--- a/Tutorial/Tutorial4.thy Sat Jan 22 16:37:00 2011 -0600
+++ b/Tutorial/Tutorial4.thy Sat Jan 22 18:59:48 2011 -0600
@@ -1,6 +1,6 @@
theory Tutorial4
-imports Tutorial1 Tutorial2 Tutorial3
+imports Tutorial1 Tutorial2
begin
section {* The CBV Reduction Relation (Small-Step Semantics) *}
@@ -21,14 +21,14 @@
declare cbv.intros[intro] cbv_star.intros[intro]
-subsection {* EXERCISE 3 *}
+subsection {* EXERCISE 11 *}
text {*
Show that cbv* is transitive, by filling the gaps in the
proof below.
*}
-lemma
+lemma cbvs3 [intro]:
assumes a: "e1 \<longrightarrow>cbv* e2" "e2 \<longrightarrow>cbv* e3"
shows "e1 \<longrightarrow>cbv* e3"
using a
@@ -46,17 +46,9 @@
show "e1 \<longrightarrow>cbv* e3" sorry
qed
-lemma cbvs3 [intro]:
- assumes a: "e1 \<longrightarrow>cbv* e2" "e2 \<longrightarrow>cbv* e3"
- shows "e1 \<longrightarrow>cbv* e3"
-using a by (induct) (auto)
-
-
-
-
text {*
- In order to help establishing the property that the CK Machine
+ 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.
*}
@@ -98,15 +90,9 @@
finally show "App (Lam [x].t) v \<longrightarrow>cbv t[x ::= v]" by simp
qed
-text {*
- The transitive closure of the cbv-reduction relation:
-*}
-
-
-
-subsection {* EXERCISE 8 *}
+subsection {* EXERCISE 12 *}
text {*
If more simple exercises are needed, then complete the following proof.
@@ -119,26 +105,22 @@
proof (induct E)
case Hole
have "t \<longrightarrow>cbv t'" by fact
- then show "\<box>\<lbrakk>t\<rbrakk> \<longrightarrow>cbv \<box>\<lbrakk>t'\<rbrakk>" by simp
+ 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
- moreover
have "t \<longrightarrow>cbv t'" by fact
- ultimately
- have "E\<lbrakk>t\<rbrakk> \<longrightarrow>cbv E\<lbrakk>t'\<rbrakk>" by simp
- then show "(CAppL E s)\<lbrakk>t\<rbrakk> \<longrightarrow>cbv (CAppL E s)\<lbrakk>t'\<rbrakk>" by auto
+
+ 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
- moreover
have a: "t \<longrightarrow>cbv t'" by fact
- ultimately
- have "E\<lbrakk>t\<rbrakk> \<longrightarrow>cbv E\<lbrakk>t'\<rbrakk>" by simp
- then show "(CAppR s E)\<lbrakk>t\<rbrakk> \<longrightarrow>cbv (CAppR s E)\<lbrakk>t'\<rbrakk>" by auto
+
+ show "(CAppR s E)\<lbrakk>t\<rbrakk> \<longrightarrow>cbv (CAppR s E)\<lbrakk>t'\<rbrakk>" sorry
qed
-section {* EXERCISE 9 *}
+section {* EXERCISE 13 *}
text {*
The point of the cbv-reduction was that we can easily relatively
@@ -167,7 +149,8 @@
text {*
It is not difficult to extend the lemma above to
- arbitrary reductions sequences of the CK machine. *}
+ arbitrary reductions sequences of the machine.
+*}
lemma machines_implies_cbvs_ctx:
assumes a: "<e, Es> \<mapsto>* <e', Es'>"
@@ -176,7 +159,7 @@
by (induct) (blast)+
text {*
- So whenever we let the CL machine start in an initial
+ 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.
*}
@@ -192,14 +175,10 @@
text {*
We now want to relate the cbv-reduction to the evaluation
- relation. For this we need two auxiliary lemmas.
+ relation. For this we need one auxiliary lemma about
+ inverting the e_App rule.
*}
-lemma eval_val:
- assumes a: "val t"
- shows "t \<Down> t"
-using a by (induct) (auto)
-
lemma e_App_elim:
assumes a: "App t1 t2 \<Down> v"
@@ -207,7 +186,7 @@
using a by (cases) (auto simp add: lam.eq_iff lam.distinct)
-subsection {* EXERCISE *}
+subsection {* EXERCISE 13 *}
text {*
Complete the first and second case in the
@@ -222,9 +201,8 @@
case (cbv1 v x t t3)
have a1: "val v" by fact
have a2: "t[x ::= v] \<Down> t3" by fact
- have a3: "Lam [x].t \<Down> Lam [x].t" by auto
- have a4: "v \<Down> v" using a1 eval_val by auto
- show "App (Lam [x].t) v \<Down> t3" using a3 a4 a2 by auto
+
+ 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
@@ -233,14 +211,15 @@
where a1: "t' \<Down> Lam [x].t''"
and a2: "t2 \<Down> v'"
and a3: "t''[x ::= v'] \<Down> t3" by (rule e_App_elim)
- have "t \<Down> Lam [x].t''" using ih a1 by auto
- then show "App t t2 \<Down> t3" using a2 a3 by auto
+
+ 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. *}
+ sequences of cbv-reductions.
+*}
lemma cbvs_eval:
assumes a: "t1 \<longrightarrow>cbv* t2" "t2 \<Down> t3"
@@ -250,30 +229,42 @@
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" "val v"
+ assumes a: "t \<longrightarrow>cbv* v"
+ and b: "val v"
shows "t \<Down> v"
-using a
-by (induct) (auto intro: eval_val cbvs_eval)
+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.
+ 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 -
- have "t1 \<longrightarrow>cbv* t2" using a machines_implies_cbvs by simp
- then show "t1 \<Down> t2" using b cbvs_implies_eval by simp
+proof -
+
+ show "t1 \<Down> t2" sorry
qed
-
-
-
end