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theory Tutorial4
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imports Tutorial1
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begin
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section {* The CBV Reduction Relation (Small-Step Semantics) *}
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text {*
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In order to help establishing the property that the CK Machine
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calculates a nomrmalform that corresponds to the evaluation
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relation, we introduce the call-by-value small-step semantics.
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*}
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inductive
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cbv :: "lam \<Rightarrow> lam \<Rightarrow> bool" ("_ \<longrightarrow>cbv _" [60, 60] 60)
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where
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cbv1: "\<lbrakk>val v; atom x \<sharp> v\<rbrakk> \<Longrightarrow> App (Lam [x].t) v \<longrightarrow>cbv t[x ::= v]"
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| cbv2[intro]: "t \<longrightarrow>cbv t' \<Longrightarrow> App t t2 \<longrightarrow>cbv App t' t2"
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| cbv3[intro]: "t \<longrightarrow>cbv t' \<Longrightarrow> App t2 t \<longrightarrow>cbv App t2 t'"
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equivariance val
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equivariance cbv
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nominal_inductive cbv
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avoids cbv1: "x"
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unfolding fresh_star_def
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by (simp_all add: lam.fresh Abs_fresh_iff fresh_Pair fresh_fact)
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text {*
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In order to satisfy the vc-condition we have to formulate
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this relation with the additional freshness constraint
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atom x \<sharp> v. Although this makes the definition vc-ompatible, it
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makes the definition less useful. We can with a little bit of
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pain show that the more restricted rule is equivalent to the
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usual rule.
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*}
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lemma subst_rename:
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assumes a: "atom y \<sharp> t"
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shows "t[x ::= s] = ((y \<leftrightarrow> x) \<bullet> t)[y ::= s]"
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using a
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by (nominal_induct t avoiding: x y s rule: lam.strong_induct)
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(auto simp add: lam.fresh fresh_at_base)
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lemma better_cbv1 [intro]:
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assumes a: "val v"
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shows "App (Lam [x].t) v \<longrightarrow>cbv t[x::=v]"
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proof -
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obtain y::"name" where fs: "atom y \<sharp> (x, t, v)" by (rule obtain_fresh)
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have "App (Lam [x].t) v = App (Lam [y].((y \<leftrightarrow> x) \<bullet> t)) v" using fs
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by (auto simp add: lam.eq_iff Abs1_eq_iff' flip_def fresh_Pair fresh_at_base)
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also have "\<dots> \<longrightarrow>cbv ((y \<leftrightarrow> x) \<bullet> t)[y ::= v]" using fs a cbv1 by auto
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also have "\<dots> = t[x ::= v]" using fs subst_rename[symmetric] by simp
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finally show "App (Lam [x].t) v \<longrightarrow>cbv t[x ::= v]" by simp
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qed
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text {*
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The transitive closure of the cbv-reduction relation:
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*}
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inductive
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"cbvs" :: "lam \<Rightarrow> lam \<Rightarrow> bool" (" _ \<longrightarrow>cbv* _" [60, 60] 60)
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where
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cbvs1[intro]: "e \<longrightarrow>cbv* e"
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| cbvs2[intro]: "\<lbrakk>e1\<longrightarrow>cbv e2; e2 \<longrightarrow>cbv* e3\<rbrakk> \<Longrightarrow> e1 \<longrightarrow>cbv* e3"
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lemma cbvs3 [intro]:
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assumes a: "e1 \<longrightarrow>cbv* e2" "e2 \<longrightarrow>cbv* e3"
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shows "e1 \<longrightarrow>cbv* e3"
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using a by (induct) (auto)
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subsection {* EXERCISE 8 *}
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text {*
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If more simple exercises are needed, then complete the following proof.
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*}
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lemma cbv_in_ctx:
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assumes a: "t \<longrightarrow>cbv t'"
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shows "E\<lbrakk>t\<rbrakk> \<longrightarrow>cbv E\<lbrakk>t'\<rbrakk>"
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using a
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proof (induct E)
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case Hole
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have "t \<longrightarrow>cbv t'" by fact
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then show "\<box>\<lbrakk>t\<rbrakk> \<longrightarrow>cbv \<box>\<lbrakk>t'\<rbrakk>" by simp
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next
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case (CAppL E s)
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have ih: "t \<longrightarrow>cbv t' \<Longrightarrow> E\<lbrakk>t\<rbrakk> \<longrightarrow>cbv E\<lbrakk>t'\<rbrakk>" by fact
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moreover
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have "t \<longrightarrow>cbv t'" by fact
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ultimately
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have "E\<lbrakk>t\<rbrakk> \<longrightarrow>cbv E\<lbrakk>t'\<rbrakk>" by simp
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then show "(CAppL E s)\<lbrakk>t\<rbrakk> \<longrightarrow>cbv (CAppL E s)\<lbrakk>t'\<rbrakk>" by auto
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next
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case (CAppR s E)
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have ih: "t \<longrightarrow>cbv t' \<Longrightarrow> E\<lbrakk>t\<rbrakk> \<longrightarrow>cbv E\<lbrakk>t'\<rbrakk>" by fact
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moreover
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have a: "t \<longrightarrow>cbv t'" by fact
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ultimately
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have "E\<lbrakk>t\<rbrakk> \<longrightarrow>cbv E\<lbrakk>t'\<rbrakk>" by simp
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then show "(CAppR s E)\<lbrakk>t\<rbrakk> \<longrightarrow>cbv (CAppR s E)\<lbrakk>t'\<rbrakk>" by auto
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qed
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section {* EXERCISE 9 *}
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text {*
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The point of the cbv-reduction was that we can easily relatively
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establish the follwoing property:
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*}
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lemma machine_implies_cbvs_ctx:
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assumes a: "<e, Es> \<mapsto> <e', Es'>"
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shows "(Es\<down>)\<lbrakk>e\<rbrakk> \<longrightarrow>cbv* (Es'\<down>)\<lbrakk>e'\<rbrakk>"
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using a
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proof (induct)
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case (m1 t1 t2 Es)
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thm machine.intros thm cbv2
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have "Es\<down>\<lbrakk>App t1 t2\<rbrakk> = (Es\<down> \<odot> CAppL \<box> t2)\<lbrakk>t1\<rbrakk>" using ctx_compose ctx_composes.simps filling.simps by simp
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then show "Es\<down>\<lbrakk>App t1 t2\<rbrakk> \<longrightarrow>cbv* ((CAppL \<box> t2) # Es)\<down>\<lbrakk>t1\<rbrakk>" using cbvs.intros by simp
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next
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case (m2 v t2 Es)
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have "val v" by fact
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have "((CAppL \<box> t2) # Es)\<down>\<lbrakk>v\<rbrakk> = (CAppR v \<box> # Es)\<down>\<lbrakk>t2\<rbrakk>" using ctx_compose ctx_composes.simps filling.simps by simp
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then show "((CAppL \<box> t2) # Es)\<down>\<lbrakk>v\<rbrakk> \<longrightarrow>cbv* (CAppR v \<box> # Es)\<down>\<lbrakk>t2\<rbrakk>" using cbvs.intros by simp
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next
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case (m3 v x t Es)
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have aa: "val v" by fact
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have "(((CAppR (Lam [x].t) \<box>) # Es)\<down>)\<lbrakk>v\<rbrakk> = Es\<down>\<lbrakk>App (Lam [x]. t) v\<rbrakk>" using ctx_compose ctx_composes.simps filling.simps by simp
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then have "(((CAppR (Lam [x].t) \<box>) # Es)\<down>)\<lbrakk>v\<rbrakk> \<longrightarrow>cbv (Es\<down>)\<lbrakk>(t[x ::= v])\<rbrakk>" using better_cbv1[OF aa] cbv_in_ctx by simp
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then show "(((CAppR (Lam [x].t) \<box>) # Es)\<down>)\<lbrakk>v\<rbrakk> \<longrightarrow>cbv* (Es\<down>)\<lbrakk>(t[x ::= v])\<rbrakk>" using cbvs.intros by blast
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qed
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text {*
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It is not difficult to extend the lemma above to
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arbitrary reductions sequences of the CK machine. *}
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lemma machines_implies_cbvs_ctx:
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assumes a: "<e, Es> \<mapsto>* <e', Es'>"
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shows "(Es\<down>)\<lbrakk>e\<rbrakk> \<longrightarrow>cbv* (Es'\<down>)\<lbrakk>e'\<rbrakk>"
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using a machine_implies_cbvs_ctx
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by (induct) (blast)+
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text {*
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So whenever we let the CL machine start in an initial
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state and it arrives at a final state, then there exists
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a corresponding cbv-reduction sequence.
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*}
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corollary machines_implies_cbvs:
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assumes a: "<e, []> \<mapsto>* <e', []>"
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shows "e \<longrightarrow>cbv* e'"
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proof -
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have "[]\<down>\<lbrakk>e\<rbrakk> \<longrightarrow>cbv* []\<down>\<lbrakk>e'\<rbrakk>"
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using a machines_implies_cbvs_ctx by blast
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then show "e \<longrightarrow>cbv* e'" by simp
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qed
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text {*
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We now want to relate the cbv-reduction to the evaluation
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relation. For this we need two auxiliary lemmas.
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*}
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lemma eval_val:
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assumes a: "val t"
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shows "t \<Down> t"
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using a by (induct) (auto)
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lemma e_App_elim:
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assumes a: "App t1 t2 \<Down> v"
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obtains x t v' where "t1 \<Down> Lam [x].t" "t2 \<Down> v'" "t[x::=v'] \<Down> v"
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using a by (cases) (auto simp add: lam.eq_iff lam.distinct)
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subsection {* EXERCISE *}
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text {*
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Complete the first and second case in the
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proof below.
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*}
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lemma cbv_eval:
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assumes a: "t1 \<longrightarrow>cbv t2" "t2 \<Down> t3"
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shows "t1 \<Down> t3"
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using a
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proof(induct arbitrary: t3)
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case (cbv1 v x t t3)
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have a1: "val v" by fact
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have a2: "t[x ::= v] \<Down> t3" by fact
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have a3: "Lam [x].t \<Down> Lam [x].t" by auto
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have a4: "v \<Down> v" using a1 eval_val by auto
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show "App (Lam [x].t) v \<Down> t3" using a3 a4 a2 by auto
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next
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case (cbv2 t t' t2 t3)
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have ih: "\<And>t3. t' \<Down> t3 \<Longrightarrow> t \<Down> t3" by fact
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have "App t' t2 \<Down> t3" by fact
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then obtain x t'' v'
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where a1: "t' \<Down> Lam [x].t''"
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and a2: "t2 \<Down> v'"
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and a3: "t''[x ::= v'] \<Down> t3" by (rule e_App_elim)
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have "t \<Down> Lam [x].t''" using ih a1 by auto
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then show "App t t2 \<Down> t3" using a2 a3 by auto
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qed (auto elim!: e_App_elim)
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text {*
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Next we extend the lemma above to arbitray initial
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sequences of cbv-reductions. *}
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lemma cbvs_eval:
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assumes a: "t1 \<longrightarrow>cbv* t2" "t2 \<Down> t3"
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shows "t1 \<Down> t3"
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using a by (induct) (auto intro: cbv_eval)
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text {*
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Finally, we can show that if from a term t we reach a value
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by a cbv-reduction sequence, then t evaluates to this value.
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*}
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lemma cbvs_implies_eval:
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assumes a: "t \<longrightarrow>cbv* v" "val v"
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shows "t \<Down> v"
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using a
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by (induct) (auto intro: eval_val cbvs_eval)
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text {*
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All facts tied together give us the desired property about
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machines.
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*}
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theorem machines_implies_eval:
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assumes a: "<t1, []> \<mapsto>* <t2, []>"
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and b: "val t2"
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shows "t1 \<Down> t2"
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proof -
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have "t1 \<longrightarrow>cbv* t2" using a machines_implies_cbvs by simp
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then show "t1 \<Down> t2" using b cbvs_implies_eval by simp
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qed
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end
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