Tutorial/Tutorial4.thy
branchNominal2-Isabelle2013
changeset 3208 da575186d492
parent 3206 fb201e383f1b
child 3209 2fb0bc0dcbf1
--- a/Tutorial/Tutorial4.thy	Tue Feb 19 05:38:46 2013 +0000
+++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
@@ -1,270 +0,0 @@
-
-theory Tutorial4
-imports Tutorial1 Tutorial2
-begin
-
-section {* The CBV Reduction Relation (Small-Step Semantics) *}
-
-
-inductive
-  cbv :: "lam \<Rightarrow> lam \<Rightarrow> bool" ("_ \<longrightarrow>cbv _" [60, 60] 60) 
-where
-  cbv1: "\<lbrakk>val v; atom x \<sharp> v\<rbrakk> \<Longrightarrow> App (Lam [x].t) v \<longrightarrow>cbv t[x ::= v]"
-| cbv2: "t \<longrightarrow>cbv t' \<Longrightarrow> App t t2 \<longrightarrow>cbv App t' t2"
-| cbv3: "t \<longrightarrow>cbv t' \<Longrightarrow> App t2 t \<longrightarrow>cbv App t2 t'"
-
-inductive 
-  "cbv_star" :: "lam \<Rightarrow> lam \<Rightarrow> bool" (" _ \<longrightarrow>cbv* _" [60, 60] 60)
-where
-  cbvs1: "e \<longrightarrow>cbv* e"
-| cbvs2: "\<lbrakk>e1\<longrightarrow>cbv e2; e2 \<longrightarrow>cbv* e3\<rbrakk> \<Longrightarrow> e1 \<longrightarrow>cbv* e3"
-
-declare cbv.intros[intro] cbv_star.intros[intro]
-
-subsection {* EXERCISE 11 *}
-
-text {*
-  Show that cbv* is transitive, by filling the gaps in the 
-  proof below.
-*}
-
-lemma cbvs3 [intro]:
-  assumes a: "e1 \<longrightarrow>cbv* e2" "e2 \<longrightarrow>cbv* e3"
-  shows "e1 \<longrightarrow>cbv* e3"
-using a 
-proof (induct) 
-  case (cbvs1 e1)
-  have asm: "e1 \<longrightarrow>cbv* e3" by fact
-  show "e1 \<longrightarrow>cbv* e3" sorry
-next
-  case (cbvs2 e1 e2 e3')
-  have "e1 \<longrightarrow>cbv e2" by fact
-  have "e2 \<longrightarrow>cbv* e3'" by fact
-  have "e3' \<longrightarrow>cbv* e3 \<Longrightarrow> e2 \<longrightarrow>cbv* e3" by fact
-  have "e3' \<longrightarrow>cbv* e3" by fact
-
-  show "e1 \<longrightarrow>cbv* e3" sorry
-qed 
-
-
-text {*
-  In order to help establishing the property that the machine
-  calculates a nomrmalform that corresponds to the evaluation 
-  relation, we introduce the call-by-value small-step semantics.
-*}
-
-
-equivariance val
-equivariance cbv
-nominal_inductive cbv
-  avoids cbv1: "x"
-  unfolding fresh_star_def
-  by (simp_all add: lam.fresh Abs_fresh_iff fresh_Pair fresh_fact)
-
-text {*
-  In order to satisfy the vc-condition we have to formulate
-  this relation with the additional freshness constraint
-  atom x \<sharp> v. Although this makes the definition vc-ompatible, it
-  makes the definition less useful. We can with a little bit of 
-  pain show that the more restricted rule is equivalent to the
-  usual rule. 
-*}
-
-lemma subst_rename: 
-  assumes a: "atom y \<sharp> t"
-  shows "t[x ::= s] = ((y \<leftrightarrow> x) \<bullet> t)[y ::= s]"
-using a 
-by (nominal_induct t avoiding: x y s rule: lam.strong_induct)
-   (auto simp add: lam.fresh fresh_at_base)
-
-
-lemma better_cbv1 [intro]: 
-  assumes a: "val v" 
-  shows "App (Lam [x].t) v \<longrightarrow>cbv t[x ::= v]"
-proof -
-  obtain y::"name" where fs: "atom y \<sharp> (x, t, v)" by (rule obtain_fresh)
-  have "App (Lam [x].t) v = App (Lam [y].((y \<leftrightarrow> x) \<bullet> t)) v" using fs
-    by (auto simp add: lam.eq_iff Abs1_eq_iff' flip_def fresh_Pair fresh_at_base)
-  also have "\<dots> \<longrightarrow>cbv ((y \<leftrightarrow> x) \<bullet> t)[y ::= v]" using fs a cbv1 by auto
-  also have "\<dots> = t[x ::= v]" using fs subst_rename[symmetric] by simp
-  finally show "App (Lam [x].t) v \<longrightarrow>cbv t[x ::= v]" by simp
-qed
-
-
-
-subsection {* EXERCISE 12 *}
-
-text {*  
-  If more simple exercises are needed, then complete the following proof. 
-*}
-
-lemma cbv_in_ctx:
-  assumes a: "t \<longrightarrow>cbv t'"
-  shows "E\<lbrakk>t\<rbrakk> \<longrightarrow>cbv E\<lbrakk>t'\<rbrakk>"
-using a
-proof (induct E)
-  case Hole
-  have "t \<longrightarrow>cbv t'" by fact
-  show "\<box>\<lbrakk>t\<rbrakk> \<longrightarrow>cbv \<box>\<lbrakk>t'\<rbrakk>" sorry
-next
-  case (CAppL E s)
-  have ih: "t \<longrightarrow>cbv t' \<Longrightarrow> E\<lbrakk>t\<rbrakk> \<longrightarrow>cbv E\<lbrakk>t'\<rbrakk>" by fact
-  have "t \<longrightarrow>cbv t'" by fact
-
-  show "(CAppL E s)\<lbrakk>t\<rbrakk> \<longrightarrow>cbv (CAppL E s)\<lbrakk>t'\<rbrakk>" sorry
-next
-  case (CAppR s E)
-  have ih: "t \<longrightarrow>cbv t' \<Longrightarrow> E\<lbrakk>t\<rbrakk> \<longrightarrow>cbv E\<lbrakk>t'\<rbrakk>" by fact
-  have a: "t \<longrightarrow>cbv t'" by fact
-  
-  show "(CAppR s E)\<lbrakk>t\<rbrakk> \<longrightarrow>cbv (CAppR s E)\<lbrakk>t'\<rbrakk>" sorry
-qed
-
-section {* EXERCISE 13 *} 
- 
-text {*
-  The point of the cbv-reduction was that we can easily relatively 
-  establish the follwoing property:
-*}
-
-lemma machine_implies_cbvs_ctx:
-  assumes a: "<e, Es> \<mapsto> <e', Es'>"
-  shows "(Es\<down>)\<lbrakk>e\<rbrakk> \<longrightarrow>cbv* (Es'\<down>)\<lbrakk>e'\<rbrakk>"
-using a 
-proof (induct)
-  case (m1 t1 t2 Es)
-
-  show "Es\<down>\<lbrakk>App t1 t2\<rbrakk> \<longrightarrow>cbv* ((CAppL \<box> t2) # Es)\<down>\<lbrakk>t1\<rbrakk>" sorry
-next
-  case (m2 v t2 Es)
-  have "val v" by fact
-
-  show "((CAppL \<box> t2) # Es)\<down>\<lbrakk>v\<rbrakk> \<longrightarrow>cbv* (CAppR v \<box> # Es)\<down>\<lbrakk>t2\<rbrakk>" sorry
-next
-  case (m3 v x t Es)
-  have "val v" by fact
- 
-  show "(((CAppR (Lam [x].t) \<box>) # Es)\<down>)\<lbrakk>v\<rbrakk> \<longrightarrow>cbv* (Es\<down>)\<lbrakk>(t[x ::= v])\<rbrakk>" sorry
-qed
-
-text {* 
-  It is not difficult to extend the lemma above to
-  arbitrary reductions sequences of the machine. 
-*}
-
-lemma machines_implies_cbvs_ctx:
-  assumes a: "<e, Es> \<mapsto>* <e', Es'>"
-  shows "(Es\<down>)\<lbrakk>e\<rbrakk> \<longrightarrow>cbv* (Es'\<down>)\<lbrakk>e'\<rbrakk>"
-using a machine_implies_cbvs_ctx 
-by (induct) (blast)+
-
-text {* 
-  So whenever we let the machine start in an initial
-  state and it arrives at a final state, then there exists
-  a corresponding cbv-reduction sequence. 
-*}
-
-corollary machines_implies_cbvs:
-  assumes a: "<e, []> \<mapsto>* <e', []>"
-  shows "e \<longrightarrow>cbv* e'"
-proof - 
-  have "[]\<down>\<lbrakk>e\<rbrakk> \<longrightarrow>cbv* []\<down>\<lbrakk>e'\<rbrakk>" 
-     using a machines_implies_cbvs_ctx by blast
-  then show "e \<longrightarrow>cbv* e'" by simp  
-qed
-
-text {*
-  We now want to relate the cbv-reduction to the evaluation
-  relation. For this we need one auxiliary lemma about
-  inverting the e_App rule. 
-*}
-
-
-lemma e_App_elim:
-  assumes a: "App t1 t2 \<Down> v"
-  obtains x t v' where "t1 \<Down> Lam [x].t" "t2 \<Down> v'" "t[x::=v'] \<Down> v"
-using a by (cases) (auto simp add: lam.eq_iff lam.distinct) 
-
-
-subsection {* EXERCISE 13 *}
-
-text {*
-  Complete the first and second case in the 
-  proof below. 
-*}
-
-lemma cbv_eval:
-  assumes a: "t1 \<longrightarrow>cbv t2" "t2 \<Down> t3"
-  shows "t1 \<Down> t3"
-using a
-proof(induct arbitrary: t3)
-  case (cbv1 v x t t3)
-  have a1: "val v" by fact
-  have a2: "t[x ::= v] \<Down> t3" by fact
-
-  show "App (Lam [x].t) v \<Down> t3" sorry
-next
-  case (cbv2 t t' t2 t3)
-  have ih: "\<And>t3. t' \<Down> t3 \<Longrightarrow> t \<Down> t3" by fact
-  have "App t' t2 \<Down> t3" by fact
-  then obtain x t'' v' 
-    where a1: "t' \<Down> Lam [x].t''" 
-      and a2: "t2 \<Down> v'" 
-      and a3: "t''[x ::= v'] \<Down> t3" by (rule e_App_elim) 
-  
-  show "App t t2 \<Down> t3" sorry
-qed (auto elim!: e_App_elim)
-
-
-text {* 
-  Next we extend the lemma above to arbitray initial
-  sequences of cbv-reductions. 
-*}
-
-lemma cbvs_eval:
-  assumes a: "t1 \<longrightarrow>cbv* t2" "t2 \<Down> t3"
-  shows "t1 \<Down> t3"
-using a by (induct) (auto intro: cbv_eval)
-
-text {* 
-  Finally, we can show that if from a term t we reach a value 
-  by a cbv-reduction sequence, then t evaluates to this value. 
-
-  This proof is not by induction. So we have to set up the proof
-  with
-
-    proof -
-    
-  in order to prevent Isabelle from applying a default introduction   
-  rule.
-*}
-
-lemma cbvs_implies_eval:
-  assumes a: "t \<longrightarrow>cbv* v" 
-  and     b: "val v"
-  shows "t \<Down> v"
-proof - 
-  have "v \<Down> v" using b eval_val by simp
-  then show "t \<Down> v" using a cbvs_eval by auto
-qed
-
-section {* EXERCISE 15 *}
-
-text {* 
-  All facts tied together give us the desired property 
-  about machines: we know that a machines transitions
-  correspond to cbvs transitions, and with the lemma
-  above they correspond to an eval judgement.
-*}
-
-theorem machines_implies_eval:
-  assumes a: "<t1, []> \<mapsto>* <t2, []>" 
-  and     b: "val t2" 
-  shows "t1 \<Down> t2"
-proof - 
-  
-  show "t1 \<Down> t2" sorry
-qed
-
-end
-