diff options
Diffstat (limited to '')
| -rw-r--r-- | ex/Methods.thy | 12 | ||||
| -rw-r--r-- | ex/Synthesis.thy | 20 |
2 files changed, 16 insertions, 16 deletions
diff --git a/ex/Methods.thy b/ex/Methods.thy index 415fbc3..c78af14 100644 --- a/ex/Methods.thy +++ b/ex/Methods.thy @@ -14,7 +14,7 @@ text "Wellformedness results, metatheorems written into the object theory using lemma assumes "A : U(i)" "B: A \<longrightarrow> U(i)" "\<And>x. x : A \<Longrightarrow> C x: B x \<longrightarrow> U(i)" shows "\<Sum>x:A. \<Prod>y:B x. \<Sum>z:C x y. \<Prod>w:A. x =\<^sub>A w : U(i)" -by (simple lems: assms) +by (routine lems: assms) lemma @@ -38,7 +38,7 @@ text "Typechecking and constructing inhabitants:" \<comment> \<open>Correctly determines the type of the pair\<close> schematic_goal "\<lbrakk>A: U(i); B: U(i); a : A; b : B\<rbrakk> \<Longrightarrow> <a, b> : ?A" -by simple +by routine \<comment> \<open>Finds pair (too easy).\<close> schematic_goal "\<lbrakk>A: U(i); B: U(i); a : A; b : B\<rbrakk> \<Longrightarrow> ?x : A \<times> B" @@ -56,19 +56,19 @@ lemma assumes "A: U(i)" "a: A" shows "(\<^bold>\<lambda>x. <x,0>)`a \<equiv> <a,0>" proof compute - show "\<And>x. x: A \<Longrightarrow> <x,0>: A \<times> \<nat>" by simple -qed (simple lems: assms) + show "\<And>x. x: A \<Longrightarrow> <x,0>: A \<times> \<nat>" by routine +qed (routine lems: assms) lemma assumes "A: U(i)" "B: A \<longrightarrow> U(i)" "a: A" "b: B(a)" shows "(\<^bold>\<lambda>x y. <x,y>)`a`b \<equiv> <a,b>" proof compute - show "\<And>x. x: A \<Longrightarrow> \<^bold>\<lambda>y. <x,y>: \<Prod>y:B(x). \<Sum>x:A. B(x)" by (simple lems: assms) + show "\<And>x. x: A \<Longrightarrow> \<^bold>\<lambda>y. <x,y>: \<Prod>y:B(x). \<Sum>x:A. B(x)" by (routine lems: assms) show "(\<^bold>\<lambda>b. <a,b>)`b \<equiv> <a, b>" proof compute - show "\<And>b. b: B(a) \<Longrightarrow> <a, b>: \<Sum>x:A. B(x)" by (simple lems: assms) + show "\<And>b. b: B(a) \<Longrightarrow> <a, b>: \<Sum>x:A. B(x)" by (routine lems: assms) qed fact qed fact diff --git a/ex/Synthesis.thy b/ex/Synthesis.thy index cff9374..a5e77ec 100644 --- a/ex/Synthesis.thy +++ b/ex/Synthesis.thy @@ -21,7 +21,7 @@ text " text "First we show that the property we want is well-defined." lemma pred_welltyped: "\<Sum>pred:\<nat>\<rightarrow>\<nat> . ((pred`0) =\<^sub>\<nat> 0) \<times> (\<Prod>n:\<nat>. (pred`(succ n)) =\<^sub>\<nat> n): U(O)" -by simple +by routine text " Now we look for an inhabitant of this type. @@ -33,7 +33,7 @@ schematic_goal "?p`0 \<equiv> 0" and "\<And>n. n: \<nat> \<Longrightarrow> (?p`( apply compute prefer 4 apply compute prefer 3 apply compute -apply (rule Nat_routine Nat_elim | assumption)+ +apply (rule Nat_routine Nat_elim | compute | assumption)+ done text " @@ -43,36 +43,36 @@ text " definition pred :: Term where "pred \<equiv> \<^bold>\<lambda>n. ind\<^sub>\<nat> (\<lambda>a b. a) 0 n" -lemma pred_type: "pred: \<nat> \<rightarrow> \<nat>" unfolding pred_def by simple +lemma pred_type: "pred: \<nat> \<rightarrow> \<nat>" unfolding pred_def by routine lemma pred_props: "<refl(0), \<^bold>\<lambda>n. refl(n)>: ((pred`0) =\<^sub>\<nat> 0) \<times> (\<Prod>n:\<nat>. (pred`(succ n)) =\<^sub>\<nat> n)" -proof (simple lems: pred_type) +proof (routine lems: pred_type) have *: "pred`0 \<equiv> 0" unfolding pred_def proof compute - show "\<And>n. n: \<nat> \<Longrightarrow> ind\<^sub>\<nat> (\<lambda>a b. a) 0 n: \<nat>" by simple + show "\<And>n. n: \<nat> \<Longrightarrow> ind\<^sub>\<nat> (\<lambda>a b. a) 0 n: \<nat>" by routine show "ind\<^sub>\<nat> (\<lambda>a b. a) 0 0 \<equiv> 0" proof compute show "\<nat>: U(O)" .. - qed simple + qed routine qed rule - then show "refl(0): (pred`0) =\<^sub>\<nat> 0" by (subst *) simple + then show "refl(0): (pred`0) =\<^sub>\<nat> 0" by (subst *) routine show "\<^bold>\<lambda>n. refl(n): \<Prod>n:\<nat>. (pred`(succ(n))) =\<^sub>\<nat> n" unfolding pred_def proof show "\<And>n. n: \<nat> \<Longrightarrow> refl(n): ((\<^bold>\<lambda>n. ind\<^sub>\<nat> (\<lambda>a b. a) 0 n)`succ(n)) =\<^sub>\<nat> n" proof compute - show "\<And>n. n: \<nat> \<Longrightarrow> ind\<^sub>\<nat> (\<lambda>a b. a) 0 n: \<nat>" by simple + show "\<And>n. n: \<nat> \<Longrightarrow> ind\<^sub>\<nat> (\<lambda>a b. a) 0 n: \<nat>" by routine show "\<And>n. n: \<nat> \<Longrightarrow> refl(n): ind\<^sub>\<nat> (\<lambda>a b. a) 0 (succ n) =\<^sub>\<nat> n" proof compute show "\<nat>: U(O)" .. - qed simple + qed routine qed rule qed rule qed theorem "<pred, <refl(0), \<^bold>\<lambda>n. refl(n)>>: \<Sum>pred:\<nat>\<rightarrow>\<nat> . ((pred`0) =\<^sub>\<nat> 0) \<times> (\<Prod>n:\<nat>. (pred`(succ n)) =\<^sub>\<nat> n)" -by (simple lems: pred_welltyped pred_type pred_props) +by (routine lems: pred_welltyped pred_type pred_props) end |