TypesType Systems
Set Warnings "-notation-overridden,-parsing,-deprecated-hint-without-locality".
From Coq Require Import Arith.Arith.
From PLF Require Import Maps.
From PLF Require Import Imp.
From PLF Require Import Smallstep.
Hint Constructors multi : core.
Typed Arithmetic Expressions
Syntax
t ::= tru
| fls
| test t then t else t
| zro
| scc t
| prd t
| iszro t
Inductive tm : Type :=
| tru : tm
| fls : tm
| test : tm → tm → tm → tm
| zro : tm
| scc : tm → tm
| prd : tm → tm
| iszro : tm → tm.
(Since it is so simple, we will not bother introducing custom
concrete syntax for this language.)
Values are tru, fls, and numeric values...
Inductive bvalue : tm → Prop :=
| bv_tru : bvalue tru
| bv_fls : bvalue fls.
Inductive nvalue : tm → Prop :=
| nv_zro : nvalue zro
| nv_scc : ∀ t, nvalue t → nvalue (scc t).
Definition value (t : tm) := bvalue t ∨ nvalue t.
Hint Constructors bvalue nvalue : core.
Hint Unfold value : core.
Operational Semantics
(ST_TestTru) | |
test tru then t1 else t2 --> t1 |
(ST_TestFls) | |
test fls then t1 else t2 --> t2 |
t1 --> t1' | (ST_Test) |
test t1 then t2 else t3 --> test t1' then t2 else t3 |
t1 --> t1' | (ST_Scc) |
scc t1 --> scc t1' |
(ST_PrdZro) | |
prd zro --> zro |
numeric value v | (ST_PrdScc) |
prd (scc v) --> v |
t1 --> t1' | (ST_Prd) |
prd t1 --> prd t1' |
(ST_IszroZro) | |
iszro zro --> tru |
numeric value v | (ST_IszroScc) |
iszro (scc v) --> fls |
t1 --> t1' | (ST_Iszro) |
iszro t1 --> iszro t1' |
Reserved Notation "t '-->' t'" (at level 40).
Inductive step : tm → tm → Prop :=
| ST_TestTru : ∀ t1 t2,
(test tru t1 t2) --> t1
| ST_TestFls : ∀ t1 t2,
(test fls t1 t2) --> t2
| ST_Test : ∀ t1 t1' t2 t3,
t1 --> t1' →
(test t1 t2 t3) --> (test t1' t2 t3)
| ST_Scc : ∀ t1 t1',
t1 --> t1' →
(scc t1) --> (scc t1')
| ST_PrdZro :
(prd zro) --> zro
| ST_PrdScc : ∀ v,
nvalue v →
(prd (scc v)) --> v
| ST_Prd : ∀ t1 t1',
t1 --> t1' →
(prd t1) --> (prd t1')
| ST_IszroZro :
(iszro zro) --> tru
| ST_IszroScc : ∀ v,
nvalue v →
(iszro (scc v)) --> fls
| ST_Iszro : ∀ t1 t1',
t1 --> t1' →
(iszro t1) --> (iszro t1')
where "t '-->' t'" := (step t t').
Hint Constructors step : core.
Notice that the step relation doesn't care about whether the
expression being stepped makes global sense -- it just checks that
the operation in the next reduction step is being applied to the
right kinds of operands. For example, the term scc tru cannot
take a step, but the almost as obviously nonsensical term
scc (test tru then tru else tru)
can take a step (once, before becoming stuck).
scc (test tru then tru else tru)
Normal Forms and Values
Notation step_normal_form := (normal_form step).
Definition stuck (t : tm) : Prop :=
step_normal_form t ∧ ¬ value t.
Hint Unfold stuck : core.
Exercise: 3 stars, standard (value_is_nf)
(Hint: You will reach a point in this proof where you need to
use an induction to reason about a term that is known to be a
numeric value. This induction can be performed either over the
term itself or over the evidence that it is a numeric value. The
proof goes through in either case, but you will find that one way
is quite a bit shorter than the other. For the sake of the
exercise, try to complete the proof both ways.) ☐
Exercise: 3 stars, standard, optional (step_deterministic)
Use value_is_nf to show that the step relation is also deterministic.Typing
In informal notation, the typing relation is often written
⊢ t \in T and pronounced "t has type T." The ⊢ symbol
is called a "turnstile." Below, we're going to see richer typing
relations where one or more additional "context" arguments are
written to the left of the turnstile. For the moment, the context
is always empty.
(T_Tru) | |
⊢ tru ∈ Bool |
(T_Fls) | |
⊢ fls ∈ Bool |
⊢ t1 ∈ Bool ⊢ t2 ∈ T ⊢ t3 ∈ T | (T_Test) |
⊢ test t1 then t2 else t3 ∈ T |
(T_Zro) | |
⊢ zro ∈ Nat |
⊢ t1 ∈ Nat | (T_Scc) |
⊢ scc t1 ∈ Nat |
⊢ t1 ∈ Nat | (T_Prd) |
⊢ prd t1 ∈ Nat |
⊢ t1 ∈ Nat | (T_IsZro) |
⊢ iszro t1 ∈ Bool |
Reserved Notation "'⊢' t '∈' T" (at level 40).
Inductive has_type : tm → ty → Prop :=
| T_Tru :
⊢ tru \in Bool
| T_Fls :
⊢ fls \in Bool
| T_Test : ∀ t1 t2 t3 T,
⊢ t1 \in Bool →
⊢ t2 \in T →
⊢ t3 \in T →
⊢ test t1 t2 t3 \in T
| T_Zro :
⊢ zro \in Nat
| T_Scc : ∀ t1,
⊢ t1 \in Nat →
⊢ scc t1 \in Nat
| T_Prd : ∀ t1,
⊢ t1 \in Nat →
⊢ prd t1 \in Nat
| T_Iszro : ∀ t1,
⊢ t1 \in Nat →
⊢ iszro t1 \in Bool
where "'⊢' t '∈' T" := (has_type t T).
Hint Constructors has_type : core.
Example has_type_1 :
⊢ test fls zro (scc zro) \in Nat.
Proof.
apply T_Test.
- apply T_Fls.
- apply T_Zro.
- apply T_Scc. apply T_Zro.
Qed.
(Since we've included all the constructors of the typing relation
in the hint database, the auto tactic can actually find this
proof automatically.)
It's important to realize that the typing relation is a
conservative (or static) approximation: it does not consider
what happens when the term is reduced -- in particular, it does
not calculate the type of its normal form.
Example has_type_not :
¬ ( ⊢ test fls zro tru \in Bool ).
Proof.
intros Contra. solve_by_inverts 2. Qed.
intros Contra. solve_by_inverts 2. Qed.
Example scc_hastype_nat__hastype_nat : ∀ t,
⊢ scc t \in Nat →
⊢ t \in Nat.
Proof.
(* FILL IN HERE *) Admitted.
☐
⊢ scc t \in Nat →
⊢ t \in Nat.
Proof.
(* FILL IN HERE *) Admitted.
☐
Canonical forms
Lemma bool_canonical : ∀ t,
⊢ t \in Bool → value t → bvalue t.
Proof.
intros t HT [Hb | Hn].
- assumption.
- destruct Hn as [ | Hs].
+ inversion HT.
+ inversion HT.
Qed.
intros t HT [Hb | Hn].
- assumption.
- destruct Hn as [ | Hs].
+ inversion HT.
+ inversion HT.
Qed.
Lemma nat_canonical : ∀ t,
⊢ t \in Nat → value t → nvalue t.
Proof.
intros t HT [Hb | Hn].
- inversion Hb; subst; inversion HT.
- assumption.
Qed.
intros t HT [Hb | Hn].
- inversion Hb; subst; inversion HT.
- assumption.
Qed.
Progress
Exercise: 3 stars, standard (finish_progress)
Complete the formal proof of the progress property. (Make sure
you understand the parts we've given of the informal proof in the
following exercise before starting -- this will save you a lot of
time.)
Proof.
intros t T HT.
induction HT; auto.
(* The cases that were obviously values, like T_Tru and
T_Fls, are eliminated immediately by auto *)
- (* T_Test *)
right. destruct IHHT1.
+ (* t1 is a value *)
apply (bool_canonical t1 HT1) in H.
destruct H.
× ∃ t2. auto.
× ∃ t3. auto.
+ (* t1 can take a step *)
destruct H as [t1' H1].
∃ (test t1' t2 t3). auto.
(* FILL IN HERE *) Admitted.
intros t T HT.
induction HT; auto.
(* The cases that were obviously values, like T_Tru and
T_Fls, are eliminated immediately by auto *)
- (* T_Test *)
right. destruct IHHT1.
+ (* t1 is a value *)
apply (bool_canonical t1 HT1) in H.
destruct H.
× ∃ t2. auto.
× ∃ t3. auto.
+ (* t1 can take a step *)
destruct H as [t1' H1].
∃ (test t1' t2 t3). auto.
(* FILL IN HERE *) Admitted.
☐
Theorem: If ⊢ t \in T, then either t is a value or else
t --> t' for some t'.
Proof: By induction on a derivation of ⊢ t \in T.
Exercise: 3 stars, advanced (finish_progress_informal)
Complete the corresponding informal proof:- If the last rule in the derivation is T_Test, then t = test t1
then t2 else t3, with ⊢ t1 \in Bool, ⊢ t2 \in T and ⊢ t3
\in T. By the IH, either t1 is a value or else t1 can step
to some t1'.
- If t1 is a value, then by the canonical forms lemmas
and the fact that ⊢ t1 \in Bool we have that t1
is a bvalue -- i.e., it is either tru or fls.
If t1 = tru, then t steps to t2 by ST_TestTru,
while if t1 = fls, then t steps to t3 by
ST_TestFls. Either way, t can step, which is what
we wanted to show.
- If t1 itself can take a step, then, by ST_Test, so can
t.
- If t1 is a value, then by the canonical forms lemmas
and the fact that ⊢ t1 \in Bool we have that t1
is a bvalue -- i.e., it is either tru or fls.
If t1 = tru, then t steps to t2 by ST_TestTru,
while if t1 = fls, then t steps to t3 by
ST_TestFls. Either way, t can step, which is what
we wanted to show.
- (* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_finish_progress_informal : option (nat×string) := None.
☐
Definition manual_grade_for_finish_progress_informal : option (nat×string) := None.
☐
Type Preservation
Exercise: 2 stars, standard (finish_preservation)
Complete the formal proof of the preservation property. (Again,
make sure you understand the informal proof fragment in the
following exercise first.)
Proof.
intros t t' T HT HE.
generalize dependent t'.
induction HT;
(* every case needs to introduce a couple of things *)
intros t' HE;
(* and we can deal with several impossible
cases all at once *)
try solve_by_invert.
- (* T_Test *) inversion HE; subst; clear HE.
+ (* ST_TESTTru *) assumption.
+ (* ST_TestFls *) assumption.
+ (* ST_Test *) apply T_Test; try assumption.
apply IHHT1; assumption.
(* FILL IN HERE *) Admitted.
intros t t' T HT HE.
generalize dependent t'.
induction HT;
(* every case needs to introduce a couple of things *)
intros t' HE;
(* and we can deal with several impossible
cases all at once *)
try solve_by_invert.
- (* T_Test *) inversion HE; subst; clear HE.
+ (* ST_TESTTru *) assumption.
+ (* ST_TestFls *) assumption.
+ (* ST_Test *) apply T_Test; try assumption.
apply IHHT1; assumption.
(* FILL IN HERE *) Admitted.
☐
Theorem: If ⊢ t \in T and t --> t', then ⊢ t' \in T.
Proof: By induction on a derivation of ⊢ t \in T.
Exercise: 3 stars, advanced (finish_preservation_informal)
Complete the following informal proof:- If the last rule in the derivation is T_Test, then t = test t1
then t2 else t3, with ⊢ t1 \in Bool, ⊢ t2 \in T and ⊢ t3
\in T.
- If the last rule was ST_TestTru, then t' = t2. But we
know that ⊢ t2 \in T, so we are done.
- If the last rule was ST_TestFls, then t' = t3. But we
know that ⊢ t3 \in T, so we are done.
- If the last rule was ST_Test, then t' = test t1' then t2
else t3, where t1 --> t1'. We know ⊢ t1 \in Bool so,
by the IH, ⊢ t1' \in Bool. The T_Test rule then gives us
⊢ test t1' then t2 else t3 \in T, as required.
- If the last rule was ST_TestTru, then t' = t2. But we
know that ⊢ t2 \in T, so we are done.
- (* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_finish_preservation_informal : option (nat×string) := None.
☐
Definition manual_grade_for_finish_preservation_informal : option (nat×string) := None.
☐
Exercise: 3 stars, standard (preservation_alternate_proof)
Now prove the same property again by induction on the evaluation derivation instead of on the typing derivation. Begin by carefully reading and thinking about the first few lines of the above proofs to make sure you understand what each one is doing. The set-up for this proof is similar, but not exactly the same.Theorem preservation' : ∀ t t' T,
⊢ t \in T →
t --> t' →
⊢ t' \in T.
Proof with eauto.
(* FILL IN HERE *) Admitted.
☐
Type Soundness
Definition multistep := (multi step).
Notation "t1 '-->*' t2" := (multistep t1 t2) (at level 40).
Corollary soundness : ∀ t t' T,
⊢ t \in T →
t -->* t' →
~(stuck t').
Proof.
intros t t' T HT P. induction P; intros [R S].
- apply progress in HT. destruct HT; auto.
- apply IHP.
+ apply preservation with (t := x); auto.
+ unfold stuck. split; auto.
Qed.
intros t t' T HT P. induction P; intros [R S].
- apply progress in HT. destruct HT; auto.
- apply IHP.
+ apply preservation with (t := x); auto.
+ unfold stuck. split; auto.
Qed.
Additional Exercises
Exercise: 2 stars, standard, especially useful (subject_expansion)
Having seen the subject reduction property, one might wonder whether the opposity property -- subject expansion -- also holds. That is, is it always the case that, if t --> t' and ⊢ t' \in T, then ⊢ t \in T? If so, prove it. If not, give a counter-example. (You do not need to prove your counter-example in Coq, but feel free to do so.)
(* Do not modify the following line: *)
Definition manual_grade_for_subject_expansion : option (nat×string) := None.
☐
Definition manual_grade_for_subject_expansion : option (nat×string) := None.
☐
Exercise: 2 stars, standard (variation1)
Suppose that we add this new rule to the typing relation:| T_SccBool : ∀ t,
⊢ t \in Bool →
⊢ scc t \in Bool
- Determinism of step
(* FILL IN HERE *)
- Progress
(* FILL IN HERE *)
- Preservation
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_variation1 : option (nat×string) := None.
☐
Definition manual_grade_for_variation1 : option (nat×string) := None.
☐
Exercise: 2 stars, standard (variation2)
Suppose, instead, that we add this new rule to the step relation:| ST_Funny1 : ∀ t2 t3,
(test tru t2 t3) --> t3
(* Do not modify the following line: *)
Definition manual_grade_for_variation2 : option (nat×string) := None.
☐
Definition manual_grade_for_variation2 : option (nat×string) := None.
☐
Exercise: 2 stars, standard, optional (variation3)
Suppose instead that we add this rule:| ST_Funny2 : ∀ t1 t2 t2' t3,
t2 --> t2' →
(test t1 t2 t3) --> (test t1 t2' t3)
☐
Exercise: 2 stars, standard, optional (variation4)
Suppose instead that we add this rule:| ST_Funny3 :
(prd fls) --> (prd (prd fls))
☐
Exercise: 2 stars, standard, optional (variation5)
Suppose instead that we add this rule:| T_Funny4 :
⊢ zro \in Bool
☐
Exercise: 2 stars, standard, optional (variation6)
Suppose instead that we add this rule:| T_Funny5 :
⊢ prd zro \in Bool
☐
Exercise: 3 stars, standard, optional (more_variations)
Make up some exercises of your own along the same lines as the ones above. Try to find ways of selectively breaking properties -- i.e., ways of changing the definitions that break just one of the properties and leave the others alone.
(* FILL IN HERE *)
☐
☐
Exercise: 1 star, standard (remove_prdzro)
The reduction rule ST_PrdZro is a bit counter-intuitive: we might feel that it makes more sense for the predecessor of zro to be undefined, rather than being defined to be zro. Can we achieve this simply by removing the rule from the definition of step? Would doing so create any problems elsewhere?
(* Do not modify the following line: *)
Definition manual_grade_for_remove_predzro : option (nat×string) := None.
☐
Definition manual_grade_for_remove_predzro : option (nat×string) := None.
☐
Exercise: 4 stars, advanced (prog_pres_bigstep)
Suppose our evaluation relation is defined in the big-step style. State appropriate analogs of the progress and preservation properties. (You do not need to prove them.)
(* Do not modify the following line: *)
Definition manual_grade_for_prog_pres_bigstep : option (nat×string) := None.
☐
Definition manual_grade_for_prog_pres_bigstep : option (nat×string) := None.
☐
(* 2021-03-18 17:24 *)