--- a/src/HOL/Metis_Examples/Abstraction.thy Thu Apr 29 18:24:52 2010 +0200
+++ b/src/HOL/Metis_Examples/Abstraction.thy Thu Apr 29 19:02:04 2010 +0200
@@ -23,13 +23,11 @@
declare [[ atp_problem_prefix = "Abstraction__Collect_triv" ]]
lemma (*Collect_triv:*) "a \<in> {x. P x} ==> P a"
-proof (neg_clausify)
-assume 0: "(a\<Colon>'a\<Colon>type) \<in> Collect (P\<Colon>'a\<Colon>type \<Rightarrow> bool)"
-assume 1: "\<not> (P\<Colon>'a\<Colon>type \<Rightarrow> bool) (a\<Colon>'a\<Colon>type)"
-have 2: "(P\<Colon>'a\<Colon>type \<Rightarrow> bool) (a\<Colon>'a\<Colon>type)"
- by (metis CollectD 0)
-show "False"
- by (metis 2 1)
+proof -
+ assume "a \<in> {x. P x}"
+ hence "a \<in> P" by (metis Collect_def)
+ hence "P a" by (metis mem_def)
+ thus "P a" by metis
qed
lemma Collect_triv: "a \<in> {x. P x} ==> P a"
@@ -38,76 +36,54 @@
declare [[ atp_problem_prefix = "Abstraction__Collect_mp" ]]
lemma "a \<in> {x. P x --> Q x} ==> a \<in> {x. P x} ==> a \<in> {x. Q x}"
- by (metis CollectI Collect_imp_eq ComplD UnE mem_Collect_eq);
- --{*34 secs*}
+ by (metis Collect_imp_eq ComplD UnE)
declare [[ atp_problem_prefix = "Abstraction__Sigma_triv" ]]
lemma "(a,b) \<in> Sigma A B ==> a \<in> A & b \<in> B a"
-proof (neg_clausify)
-assume 0: "(a\<Colon>'a\<Colon>type, b\<Colon>'b\<Colon>type) \<in> Sigma (A\<Colon>'a\<Colon>type set) (B\<Colon>'a\<Colon>type \<Rightarrow> 'b\<Colon>type set)"
-assume 1: "(a\<Colon>'a\<Colon>type) \<notin> (A\<Colon>'a\<Colon>type set) \<or> (b\<Colon>'b\<Colon>type) \<notin> (B\<Colon>'a\<Colon>type \<Rightarrow> 'b\<Colon>type set) a"
-have 2: "(a\<Colon>'a\<Colon>type) \<in> (A\<Colon>'a\<Colon>type set)"
- by (metis SigmaD1 0)
-have 3: "(b\<Colon>'b\<Colon>type) \<in> (B\<Colon>'a\<Colon>type \<Rightarrow> 'b\<Colon>type set) (a\<Colon>'a\<Colon>type)"
- by (metis SigmaD2 0)
-have 4: "(b\<Colon>'b\<Colon>type) \<notin> (B\<Colon>'a\<Colon>type \<Rightarrow> 'b\<Colon>type set) (a\<Colon>'a\<Colon>type)"
- by (metis 1 2)
-show "False"
- by (metis 3 4)
+proof -
+ assume A1: "(a, b) \<in> Sigma A B"
+ hence F1: "b \<in> B a" by (metis mem_Sigma_iff)
+ have F2: "a \<in> A" by (metis A1 mem_Sigma_iff)
+ have "b \<in> B a" by (metis F1)
+ thus "a \<in> A \<and> b \<in> B a" by (metis F2)
qed
lemma Sigma_triv: "(a,b) \<in> Sigma A B ==> a \<in> A & b \<in> B a"
by (metis SigmaD1 SigmaD2)
declare [[ atp_problem_prefix = "Abstraction__Sigma_Collect" ]]
-lemma "(a,b) \<in> (SIGMA x: A. {y. x = f y}) ==> a \<in> A & a = f b"
-(*???metis says this is satisfiable!
+lemma "(a, b) \<in> (SIGMA x:A. {y. x = f y}) \<Longrightarrow> a \<in> A \<and> a = f b"
+(* Metis says this is satisfiable!
by (metis CollectD SigmaD1 SigmaD2)
*)
by (meson CollectD SigmaD1 SigmaD2)
-(*single-step*)
-lemma "(a,b) \<in> (SIGMA x: A. {y. x = f y}) ==> a \<in> A & a = f b"
-by (metis SigmaD1 SigmaD2 insert_def singleton_conv2 Un_empty_right vimage_Collect_eq vimage_def vimage_singleton_eq)
-
+lemma "(a, b) \<in> (SIGMA x:A. {y. x = f y}) \<Longrightarrow> a \<in> A \<and> a = f b"
+by (metis mem_Sigma_iff singleton_conv2 vimage_Collect_eq vimage_singleton_eq)
-lemma "(a,b) \<in> (SIGMA x: A. {y. x = f y}) ==> a \<in> A & a = f b"
-proof (neg_clausify)
-assume 0: "(a\<Colon>'a\<Colon>type, b\<Colon>'b\<Colon>type)
-\<in> Sigma (A\<Colon>'a\<Colon>type set)
- (COMBB Collect (COMBC (COMBB COMBB op =) (f\<Colon>'b\<Colon>type \<Rightarrow> 'a\<Colon>type)))"
-assume 1: "(a\<Colon>'a\<Colon>type) \<notin> (A\<Colon>'a\<Colon>type set) \<or> a \<noteq> (f\<Colon>'b\<Colon>type \<Rightarrow> 'a\<Colon>type) (b\<Colon>'b\<Colon>type)"
-have 2: "(a\<Colon>'a\<Colon>type) \<in> (A\<Colon>'a\<Colon>type set)"
- by (metis 0 SigmaD1)
-have 3: "(b\<Colon>'b\<Colon>type)
-\<in> COMBB Collect (COMBC (COMBB COMBB op =) (f\<Colon>'b\<Colon>type \<Rightarrow> 'a\<Colon>type)) (a\<Colon>'a\<Colon>type)"
- by (metis 0 SigmaD2)
-have 4: "(b\<Colon>'b\<Colon>type) \<in> Collect (COMBB (op = (a\<Colon>'a\<Colon>type)) (f\<Colon>'b\<Colon>type \<Rightarrow> 'a\<Colon>type))"
- by (metis 3)
-have 5: "(f\<Colon>'b\<Colon>type \<Rightarrow> 'a\<Colon>type) (b\<Colon>'b\<Colon>type) \<noteq> (a\<Colon>'a\<Colon>type)"
- by (metis 1 2)
-have 6: "(f\<Colon>'b\<Colon>type \<Rightarrow> 'a\<Colon>type) (b\<Colon>'b\<Colon>type) = (a\<Colon>'a\<Colon>type)"
- by (metis 4 vimage_singleton_eq insert_def singleton_conv2 Un_empty_right vimage_Collect_eq vimage_def)
-show "False"
- by (metis 5 6)
+lemma "(a, b) \<in> (SIGMA x:A. {y. x = f y}) \<Longrightarrow> a \<in> A \<and> a = f b"
+proof -
+ assume A1: "(a, b) \<in> (SIGMA x:A. {y. x = f y})"
+ have F1: "\<forall>u. {u} = op = u" by (metis singleton_conv2 Collect_def)
+ have F2: "\<forall>x w. (\<lambda>R. w (x R)) = x -` w" by (metis vimage_Collect_eq Collect_def)
+ have F3: "\<forall>v w y. v \<in> w -` op = y \<longrightarrow> w v = y" by (metis F1 vimage_singleton_eq)
+ have F4: "b \<in> {R. a = f R}" by (metis A1 mem_Sigma_iff)
+ have F5: "a \<in> A" by (metis A1 mem_Sigma_iff)
+ have "b \<in> f -` op = a" by (metis F2 F4 Collect_def)
+ hence "f b = a" by (metis F3)
+ thus "a \<in> A \<and> a = f b" by (metis F5)
qed
-(*Alternative structured proof, untyped*)
-lemma "(a,b) \<in> (SIGMA x: A. {y. x = f y}) ==> a \<in> A & a = f b"
-proof (neg_clausify)
-assume 0: "(a, b) \<in> Sigma A (COMBB Collect (COMBC (COMBB COMBB op =) f))"
-have 1: "b \<in> Collect (COMBB (op = a) f)"
- by (metis 0 SigmaD2)
-have 2: "f b = a"
- by (metis 1 vimage_Collect_eq singleton_conv2 insert_def Un_empty_right vimage_singleton_eq vimage_def)
-assume 3: "a \<notin> A \<or> a \<noteq> f b"
-have 4: "a \<in> A"
- by (metis 0 SigmaD1)
-have 5: "f b \<noteq> a"
- by (metis 4 3)
-show "False"
- by (metis 5 2)
+(* Alternative structured proof *)
+lemma "(a, b) \<in> (SIGMA x:A. {y. x = f y}) \<Longrightarrow> a \<in> A \<and> a = f b"
+proof -
+ assume A1: "(a, b) \<in> (SIGMA x:A. {y. x = f y})"
+ hence F1: "a \<in> A" by (metis mem_Sigma_iff)
+ have "b \<in> {R. a = f R}" by (metis A1 mem_Sigma_iff)
+ hence F2: "b \<in> (\<lambda>R. a = f R)" by (metis Collect_def)
+ hence "a = f b" by (unfold mem_def)
+ thus "a \<in> A \<and> a = f b" by (metis F1)
qed
@@ -116,56 +92,40 @@
by (metis Collect_mem_eq SigmaD2)
lemma "(cl,f) \<in> CLF ==> CLF = (SIGMA cl: CL.{f. f \<in> pset cl}) ==> f \<in> pset cl"
-proof (neg_clausify)
-assume 0: "(cl, f) \<in> CLF"
-assume 1: "CLF = Sigma CL (COMBB Collect (COMBB (COMBC op \<in>) pset))"
-assume 2: "f \<notin> pset cl"
-have 3: "\<And>X1 X2. X2 \<in> COMBB Collect (COMBB (COMBC op \<in>) pset) X1 \<or> (X1, X2) \<notin> CLF"
- by (metis SigmaD2 1)
-have 4: "\<And>X1 X2. X2 \<in> pset X1 \<or> (X1, X2) \<notin> CLF"
- by (metis 3 Collect_mem_eq)
-have 5: "(cl, f) \<notin> CLF"
- by (metis 2 4)
-show "False"
- by (metis 5 0)
+proof -
+ assume A1: "(cl, f) \<in> CLF"
+ assume A2: "CLF = (SIGMA cl:CL. {f. f \<in> pset cl})"
+ have F1: "\<forall>v. (\<lambda>R. R \<in> v) = v" by (metis Collect_mem_eq Collect_def)
+ have "\<forall>v u. (u, v) \<in> CLF \<longrightarrow> v \<in> {R. R \<in> pset u}" by (metis A2 mem_Sigma_iff)
+ hence "\<forall>v u. (u, v) \<in> CLF \<longrightarrow> v \<in> pset u" by (metis F1 Collect_def)
+ hence "f \<in> pset cl" by (metis A1)
+ thus "f \<in> pset cl" by metis
qed
declare [[ atp_problem_prefix = "Abstraction__Sigma_Collect_Pi" ]]
lemma
"(cl,f) \<in> (SIGMA cl: CL. {f. f \<in> pset cl \<rightarrow> pset cl}) ==>
f \<in> pset cl \<rightarrow> pset cl"
-proof (neg_clausify)
-assume 0: "f \<notin> Pi (pset cl) (COMBK (pset cl))"
-assume 1: "(cl, f)
-\<in> Sigma CL
- (COMBB Collect
- (COMBB (COMBC op \<in>) (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)
+proof -
+ assume A1: "(cl, f) \<in> (SIGMA cl:CL. {f. f \<in> pset cl \<rightarrow> pset cl})"
+ have F1: "\<forall>v. (\<lambda>R. R \<in> v) = v" by (metis Collect_mem_eq Collect_def)
+ have "f \<in> {R. R \<in> pset cl \<rightarrow> pset cl}" using A1 by simp
+ hence "f \<in> pset cl \<rightarrow> pset cl" by (metis F1 Collect_def)
+ thus "f \<in> pset cl \<rightarrow> pset cl" by metis
qed
-
declare [[ atp_problem_prefix = "Abstraction__Sigma_Collect_Int" ]]
lemma
"(cl,f) \<in> (SIGMA cl: CL. {f. f \<in> pset cl \<inter> cl}) ==>
f \<in> pset cl \<inter> cl"
-proof (neg_clausify)
-assume 0: "(cl, f)
-\<in> Sigma CL
- (COMBB Collect (COMBB (COMBC op \<in>) (COMBS (COMBB op \<inter> pset) COMBI)))"
-assume 1: "f \<notin> pset cl \<inter> cl"
-have 2: "f \<in> COMBB Collect (COMBB (COMBC op \<in>) (COMBS (COMBB op \<inter> pset) COMBI)) cl"
- by (insert 0, simp add: COMBB_def)
-(* by (metis SigmaD2 0) ??doesn't terminate*)
-have 3: "f \<in> COMBS (COMBB op \<inter> pset) COMBI cl"
- by (metis 2 Collect_mem_eq)
-have 4: "f \<notin> cl \<inter> pset cl"
- by (metis 1 Int_commute)
-have 5: "f \<in> cl \<inter> pset cl"
- by (metis 3 Int_commute)
-show "False"
- by (metis 5 4)
+proof -
+ assume A1: "(cl, f) \<in> (SIGMA cl:CL. {f. f \<in> pset cl \<inter> cl})"
+ have F1: "\<forall>v. (\<lambda>R. R \<in> v) = v" by (metis Collect_mem_eq Collect_def)
+ have "f \<in> {R. R \<in> pset cl \<inter> cl}" using A1 by simp
+ hence "f \<in> Id_on cl `` pset cl" by (metis F1 Int_commute Image_Id_on Collect_def)
+ hence "f \<in> Id_on cl `` pset cl" by metis
+ hence "f \<in> cl \<inter> pset cl" by (metis Image_Id_on)
+ thus "f \<in> pset cl \<inter> cl" by (metis Int_commute)
qed
@@ -181,19 +141,13 @@
f \<in> pset cl \<inter> cl"
by auto
-(*??no longer terminates, with combinators
-by (metis Collect_mem_eq Int_def SigmaD2 UnCI Un_absorb1)
- --{*@{text Int_def} is redundant*}
-*)
declare [[ atp_problem_prefix = "Abstraction__CLF_eq_Collect_Int" ]]
lemma "(cl,f) \<in> CLF ==>
CLF = (SIGMA cl: CL. {f. f \<in> pset cl \<inter> cl}) ==>
f \<in> pset cl \<inter> cl"
by auto
-(*??no longer terminates, with combinators
-by (metis Collect_mem_eq Int_commute SigmaD2)
-*)
+
declare [[ atp_problem_prefix = "Abstraction__CLF_subset_Collect_Pi" ]]
lemma
@@ -201,9 +155,7 @@
CLF \<subseteq> (SIGMA cl': CL. {f. f \<in> pset cl' \<rightarrow> pset cl'}) ==>
f \<in> pset cl \<rightarrow> pset cl"
by fast
-(*??no longer terminates, with combinators
-by (metis Collect_mem_eq SigmaD2 subsetD)
-*)
+
declare [[ atp_problem_prefix = "Abstraction__CLF_eq_Collect_Pi" ]]
lemma
@@ -211,9 +163,7 @@
CLF = (SIGMA cl: CL. {f. f \<in> pset cl \<rightarrow> pset cl}) ==>
f \<in> pset cl \<rightarrow> pset cl"
by auto
-(*??no longer terminates, with combinators
-by (metis Collect_mem_eq SigmaD2 contra_subsetD equalityE)
-*)
+
declare [[ atp_problem_prefix = "Abstraction__CLF_eq_Collect_Pi_mono" ]]
lemma
@@ -225,37 +175,33 @@
declare [[ atp_problem_prefix = "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
+ apply (metis map_is_Nil_conv zip.simps(1))
+by auto
declare [[ atp_problem_prefix = "Abstraction__map_eq_zipB" ]]
lemma "map (%w. (w -> w, w \<times> w)) xs =
zip (map (%w. w -> w) xs) (map (%w. w \<times> w) xs)"
apply (induct xs)
-(*sledgehammer*)
-apply auto
-done
+ apply (metis Nil_is_map_conv zip_Nil)
+by auto
declare [[ atp_problem_prefix = "Abstraction__image_evenA" ]]
-lemma "(%x. Suc(f x)) ` {x. even x} <= A ==> (\<forall>x. even x --> Suc(f x) \<in> A)";
-(*sledgehammer*)
-by auto
+lemma "(%x. Suc(f x)) ` {x. even x} <= A ==> (\<forall>x. even x --> Suc(f x) \<in> A)"
+by (metis Collect_def image_subset_iff mem_def)
declare [[ atp_problem_prefix = "Abstraction__image_evenB" ]]
lemma "(%x. f (f x)) ` ((%x. Suc(f x)) ` {x. even x}) <= A
==> (\<forall>x. even x --> f (f (Suc(f x))) \<in> A)";
-(*sledgehammer*)
-by auto
+by (metis Collect_def imageI image_image image_subset_iff mem_def)
declare [[ atp_problem_prefix = "Abstraction__image_curry" ]]
lemma "f \<in> (%u v. b \<times> u \<times> v) ` A ==> \<forall>u v. P (b \<times> u \<times> v) ==> P(f y)"
-(*sledgehammer*)
+(*sledgehammer*)
by auto
declare [[ atp_problem_prefix = "Abstraction__image_TimesA" ]]
lemma image_TimesA: "(%(x,y). (f x, g y)) ` (A \<times> B) = (f`A) \<times> (g`B)"
-(*sledgehammer*)
+(*sledgehammer*)
apply (rule equalityI)
(***Even the two inclusions are far too difficult
using [[ atp_problem_prefix = "Abstraction__image_TimesA_simpler"]]
@@ -283,15 +229,15 @@
declare [[ atp_problem_prefix = "Abstraction__image_TimesB" ]]
lemma image_TimesB:
- "(%(x,y,z). (f x, g y, h z)) ` (A \<times> B \<times> C) = (f`A) \<times> (g`B) \<times> (h`C)"
-(*sledgehammer*)
+ "(%(x,y,z). (f x, g y, h z)) ` (A \<times> B \<times> C) = (f`A) \<times> (g`B) \<times> (h`C)"
+(*sledgehammer*)
by force
declare [[ atp_problem_prefix = "Abstraction__image_TimesC" ]]
lemma image_TimesC:
"(%(x,y). (x \<rightarrow> x, y \<times> y)) ` (A \<times> B) =
((%x. x \<rightarrow> x) ` A) \<times> ((%y. y \<times> y) ` B)"
-(*sledgehammer*)
+(*sledgehammer*)
by auto
end
--- a/src/HOL/Metis_Examples/set.thy Thu Apr 29 18:24:52 2010 +0200
+++ b/src/HOL/Metis_Examples/set.thy Thu Apr 29 19:02:04 2010 +0200
@@ -8,24 +8,21 @@
imports Main
begin
+sledgehammer_params [isar_proof, debug, overlord]
+
lemma "EX x X. ALL y. EX z Z. (~P(y,y) | P(x,x) | ~S(z,x)) &
(S(x,y) | ~S(y,z) | Q(Z,Z)) &
(Q(X,y) | ~Q(y,Z) | S(X,X))"
by metis
-(*??But metis can't prove the single-step version...*)
-
-
lemma "P(n::nat) ==> ~P(0) ==> n ~= 0"
by metis
sledgehammer_params [shrink_factor = 1]
-
(*multiple versions of this example*)
lemma (*equal_union: *)
- "(X = Y \<union> Z) =
- (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))"
+ "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))"
proof (neg_clausify)
fix x
assume 0: "Y \<subseteq> X \<or> X = Y \<union> Z"
@@ -269,15 +266,14 @@
"P (f b) \<Longrightarrow> \<exists>s A. (\<forall>x \<in> A. P x) \<and> f s \<in> A"
"P (f b) \<Longrightarrow> \<exists>s A. (\<forall>x \<in> A. P x) \<and> f s \<in> A"
"\<exists>A. a \<notin> A"
- "(\<forall>C. (0, 0) \<in> C \<and> (\<forall>x y. (x, y) \<in> C \<longrightarrow> (Suc x, Suc y) \<in> C) \<longrightarrow> (n, m) \<in> C) \<and> Q n \<longrightarrow> Q m"
-apply (metis atMost_iff)
-apply (metis emptyE)
-apply (metis insert_iff singletonE)
+ "(\<forall>C. (0, 0) \<in> C \<and> (\<forall>x y. (x, y) \<in> C \<longrightarrow> (Suc x, Suc y) \<in> C) \<longrightarrow> (n, m) \<in> C) \<and> Q n \<longrightarrow> Q m"
+apply (metis all_not_in_conv)+
+apply (metis mem_def)
apply (metis insertCI singletonE zless_le)
apply (metis Collect_def Collect_mem_eq)
apply (metis Collect_def Collect_mem_eq)
apply (metis DiffE)
-apply (metis pair_in_Id_conv)
+apply (metis pair_in_Id_conv)
done
end