Theory Abstraction

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theory Abstraction
imports FuncSet
begin

(*  Title:      HOL/MetisExamples/Abstraction.thy
    ID:         $Id: Abstraction.thy,v 1.7 2007/10/04 10:32:59 paulson Exp $
    Author:     Lawrence C Paulson, Cambridge University Computer Laboratory

Testing the metis method
*)

theory Abstraction imports FuncSet
begin

(*For Christoph Benzmueller*)
lemma "x<1 & ((op=) = (op=)) ==> ((op=) = (op=)) & (x<(2::nat))";
  by (metis One_nat_def less_Suc0 not_less0 not_less_eq numeral_2_eq_2)

(*this is a theorem, but we can't prove it unless ext is applied explicitly
lemma "(op=) = (%x y. y=x)"
*)

consts
  monotone :: "['a => 'a, 'a set, ('a *'a)set] => bool"
  pset  :: "'a set => 'a set"
  order :: "'a set => ('a * 'a) set"

ML{*ResAtp.problem_name := "Abstraction__Collect_triv"*}
lemma (*Collect_triv:*) "a ∈ {x. P x} ==> P a"
proof (neg_clausify)
assume 0: "(a::'a::type) ∈ Collect (P::'a::type => bool)"
assume 1: "¬ (P::'a::type => bool) (a::'a::type)"
have 2: "(P::'a::type => bool) (a::'a::type)"
  by (metis CollectD 0)
show "False"
  by (metis 2 1)
qed

lemma Collect_triv: "a ∈ {x. P x} ==> P a"
by (metis mem_Collect_eq)


ML{*ResAtp.problem_name := "Abstraction__Collect_mp"*}
lemma "a ∈ {x. P x --> Q x} ==> a ∈ {x. P x} ==> a ∈ {x. Q x}"
  by (metis CollectI Collect_imp_eq ComplD UnE mem_Collect_eq);
  --{*34 secs*}

ML{*ResAtp.problem_name := "Abstraction__Sigma_triv"*}
lemma "(a,b) ∈ Sigma A B ==> a ∈ A & b ∈ B a"
proof (neg_clausify)
assume 0: "(a::'a::type, b::'b::type) ∈ Sigma (A::'a::type set) (B::'a::type => 'b::type set)"
assume 1: "(a::'a::type) ∉ (A::'a::type set) ∨ (b::'b::type) ∉ (B::'a::type => 'b::type set) a"
have 2: "(a::'a::type) ∈ (A::'a::type set)"
  by (metis SigmaD1 0)
have 3: "(b::'b::type) ∈ (B::'a::type => 'b::type set) (a::'a::type)"
  by (metis SigmaD2 0)
have 4: "(b::'b::type) ∉ (B::'a::type => 'b::type set) (a::'a::type)"
  by (metis 1 2)
show "False"
  by (metis 3 4)
qed

lemma Sigma_triv: "(a,b) ∈ Sigma A B ==> a ∈ A & b ∈ B a"
by (metis SigmaD1 SigmaD2)

ML{*ResAtp.problem_name := "Abstraction__Sigma_Collect"*}
lemma "(a,b) ∈ (SIGMA x: A. {y. x = f y}) ==> a ∈ A & a = f b"
(*???metis cannot prove this
by (metis CollectD SigmaD1 SigmaD2 UN_eq)
Also, UN_eq is unnecessary*)
by (meson CollectD SigmaD1 SigmaD2)


(*single-step*)
lemma "(a,b) ∈ (SIGMA x: A. {y. x = f y}) ==> a ∈ A & a = f b"
by (metis SigmaD1 SigmaD2 insert_def singleton_conv2 union_empty2 vimage_Collect_eq vimage_def vimage_singleton_eq)


lemma "(a,b) ∈ (SIGMA x: A. {y. x = f y}) ==> a ∈ A & a = f b"
proof (neg_clausify)
assume 0: "(a::'a::type, b::'b::type)
∈ Sigma (A::'a::type set)
   (COMBB Collect (COMBC (COMBB COMBB op =) (f::'b::type => 'a::type)))"
assume 1: "(a::'a::type) ∉ (A::'a::type set) ∨ a ≠ (f::'b::type => 'a::type) (b::'b::type)"
have 2: "(a::'a::type) ∈ (A::'a::type set)"
  by (metis 0 SigmaD1)
have 3: "(b::'b::type)
∈ COMBB Collect (COMBC (COMBB COMBB op =) (f::'b::type => 'a::type)) (a::'a::type)"
  by (metis 0 SigmaD2) 
have 4: "(b::'b::type) ∈ Collect (COMBB (op = (a::'a::type)) (f::'b::type => 'a::type))"
  by (metis 3)
have 5: "(f::'b::type => 'a::type) (b::'b::type) ≠ (a::'a::type)"
  by (metis 1 2)
have 6: "(f::'b::type => 'a::type) (b::'b::type) = (a::'a::type)"
  by (metis 4 vimage_singleton_eq insert_def singleton_conv2 union_empty2 vimage_Collect_eq vimage_def)
show "False"
  by (metis 5 6)
qed

(*Alternative structured proof, untyped*)
lemma "(a,b) ∈ (SIGMA x: A. {y. x = f y}) ==> a ∈ A & a = f b"
proof (neg_clausify)
assume 0: "(a, b) ∈ Sigma A (COMBB Collect (COMBC (COMBB COMBB op =) f))"
have 1: "b ∈ Collect (COMBB (op = a) f)"
  by (metis 0 SigmaD2)
have 2: "f b = a"
  by (metis 1 vimage_Collect_eq singleton_conv2 insert_def union_empty2 vimage_singleton_eq vimage_def)
assume 3: "a ∉ A ∨ a ≠ f b"
have 4: "a ∈ A"
  by (metis 0 SigmaD1)
have 5: "f b ≠ a"
  by (metis 4 3)
show "False"
  by (metis 5 2)
qed


ML{*ResAtp.problem_name := "Abstraction__CLF_eq_in_pp"*}
lemma "(cl,f) ∈ CLF ==> CLF = (SIGMA cl: CL.{f. f ∈ pset cl}) ==> f ∈ pset cl"
by (metis Collect_mem_eq SigmaD2)

lemma "(cl,f) ∈ CLF ==> CLF = (SIGMA cl: CL.{f. f ∈ pset cl}) ==> f ∈ pset cl"
proof (neg_clausify)
assume 0: "(cl, f) ∈ CLF"
assume 1: "CLF = Sigma CL (COMBB Collect (COMBB (COMBC op ∈) pset))"
assume 2: "f ∉ pset cl"
have 3: "!!X1 X2. X2 ∈ COMBB Collect (COMBB (COMBC op ∈) pset) X1 ∨ (X1, X2) ∉ CLF"
  by (metis SigmaD2 1)
have 4: "!!X1 X2. X2 ∈ pset X1 ∨ (X1, X2) ∉ CLF"
  by (metis 3 Collect_mem_eq)
have 5: "(cl, f) ∉ CLF"
  by (metis 2 4)
show "False"
  by (metis 5 0)
qed

ML{*ResAtp.problem_name := "Abstraction__Sigma_Collect_Pi"*}
lemma
    "(cl,f) ∈ (SIGMA cl: CL. {f. f ∈ pset cl -> pset cl}) ==> 
    f ∈ pset cl -> pset cl"
proof (neg_clausify)
assume 0: "f ∉ Pi (pset cl) (COMBK (pset cl))"
assume 1: "(cl, f)
∈ Sigma CL
   (COMBB Collect
     (COMBB (COMBC op ∈) (COMBS (COMBB Pi pset) (COMBB COMBK pset))))"
show "False"
(*  by (metis 0 Collect_mem_eq SigmaD2 1) ??doesn't terminate*)
  by (insert 0 1, simp add: COMBB_def COMBS_def COMBC_def)
qed


ML{*ResAtp.problem_name := "Abstraction__Sigma_Collect_Int"*}
lemma
    "(cl,f) ∈ (SIGMA cl: CL. {f. f ∈ pset cl ∩ cl}) ==>
   f ∈ pset cl ∩ cl"
proof (neg_clausify)
assume 0: "(cl, f)
∈ Sigma CL
   (COMBB Collect (COMBB (COMBC op ∈) (COMBS (COMBB op ∩ pset) COMBI)))"
assume 1: "f ∉ pset cl ∩ cl"
have 2: "f ∈ COMBB Collect (COMBB (COMBC op ∈) (COMBS (COMBB op ∩ pset) COMBI)) cl" 
  by (insert 0, simp add: COMBB_def) 
(*  by (metis SigmaD2 0)  ??doesn't terminate*)
have 3: "f ∈ COMBS (COMBB op ∩ pset) COMBI cl"
  by (metis 2 Collect_mem_eq)
have 4: "f ∉ cl ∩ pset cl"
  by (metis 1 Int_commute)
have 5: "f ∈ cl ∩ pset cl"
  by (metis 3 Int_commute)
show "False"
  by (metis 5 4)
qed


ML{*ResAtp.problem_name := "Abstraction__Sigma_Collect_Pi_mono"*}
lemma
    "(cl,f) ∈ (SIGMA cl: CL. {f. f ∈ pset cl -> pset cl & monotone f (pset cl) (order cl)}) ==>
   (f ∈ pset cl -> pset cl)  &  (monotone f (pset cl) (order cl))"
by auto

ML{*ResAtp.problem_name := "Abstraction__CLF_subset_Collect_Int"*}
lemma "(cl,f) ∈ CLF ==> 
   CLF ⊆ (SIGMA cl: CL. {f. f ∈ pset cl ∩ cl}) ==>
   f ∈ pset cl ∩ cl"
by auto
(*??no longer terminates, with combinators
by (metis Collect_mem_eq Int_def SigmaD2 UnCI Un_absorb1)
  --{*@{text Int_def} is redundant}
*)

ML{*ResAtp.problem_name := "Abstraction__CLF_eq_Collect_Int"*}
lemma "(cl,f) ∈ CLF ==> 
   CLF = (SIGMA cl: CL. {f. f ∈ pset cl ∩ cl}) ==>
   f ∈ pset cl ∩ cl"
by auto
(*??no longer terminates, with combinators
by (metis Collect_mem_eq Int_commute SigmaD2)
*)

ML{*ResAtp.problem_name := "Abstraction__CLF_subset_Collect_Pi"*}
lemma 
   "(cl,f) ∈ CLF ==> 
    CLF ⊆ (SIGMA cl': CL. {f. f ∈ pset cl' -> pset cl'}) ==> 
    f ∈ pset cl -> pset cl"
by auto
(*??no longer terminates, with combinators
by (metis Collect_mem_eq SigmaD2 subsetD)
*)

ML{*ResAtp.problem_name := "Abstraction__CLF_eq_Collect_Pi"*}
lemma 
  "(cl,f) ∈ CLF ==> 
   CLF = (SIGMA cl: CL. {f. f ∈ pset cl -> pset cl}) ==> 
   f ∈ pset cl -> pset cl"
by auto
(*??no longer terminates, with combinators
by (metis Collect_mem_eq SigmaD2 contra_subsetD equalityE)
*)

ML{*ResAtp.problem_name := "Abstraction__CLF_eq_Collect_Pi_mono"*}
lemma 
  "(cl,f) ∈ CLF ==> 
   CLF = (SIGMA cl: CL. {f. f ∈ pset cl -> pset cl & monotone f (pset cl) (order cl)}) ==>
   (f ∈ pset cl -> pset cl)  &  (monotone f (pset cl) (order cl))"
by auto

ML{*ResAtp.problem_name := "Abstraction__map_eq_zipA"*}
lemma "map (%x. (f x, g x)) xs = zip (map f xs) (map g xs)"
apply (induct xs)
(*sledgehammer*)  
apply auto
done

ML{*ResAtp.problem_name := "Abstraction__map_eq_zipB"*}
lemma "map (%w. (w -> w, w × w)) xs = 
       zip (map (%w. w -> w) xs) (map (%w. w × w) xs)"
apply (induct xs)
(*sledgehammer*)  
apply auto
done

ML{*ResAtp.problem_name := "Abstraction__image_evenA"*}
lemma "(%x. Suc(f x)) ` {x. even x} <= A ==> (∀x. even x --> Suc(f x) ∈ A)";
(*sledgehammer*)  
by auto

ML{*ResAtp.problem_name := "Abstraction__image_evenB"*}
lemma "(%x. f (f x)) ` ((%x. Suc(f x)) ` {x. even x}) <= A 
       ==> (∀x. even x --> f (f (Suc(f x))) ∈ A)";
(*sledgehammer*)  
by auto

ML{*ResAtp.problem_name := "Abstraction__image_curry"*}
lemma "f ∈ (%u v. b × u × v) ` A ==> ∀u v. P (b × u × v) ==> P(f y)" 
(*sledgehammer*)  
by auto

ML{*ResAtp.problem_name := "Abstraction__image_TimesA"*}
lemma image_TimesA: "(%(x,y). (f x, g y)) ` (A × B) = (f`A) × (g`B)"
(*sledgehammer*) 
apply (rule equalityI)
(***Even the two inclusions are far too difficult
ML{*ResAtp.problem_name := "Abstraction__image_TimesA_simpler"*}
***)
apply (rule subsetI)
apply (erule imageE)
(*V manages from here with help: Abstraction__image_TimesA_simpler_1_b.p*)
apply (erule ssubst)
apply (erule SigmaE)
(*V manages from here: Abstraction__image_TimesA_simpler_1_a.p*)
apply (erule ssubst)
apply (subst split_conv)
apply (rule SigmaI) 
apply (erule imageI) +
txt{*subgoal 2*}
apply (clarify );
apply (simp add: );  
apply (rule rev_image_eqI)  
apply (blast intro: elim:); 
apply (simp add: );
done

(*Given the difficulty of the previous problem, these two are probably
impossible*)

ML{*ResAtp.problem_name := "Abstraction__image_TimesB"*}
lemma image_TimesB:
    "(%(x,y,z). (f x, g y, h z)) ` (A × B × C) = (f`A) × (g`B) × (h`C)" 
(*sledgehammer*) 
by force

ML{*ResAtp.problem_name := "Abstraction__image_TimesC"*}
lemma image_TimesC:
    "(%(x,y). (x -> x, y × y)) ` (A × B) = 
     ((%x. x -> x) ` A) × ((%y. y × y) ` B)" 
(*sledgehammer*) 
by auto

end

lemma

  x < 1 ∧ op = = op = ==> op = = op = ∧ x < 2

lemma

  a ∈ {x. P x} ==> P a

lemma Collect_triv:

  a ∈ {x. P x} ==> P a

lemma

  [| a ∈ {x. P x --> Q x}; a ∈ {x. P x} |] ==> a ∈ {x. Q x}

lemma

  (a, b) ∈ Sigma A B ==> aAbB a

lemma Sigma_triv:

  (a, b) ∈ Sigma A B ==> aAbB a

lemma

  (a, b) ∈ (SIGMA x:A. {y. x = f y}) ==> aAa = f b

lemma

  (a, b) ∈ (SIGMA x:A. {y. x = f y}) ==> aAa = f b

lemma

  (a, b) ∈ (SIGMA x:A. {y. x = f y}) ==> aAa = f b

lemma

  (a, b) ∈ (SIGMA x:A. {y. x = f y}) ==> aAa = f b

lemma

  [| (cl, f) ∈ CLF; CLF = (SIGMA cl:CL. {f. f ∈ pset cl}) |] ==> f ∈ pset cl

lemma

  [| (cl, f) ∈ CLF; CLF = (SIGMA cl:CL. {f. f ∈ pset cl}) |] ==> f ∈ pset cl

lemma

  (cl, f) ∈ (SIGMA cl:CL. {f. f ∈ pset cl -> pset cl}) ==> f ∈ pset cl -> pset cl

lemma

  (cl, f) ∈ (SIGMA cl:CL. {f. f ∈ pset clcl}) ==> f ∈ pset clcl

lemma

  (cl, f)
  ∈ (SIGMA cl:CL.
        {f : pset cl -> pset cl. monotone f (pset cl) (Abstraction.order cl)})
  ==> f ∈ pset cl -> pset cl ∧ monotone f (pset cl) (Abstraction.order cl)

lemma

  [| (cl, f) ∈ CLF; CLF  (SIGMA cl:CL. {f. f ∈ pset clcl}) |]
  ==> f ∈ pset clcl

lemma

  [| (cl, f) ∈ CLF; CLF = (SIGMA cl:CL. {f. f ∈ pset clcl}) |]
  ==> f ∈ pset clcl

lemma

  [| (cl, f) ∈ CLF; CLF  (SIGMA cl':CL. {f. f ∈ pset cl' -> pset cl'}) |]
  ==> f ∈ pset cl -> pset cl

lemma

  [| (cl, f) ∈ CLF; CLF = (SIGMA cl:CL. {f. f ∈ pset cl -> pset cl}) |]
  ==> f ∈ pset cl -> pset cl

lemma

  [| (cl, f) ∈ CLF;
     CLF =
     (SIGMA cl:CL.
         {f : pset cl -> pset cl. monotone f (pset cl) (Abstraction.order cl)}) |]
  ==> f ∈ pset cl -> pset cl ∧ monotone f (pset cl) (Abstraction.order cl)

lemma

  map (λx. (f x, g x)) xs = zip (map f xs) (map g xs)

lemma

  map (λw. (w -> w, w × w)) xs = zip (map (λw. w -> w) xs) (map (λw. w × w) xs)

lemma

  x. Suc (f x)) ` {x. even x}  A ==> ∀x. even x --> Suc (f x) ∈ A

lemma

  x. f (f x)) ` (λx. Suc (f x)) ` {x. even x}  A
  ==> ∀x. even x --> f (f (Suc (f x))) ∈ A

lemma

  [| f ∈ (λu v. b × u × v) ` A; ∀u v. P (b × u × v) |] ==> P (f y)

lemma image_TimesA:

  (λ(x, y). (f x, g y)) ` (A × B) = f ` A × g ` B

lemma image_TimesB:

  (λ(x, y, z). (f x, g y, h z)) ` (A × B × C) = f ` A × g ` B × h ` C

lemma image_TimesC:

  (λ(x, y). (x -> x, y × y)) ` (A × B) = (λx. x -> x) ` A ×y. y × y) ` B