"
-proof
- assume "Abs_quot x = Abs_quot y"
- hence "Rep_quot (Abs_quot x) = Rep_quot (Abs_quot y)" by simp
- also assume "x \<in> quot" hence "Rep_quot (Abs_quot x) = x" by (rule Abs_quot_inverse)
- also assume "y \<in> quot" hence "Rep_quot (Abs_quot y) = y" by (rule Abs_quot_inverse)
- finally show "x = y" .
-next
- assume "x = y"
- thus "Abs_quot x = Abs_quot y" by simp
-qed
-
-theorem Rep_quot_induct: "y \<in> quot ==> (!!x. P (Rep_quot x)) ==> P y"
-proof -
- assume "!!x. P (Rep_quot x)" hence "P (Rep_quot (Abs_quot y))" .
- also assume "y \<in> quot" hence "Rep_quot (Abs_quot y) = y" by (rule Abs_quot_inverse)
- finally show "P y" .
-qed
-
-theorem Abs_quot_induct: "(!!y. y \<in> quot ==> P (Abs_quot y)) ==> P x"
-proof -
- assume r: "!!y. y \<in> quot ==> P (Abs_quot y)"
- have "Rep_quot x \<in> quot" by (rule Rep_quot)
- hence "P (Abs_quot (Rep_quot x))" by (rule r)
- also have "Abs_quot (Rep_quot x) = x" by (rule Rep_quot_inverse)
- finally show "P x" .
-qed
-
text {*
\medskip Abstracted equivalence classes are the canonical
representation of elements of a quotient type.
@@ -217,13 +172,11 @@
"\<lfloor>a\<rfloor> == Abs_quot {x. a \<sim> x}"
theorem quot_rep: "\<exists>a. A = \<lfloor>a\<rfloor>"
-proof (unfold eqv_class_def)
- show "\<exists>a. A = Abs_quot {x. a \<sim> x}"
- proof (induct A rule: Abs_quot_induct)
- fix R assume "R \<in> quot"
- hence "\<exists>a. R = {x. a \<sim> x}" by blast
- thus "\<exists>a. Abs_quot R = Abs_quot {x. a \<sim> x}" by blast
- qed
+proof (cases A)
+ fix R assume R: "A = Abs_quot R"
+ assume "R \<in> quot" hence "\<exists>a. R = {x. a \<sim> x}" by blast
+ with R have "\<exists>a. A = Abs_quot {x. a \<sim> x}" by blast
+ thus ?thesis by (unfold eqv_class_def)
qed
lemma quot_cases [case_names rep, cases type: quot]:
@@ -302,10 +255,10 @@
show "a \<in> domain" ..
qed
-theorem pick_inverse: "\<lfloor>pick A\<rfloor> = (A::'a::equiv quot)" (* FIXME tune proof *)
+theorem pick_inverse: "\<lfloor>pick A\<rfloor> = (A::'a::equiv quot)"
proof (cases A)
fix a assume a: "A = \<lfloor>a\<rfloor>"
- hence "pick A \<sim> a" by (simp only: pick_eqv)
+ hence "pick A \<sim> a" by simp
hence "\<lfloor>pick A\<rfloor> = \<lfloor>a\<rfloor>" by simp
with a show ?thesis by simp
qed