QuotScript.thy
author Cezary Kaliszyk <kaliszyk@in.tum.de>
Wed, 14 Oct 2009 16:23:32 +0200
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permissions -rw-r--r--
Manually regularized list_induct2
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theory QuotScript
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imports Main
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begin
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definition 
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  "EQUIV E \<equiv> \<forall>x y. E x y = (E x = E y)" 
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definition
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  "REFL E \<equiv> \<forall>x. E x x"
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definition 
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  "SYM E \<equiv> \<forall>x y. E x y \<longrightarrow> E y x"
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definition
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  "TRANS E \<equiv> \<forall>x y z. E x y \<and> E y z \<longrightarrow> E x z"
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lemma EQUIV_REFL_SYM_TRANS:
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  shows "EQUIV E = (REFL E \<and> SYM E \<and> TRANS E)"
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unfolding EQUIV_def REFL_def SYM_def TRANS_def expand_fun_eq
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by (blast)
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definition
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  "PART_EQUIV E \<equiv> (\<exists>x. E x x) \<and> (\<forall>x y. E x y = (E x x \<and> E y y \<and> (E x = E y)))"
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lemma EQUIV_IMP_PART_EQUIV:
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  assumes a: "EQUIV E"
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  shows "PART_EQUIV E"
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using a unfolding EQUIV_def PART_EQUIV_def
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by auto
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definition
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  "QUOTIENT E Abs Rep \<equiv> (\<forall>a. Abs (Rep a) = a) \<and> 
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                        (\<forall>a. E (Rep a) (Rep a)) \<and> 
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                        (\<forall>r s. E r s = (E r r \<and> E s s \<and> (Abs r = Abs s)))"
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lemma QUOTIENT_ABS_REP:
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  assumes a: "QUOTIENT E Abs Rep"
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  shows "Abs (Rep a) = a" 
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using a unfolding QUOTIENT_def
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by simp
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lemma QUOTIENT_REP_REFL:
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  assumes a: "QUOTIENT E Abs Rep"
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  shows "E (Rep a) (Rep a)" 
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using a unfolding QUOTIENT_def
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by blast
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lemma QUOTIENT_REL:
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  assumes a: "QUOTIENT E Abs Rep"
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  shows " E r s = (E r r \<and> E s s \<and> (Abs r = Abs s))"
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using a unfolding QUOTIENT_def
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by blast
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lemma QUOTIENT_REL_ABS:
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  assumes a: "QUOTIENT E Abs Rep"
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  shows "E r s \<Longrightarrow> Abs r = Abs s"
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using a unfolding QUOTIENT_def
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by blast
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lemma QUOTIENT_REL_ABS_EQ:
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  assumes a: "QUOTIENT E Abs Rep"
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  shows "E r r \<Longrightarrow> E s s \<Longrightarrow> E r s = (Abs r = Abs s)"
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using a unfolding QUOTIENT_def
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by blast
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lemma QUOTIENT_REL_REP:
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  assumes a: "QUOTIENT E Abs Rep"
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  shows "E (Rep a) (Rep b) = (a = b)"
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using a unfolding QUOTIENT_def
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by metis
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lemma QUOTIENT_REP_ABS:
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  assumes a: "QUOTIENT E Abs Rep"
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  shows "E r r \<Longrightarrow> E (Rep (Abs r)) r"
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using a unfolding QUOTIENT_def
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by blast
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lemma IDENTITY_EQUIV:
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  shows "EQUIV (op =)"
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unfolding EQUIV_def
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by auto
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lemma IDENTITY_QUOTIENT:
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  shows "QUOTIENT (op =) (\<lambda>x. x) (\<lambda>x. x)"
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unfolding QUOTIENT_def
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by blast
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lemma QUOTIENT_SYM:
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  assumes a: "QUOTIENT E Abs Rep"
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  shows "SYM E"
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using a unfolding QUOTIENT_def SYM_def
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by metis
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lemma QUOTIENT_TRANS:
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  assumes a: "QUOTIENT E Abs Rep"
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  shows "TRANS E"
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using a unfolding QUOTIENT_def TRANS_def
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by metis
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fun
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  prod_rel
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where
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  "prod_rel r1 r2 = (\<lambda>(a,b) (c,d). r1 a c \<and> r2 b d)"
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fun
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  fun_map 
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where
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  "fun_map f g h x = g (h (f x))"
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abbreviation
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  fun_map_syn ("_ ---> _")
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where
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  "f ---> g \<equiv> fun_map f g"  
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lemma FUN_MAP_I:
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  shows "(\<lambda>x. x ---> \<lambda>x. x) = (\<lambda>x. x)"
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by (simp add: expand_fun_eq)
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lemma IN_FUN:
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  shows "x \<in> ((f ---> g) s) = g (f x \<in> s)"
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by (simp add: mem_def)
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fun
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  FUN_REL 
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where
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  "FUN_REL E1 E2 f g = (\<forall>x y. E1 x y \<longrightarrow> E2 (f x) (g y))"
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abbreviation
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  FUN_REL_syn ("_ ===> _")
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where
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  "E1 ===> E2 \<equiv> FUN_REL E1 E2"  
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lemma FUN_REL_EQ:
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  "(op =) ===> (op =) = (op =)"
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by (simp add: expand_fun_eq)
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lemma FUN_QUOTIENT:
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  assumes q1: "QUOTIENT R1 abs1 rep1"
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  and     q2: "QUOTIENT R2 abs2 rep2"
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  shows "QUOTIENT (R1 ===> R2) (rep1 ---> abs2) (abs1 ---> rep2)"
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proof -
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  have "\<forall>a. (rep1 ---> abs2) ((abs1 ---> rep2) a) = a"
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    apply(simp add: expand_fun_eq)
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    using q1 q2
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    apply(simp add: QUOTIENT_def)
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    done
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  moreover
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  have "\<forall>a. (R1 ===> R2) ((abs1 ---> rep2) a) ((abs1 ---> rep2) a)"
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    apply(auto)
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    using q1 q2 unfolding QUOTIENT_def
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    apply(metis)
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    done
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  moreover
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  have "\<forall>r s. (R1 ===> R2) r s = ((R1 ===> R2) r r \<and> (R1 ===> R2) s s \<and> 
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        (rep1 ---> abs2) r  = (rep1 ---> abs2) s)"
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    apply(auto simp add: expand_fun_eq)
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    using q1 q2 unfolding QUOTIENT_def
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    apply(metis)
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    using q1 q2 unfolding QUOTIENT_def
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    apply(metis)
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    using q1 q2 unfolding QUOTIENT_def
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    apply(metis)
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    using q1 q2 unfolding QUOTIENT_def
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    apply(metis)
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    done
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  ultimately
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  show "QUOTIENT (R1 ===> R2) (rep1 ---> abs2) (abs1 ---> rep2)"
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    unfolding QUOTIENT_def by blast
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qed
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definition
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  Respects
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where
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  "Respects R x \<equiv> (R x x)"
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lemma IN_RESPECTS:
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  shows "(x \<in> Respects R) = R x x"
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unfolding mem_def Respects_def by simp
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lemma RESPECTS_THM:
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  shows "Respects (R1 ===> R2) f = (\<forall>x y. R1 x y \<longrightarrow> R2 (f x) (f y))"
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unfolding Respects_def
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by (simp add: expand_fun_eq) 
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lemma RESPECTS_MP:
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  assumes a: "Respects (R1 ===> R2) f"
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  and     b: "R1 x y"
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  shows "R2 (f x) (f y)"
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using a b unfolding Respects_def
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by simp
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lemma RESPECTS_REP_ABS:
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  assumes a: "QUOTIENT R1 Abs1 Rep1"
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  and     b: "Respects (R1 ===> R2) f"
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  and     c: "R1 x x"
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  shows "R2 (f (Rep1 (Abs1 x))) (f x)"
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using a b[simplified RESPECTS_THM] c unfolding QUOTIENT_def
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by blast
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lemma RESPECTS_o:
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  assumes a: "Respects (R2 ===> R3) f"
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  and     b: "Respects (R1 ===> R2) g"
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  shows "Respects (R1 ===> R3) (f o g)"
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using a b unfolding Respects_def
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by simp
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(*
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definition
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  "RES_EXISTS_EQUIV R P \<equiv> (\<exists>x \<in> Respects R. P x) \<and> 
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                          (\<forall>x\<in> Respects R. \<forall>y\<in> Respects R. P x \<and> P y \<longrightarrow> R x y)"
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*)
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lemma FUN_REL_EQ_REL:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  shows "(R1 ===> R2) f g = ((Respects (R1 ===> R2) f) \<and> (Respects (R1 ===> R2) g) 
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                             \<and> ((Rep1 ---> Abs2) f = (Rep1 ---> Abs2) g))"
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using FUN_QUOTIENT[OF q1 q2] unfolding Respects_def QUOTIENT_def expand_fun_eq
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by blast
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(* q1 and q2 not used; see next lemma *)
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lemma FUN_REL_MP:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  shows "(R1 ===> R2) f g \<Longrightarrow> R1 x y \<Longrightarrow> R2 (f x) (g y)"
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by simp
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lemma FUN_REL_IMP:
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  shows "(R1 ===> R2) f g \<Longrightarrow> R1 x y \<Longrightarrow> R2 (f x) (g y)"
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by simp
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lemma FUN_REL_EQUALS:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  and     r1: "Respects (R1 ===> R2) f"
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  and     r2: "Respects (R1 ===> R2) g" 
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  shows "((Rep1 ---> Abs2) f = (Rep1 ---> Abs2) g) = (\<forall>x y. R1 x y \<longrightarrow> R2 (f x) (g y))"
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apply(rule_tac iffI)
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using FUN_QUOTIENT[OF q1 q2] r1 r2 unfolding QUOTIENT_def Respects_def
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apply(metis FUN_REL_IMP)
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using r1 unfolding Respects_def expand_fun_eq
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apply(simp (no_asm_use))
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apply(metis QUOTIENT_REL[OF q2] QUOTIENT_REL_REP[OF q1])
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done
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(* ask Peter: FUN_REL_IMP used twice *) 
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lemma FUN_REL_IMP2:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  and     r1: "Respects (R1 ===> R2) f"
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  and     r2: "Respects (R1 ===> R2) g" 
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  and     a:  "(Rep1 ---> Abs2) f = (Rep1 ---> Abs2) g"
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  shows "R1 x y \<Longrightarrow> R2 (f x) (g y)"
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using q1 q2 r1 r2 a
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by (simp add: FUN_REL_EQUALS)
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lemma EQUALS_PRS:
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  assumes q: "QUOTIENT R Abs Rep"
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  shows "(x = y) = R (Rep x) (Rep y)"
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by (simp add: QUOTIENT_REL_REP[OF q]) 
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lemma EQUALS_RSP:
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  assumes q: "QUOTIENT R Abs Rep"
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  and     a: "R x1 x2" "R y1 y2"
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  shows "R x1 y1 = R x2 y2"
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using QUOTIENT_SYM[OF q] QUOTIENT_TRANS[OF q] unfolding SYM_def TRANS_def
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using a by blast
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lemma LAMBDA_PRS:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  shows "(\<lambda>x. f x) = (Rep1 ---> Abs2) (\<lambda>x. Rep2 (f (Abs1 x)))"
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unfolding expand_fun_eq
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using QUOTIENT_ABS_REP[OF q1] QUOTIENT_ABS_REP[OF q2]
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by simp
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lemma LAMBDA_PRS1:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  shows "(\<lambda>x. f x) = (Rep1 ---> Abs2) (\<lambda>x. (Abs1 ---> Rep2) f x)"
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unfolding expand_fun_eq
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by (subst LAMBDA_PRS[OF q1 q2]) (simp)
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(* Ask Peter: assumption q1 and q2 not used and lemma is the 'identity' *)
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lemma LAMBDA_RSP:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  and     a: "(R1 ===> R2) f1 f2"
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  shows "(R1 ===> R2) (\<lambda>x. f1 x) (\<lambda>y. f2 y)"
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by (rule a)
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(* ASK Peter about next four lemmas in quotientScript
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lemma ABSTRACT_PRS:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  shows "f = (Rep1 ---> Abs2) ???"
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*)
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lemma LAMBDA_REP_ABS_RSP:
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  assumes r1: "\<And>r r'. R1 r r' \<Longrightarrow>R1 r (Rep1 (Abs1 r'))"
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  and     r2: "\<And>r r'. R2 r r' \<Longrightarrow>R2 r (Rep2 (Abs2 r'))"
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  shows "(R1 ===> R2) f1 f2 \<Longrightarrow> (R1 ===> R2) f1 ((Abs1 ---> Rep2) ((Rep1 ---> Abs2) f2))"
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using r1 r2 by auto
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lemma REP_ABS_RSP:
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  assumes q: "QUOTIENT R Abs Rep"
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  and     a: "R x1 x2"
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  shows "R x1 (Rep (Abs x2))"
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using a
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by (metis QUOTIENT_REL[OF q] QUOTIENT_ABS_REP[OF q] QUOTIENT_REP_REFL[OF q])
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(* ----------------------------------------------------- *)
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(* Quantifiers: FORALL, EXISTS, EXISTS_UNIQUE,           *)
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(*              RES_FORALL, RES_EXISTS, RES_EXISTS_EQUIV *)
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(* ----------------------------------------------------- *)
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(* what is RES_FORALL *)
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(*--`!R (abs:'a -> 'b) rep. QUOTIENT R abs rep ==>
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         !f. $! f = RES_FORALL (respects R) ((abs --> I) f)`--*)
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(*as peter here *)
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(* bool theory: COND, LET *)
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lemma IF_PRS:
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  assumes q: "QUOTIENT R Abs Rep"
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  shows "If a b c = Abs (If a (Rep b) (Rep c))"
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using QUOTIENT_ABS_REP[OF q] by auto
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(* ask peter: no use of q *)
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lemma IF_RSP:
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  assumes q: "QUOTIENT R Abs Rep"
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  and     a: "a1 = a2" "R b1 b2" "R c1 c2"
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  shows "R (If a1 b1 c1) (If a2 b2 c2)"
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using a by auto
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lemma LET_PRS:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  shows "Let x f = Abs2 (Let (Rep1 x) ((Abs1 ---> Rep2) f))"
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using QUOTIENT_ABS_REP[OF q1] QUOTIENT_ABS_REP[OF q2] by auto
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lemma LET_RSP:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  and     a1: "(R1 ===> R2) f g"
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  and     a2: "R1 x y"
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  shows "R2 (Let x f) (Let y g)"
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using FUN_REL_MP[OF q1 q2 a1] a2
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by auto
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(* ask peter what are literal_case *)
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(* literal_case_PRS *)
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(* literal_case_RSP *)
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(* FUNCTION APPLICATION *)
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lemma APPLY_PRS:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  shows "f x = Abs2 (((Abs1 ---> Rep2) f) (Rep1 x))"
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using QUOTIENT_ABS_REP[OF q1] QUOTIENT_ABS_REP[OF q2] by auto
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(* ask peter: no use of q1 q2 *)
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lemma APPLY_RSP:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  and     a: "(R1 ===> R2) f g" "R1 x y"
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  shows "R2 (f x) (g y)"
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using a by (rule FUN_REL_IMP)
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(* combinators: I, K, o, C, W *)
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lemma I_PRS:
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  assumes q: "QUOTIENT R Abs Rep"
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  shows "(\<lambda>x. x) e = Abs ((\<lambda> x. x) (Rep e))"
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using QUOTIENT_ABS_REP[OF q] by auto
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lemma I_RSP:
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  assumes q: "QUOTIENT R Abs Rep"
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  and     a: "R e1 e2"
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  shows "R ((\<lambda>x. x) e1) ((\<lambda> x. x) e2)"
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using a by auto
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lemma o_PRS:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  and     q3: "QUOTIENT R3 Abs3 Rep3"
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  shows "f o g = (Rep1 ---> Abs3) (((Abs2 ---> Rep3) f) o ((Abs1 ---> Rep2) g))"
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using QUOTIENT_ABS_REP[OF q1] QUOTIENT_ABS_REP[OF q2] QUOTIENT_ABS_REP[OF q3]
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unfolding o_def expand_fun_eq
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by simp
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lemma o_RSP:
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  assumes q1: "QUOTIENT R1 Abs1 Rep1"
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  and     q2: "QUOTIENT R2 Abs2 Rep2"
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  and     q3: "QUOTIENT R3 Abs3 Rep3"
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  and     a1: "(R2 ===> R3) f1 f2"
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  and     a2: "(R1 ===> R2) g1 g2"
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  shows "(R1 ===> R3) (f1 o g1) (f2 o g2)"
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using a1 a2 unfolding o_def expand_fun_eq
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by (auto)
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93
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lemma equiv_res_forall:
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  fixes P :: "'a \<Rightarrow> bool"
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  assumes a: "EQUIV E"
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  shows "Ball (Respects E) P = (All P)"
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  using a by (metis EQUIV_def IN_RESPECTS a)
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lemma equiv_res_exists:
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  fixes P :: "'a \<Rightarrow> bool"
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  assumes a: "EQUIV E"
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  shows "Bex (Respects E) P = (Ex P)"
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  using a by (metis EQUIV_def IN_RESPECTS a)
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end