# TypesType Systems

*type systems*— static program analyses that classify expressions according to the "shapes" of their results. We'll begin with a typed version of the simplest imaginable language, to introduce the basic ideas of types and typing rules and the fundamental theorems about type systems:

*type preservation*and

*progress*. In chapter Stlc we'll move on to the

*simply typed lambda-calculus*, which lives at the core of every modern functional programming language (including Coq!).

Set Warnings "-notation-overridden,-parsing".

From Coq Require Import Arith.Arith.

From PLF Require Import Maps.

From PLF Require Import Imp.

From PLF Require Import Smallstep.

Hint Constructors multi.

From Coq Require Import Arith.Arith.

From PLF Require Import Maps.

From PLF Require Import Imp.

From PLF Require Import Smallstep.

Hint Constructors multi.

# Typed Arithmetic Expressions

## Syntax

t ::= tru

| fls

| test t then t else t

| zro

| scc t

| prd t

| iszro t

And here it is formally:
| 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.

| tru : tm

| fls : tm

| test : tm → tm → tm → tm

| zro : tm

| scc : tm → tm

| prd : tm → tm

| iszro : tm → tm.

*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.

Hint Unfold value.

Hint Unfold update.

| 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.

Hint Unfold value.

Hint Unfold update.

## Operational Semantics

(ST_TestTru) | |

test tru then t_{1} else t_{2} --> t_{1} |

(ST_TestFls) | |

test fls then t_{1} else t_{2} --> t_{2} |

t_{1} --> t_{1}' |
(ST_Test) |

test t_{1} then t_{2} else t_{3} --> test t_{1}' then t_{2} else t_{3} |

t_{1} --> t_{1}' |
(ST_Scc) |

scc t_{1} --> scc t_{1}' |

(ST_PrdZro) | |

prd zro --> zro |

numeric value v_{1} |
(ST_PrdScc) |

prd (scc v_{1}) --> v_{1} |

t_{1} --> t_{1}' |
(ST_Prd) |

prd t_{1} --> prd t_{1}' |

(ST_IszroZro) | |

iszro zro --> tru |

numeric value v_{1} |
(ST_IszroScc) |

iszro (scc v_{1}) --> fls |

t_{1} --> t_{1}' |
(ST_Iszro) |

iszro t_{1} --> iszro t_{1}' |

Reserved Notation "t

Inductive step : tm → tm → Prop :=

| ST_TestTru : ∀t

(test tru t

| ST_TestFls : ∀t

(test fls t

| ST_Test : ∀t

t

(test t

| ST_Scc : ∀t

t

(scc t

| ST_PrdZro :

(prd zro) --> zro

| ST_PrdScc : ∀t

nvalue t

(prd (scc t

| ST_Prd : ∀t

t

(prd t

| ST_IszroZro :

(iszro zro) --> tru

| ST_IszroScc : ∀t

nvalue t

(iszro (scc t

| ST_Iszro : ∀t

t

(iszro t

where "t

Hint Constructors step.

_{1}'-->' t_{2}" (at level 40).Inductive step : tm → tm → Prop :=

| ST_TestTru : ∀t

_{1}t_{2},(test tru t

_{1}t_{2}) --> t_{1}| ST_TestFls : ∀t

_{1}t_{2},(test fls t

_{1}t_{2}) --> t_{2}| ST_Test : ∀t

_{1}t_{1}' t_{2}t_{3},t

_{1}--> t_{1}' →(test t

_{1}t_{2}t_{3}) --> (test t_{1}' t_{2}t_{3})| ST_Scc : ∀t

_{1}t_{1}',t

_{1}--> t_{1}' →(scc t

_{1}) --> (scc t_{1}')| ST_PrdZro :

(prd zro) --> zro

| ST_PrdScc : ∀t

_{1},nvalue t

_{1}→(prd (scc t

_{1})) --> t_{1}| ST_Prd : ∀t

_{1}t_{1}',t

_{1}--> t_{1}' →(prd t

_{1}) --> (prd t_{1}')| ST_IszroZro :

(iszro zro) --> tru

| ST_IszroScc : ∀t

_{1},nvalue t

_{1}→(iszro (scc t

_{1})) --> fls| ST_Iszro : ∀t

_{1}t_{1}',t

_{1}--> t_{1}' →(iszro t

_{1}) --> (iszro t_{1}')where "t

_{1}'-->' t_{2}" := (step t_{1}t_{2}).Hint Constructors step.

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).
## Normal Forms and Values

*stuck*.

Notation step_normal_form := (normal_form step).

Definition stuck (t : tm) : Prop :=

step_normal_form t ∧ ¬value t.

Hint Unfold stuck.

Definition stuck (t : tm) : Prop :=

step_normal_form t ∧ ¬value t.

Hint Unfold stuck.

Example some_term_is_stuck :

∃t, stuck t.

Proof.

(* FILL IN HERE *) Admitted.

☐
∃t, stuck t.

Proof.

(* FILL IN HERE *) Admitted.

*not*the same in this language, the set of values is a subset of the set of normal forms. This is important because it shows we did not accidentally define things so that some value could still take a step.

#### Exercise: 3 stars, standard (value_is_nf)

Lemma value_is_nf : ∀t,

value t → step_normal_form t.

value t → step_normal_form t.

Proof.

(* FILL IN HERE *) Admitted.

(* FILL IN HERE *) Admitted.

(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.
Theorem step_deterministic:

deterministic step.

Proof with eauto.

(* FILL IN HERE *) Admitted.

☐
deterministic step.

Proof with eauto.

(* FILL IN HERE *) Admitted.

## Typing

*want*to have a meaning. We can easily exclude such ill-typed terms by defining a

*typing relation*that relates terms to the types (either numeric or boolean) of their final results.

Inductive ty : Type :=

| Bool : ty

| Nat : ty.

| Bool : ty

| Nat : ty.

In informal notation, the typing relation is often written
⊢ t ∈ 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 |

⊢ t_{1} ∈ Bool ⊢ t_{2} ∈ T ⊢ t_{3} ∈ T |
(T_Test) |

⊢ test t_{1} then t_{2} else t_{3} ∈ T |

(T_Zro) | |

⊢ zro ∈ Nat |

⊢ t_{1} ∈ Nat |
(T_Scc) |

⊢ scc t_{1} ∈ Nat |

⊢ t_{1} ∈ Nat |
(T_Prd) |

⊢ prd t_{1} ∈ Nat |

⊢ t_{1} ∈ Nat |
(T_IsZro) |

⊢ iszro t_{1} ∈ Bool |

Reserved Notation "'⊢' t '∈' T" (at level 40).

Inductive has_type : tm → ty → Prop :=

| T_Tru :

⊢ tru ∈ Bool

| T_Fls :

⊢ fls ∈ Bool

| T_Test : ∀t

⊢ t

⊢ t

⊢ t

⊢ test t

| T_Zro :

⊢ zro ∈ Nat

| T_Scc : ∀t

⊢ t

⊢ scc t

| T_Prd : ∀t

⊢ t

⊢ prd t

| T_Iszro : ∀t

⊢ t

⊢ iszro t

where "'⊢' t '∈' T" := (has_type t T).

Hint Constructors has_type.

Example has_type_1 :

⊢ test fls zro (scc zro) ∈ Nat.

Proof.

apply T_Test.

- apply T_Fls.

- apply T_Zro.

- apply T_Scc.

+ apply T_Zro.

Qed.

Inductive has_type : tm → ty → Prop :=

| T_Tru :

⊢ tru ∈ Bool

| T_Fls :

⊢ fls ∈ Bool

| T_Test : ∀t

_{1}t_{2}t_{3}T,⊢ t

_{1}∈ Bool →⊢ t

_{2}∈ T →⊢ t

_{3}∈ T →⊢ test t

_{1}t_{2}t_{3}∈ T| T_Zro :

⊢ zro ∈ Nat

| T_Scc : ∀t

_{1},⊢ t

_{1}∈ Nat →⊢ scc t

_{1}∈ Nat| T_Prd : ∀t

_{1},⊢ t

_{1}∈ Nat →⊢ prd t

_{1}∈ Nat| T_Iszro : ∀t

_{1},⊢ t

_{1}∈ Nat →⊢ iszro t

_{1}∈ Boolwhere "'⊢' t '∈' T" := (has_type t T).

Hint Constructors has_type.

Example has_type_1 :

⊢ test fls zro (scc zro) ∈ 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 ∈ Bool ).

¬( ⊢ test fls zro tru ∈ 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 ∈ Nat →

⊢ t ∈ Nat.

Proof.

(* FILL IN HERE *) Admitted.

☐
⊢ scc t ∈ Nat →

⊢ t ∈ Nat.

Proof.

(* FILL IN HERE *) Admitted.

### Canonical forms

Lemma bool_canonical : ∀t,

⊢ t ∈ Bool → value t → bvalue t.

Lemma nat_canonical : ∀t,

⊢ t ∈ Nat → value t → nvalue t.

⊢ t ∈ Bool → value t → bvalue t.

Proof.

intros t HT [Hb | Hn].

- assumption.

- induction Hn; inversion HT; auto.

Qed.

intros t HT [Hb | Hn].

- assumption.

- induction Hn; inversion HT; auto.

Qed.

Lemma nat_canonical : ∀t,

⊢ t ∈ 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

*progress*.

#### Exercise: 3 stars, standard (finish_progress)

Theorem progress : ∀t T,

⊢ t ∈ T →

value t ∨ ∃t', t --> t'.

⊢ t ∈ T →

value t ∨ ∃t', t --> t'.

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 with auto.

intros t T HT.

induction HT...

(* The cases that were obviously values, like T_Tru and

T_Fls, were eliminated immediately by auto *)

- (* T_Test *)

right. inversion IHHT1; clear IHHT1.

+ (* t

apply (bool_canonical t

inversion H; subst; clear H.

∃t

∃t

+ (* t

inversion H as [t

∃(test t

(* FILL IN HERE *) Admitted.

intros t T HT.

induction HT...

(* The cases that were obviously values, like T_Tru and

T_Fls, were eliminated immediately by auto *)

- (* T_Test *)

right. inversion IHHT1; clear IHHT1.

+ (* t

_{1}is a value *)apply (bool_canonical t

_{1}HT_{1}) in H.inversion H; subst; clear H.

∃t

_{2}...∃t

_{3}...+ (* t

_{1}can take a step *)inversion H as [t

_{1}' H_{1}].∃(test t

_{1}' t_{2}t_{3})...(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, advanced (finish_progress_informal)

Complete the corresponding informal proof:*Theorem*: If ⊢ t ∈ T, then either t is a value or else t --> t' for some t'.

*Proof*: By induction on a derivation of ⊢ t ∈ T.

- If the last rule in the derivation is T_Test, then t = test t
_{1}then t_{2}else t_{3}, with ⊢ t_{1}∈ Bool, ⊢ t_{2}∈ T and ⊢ t_{3}∈ T. By the IH, either t_{1}is a value or else t_{1}can step to some t_{1}'.- If t
_{1}is a value, then by the canonical forms lemmas and the fact that ⊢ t_{1}∈ Bool we have that t_{1}is a bvalue — i.e., it is either tru or fls. If t_{1}= tru, then t steps to t_{2}by ST_TestTru, while if t_{1}= fls, then t steps to t_{3}by ST_TestFls. Either way, t can step, which is what we wanted to show. - If t
_{1}itself can take a step, then, by ST_Test, so can t.

- If t
- (* 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.

*all*normal forms were values. Here a term can be stuck, but only if it is ill typed.

## Type Preservation

#### Exercise: 2 stars, standard (finish_preservation)

Theorem preservation : ∀t t' T,

⊢ t ∈ T →

t --> t' →

⊢ t' ∈ T.

⊢ t ∈ T →

t --> t' →

⊢ t' ∈ T.

Complete the formal proof of the preservation property. (Again,
make sure you understand the informal proof fragment in the
following exercise first.)

Proof with auto.

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.

#### Exercise: 3 stars, advanced (finish_preservation_informal)

Complete the following informal proof:*Theorem*: If ⊢ t ∈ T and t --> t', then ⊢ t' ∈ T.

*Proof*: By induction on a derivation of ⊢ t ∈ T.

- If the last rule in the derivation is T_Test, then t = test t
_{1}then t_{2}else t_{3}, with ⊢ t_{1}∈ Bool, ⊢ t_{2}∈ T and ⊢ t_{3}∈ T.- If the last rule was ST_TestTru, then t' = t
_{2}. But we know that ⊢ t_{2}∈ T, so we are done. - If the last rule was ST_TestFls, then t' = t
_{3}. But we know that ⊢ t_{3}∈ T, so we are done. - If the last rule was ST_Test, then t' = test t
_{1}' then t_{2}else t_{3}, where t_{1}--> t_{1}'. We know ⊢ t_{1}∈ Bool so, by the IH, ⊢ t_{1}' ∈ Bool. The T_Test rule then gives us ⊢ test t_{1}' then t_{2}else t_{3}∈ T, as required.

- If the last rule was ST_TestTru, then t' = t
- (* 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 ∈ T →

t --> t' →

⊢ t' ∈ T.

Proof with eauto.

(* FILL IN HERE *) Admitted.

☐
⊢ t ∈ T →

t --> t' →

⊢ t' ∈ T.

Proof with eauto.

(* FILL IN HERE *) Admitted.

*subject reduction*, because it tells us what happens when the "subject" of the typing relation is reduced. This terminology comes from thinking of typing statements as sentences, where the term is the subject and the type is the predicate.

## Type Soundness

Definition multistep := (multi step).

Notation "t

Corollary soundness : ∀t t' T,

⊢ t ∈ T →

t -->* t' →

~(stuck t').

Notation "t

_{1}'-->*' t_{2}" := (multistep t_{1}t_{2}) (at level 40).Corollary soundness : ∀t t' T,

⊢ t ∈ T →

t -->* t' →

~(stuck t').

Proof.

intros t t' T HT P. induction P; intros [R S].

destruct (progress x T HT); auto.

apply IHP. apply (preservation x y T HT H).

unfold stuck. split; auto. Qed.

intros t t' T HT P. induction P; intros [R S].

destruct (progress x T HT); auto.

apply IHP. apply (preservation x y T HT H).

unfold stuck. split; auto. Qed.

## Additional Exercises

#### Exercise: 2 stars, standard, recommended (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' ∈ T, then ⊢ t ∈ 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 ∈ Bool →

⊢ scc t ∈ Bool

Which of the following properties remain true in the presence of
this rule? For each one, write either "remains true" or
else "becomes false." If a property becomes false, give a
counterexample.
⊢ t ∈ Bool →

⊢ scc t ∈ 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 : ∀t

(test tru t

Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
(* FILL IN HERE *)_{2}t_{3},(test tru t

_{2}t_{3}) --> t_{3}
(* 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 : ∀t

t

(test t

Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
(* FILL IN HERE *)_{1}t_{2}t_{2}' t_{3},t

_{2}--> t_{2}' →(test t

_{1}t_{2}t_{3}) --> (test t_{1}t_{2}' t_{3})☐

#### Exercise: 2 stars, standard, optional (variation4)

Suppose instead that we add this rule:
| ST_Funny3 :

(prd fls) --> (prd (prd fls))

Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
(* FILL IN HERE *)(prd fls) --> (prd (prd fls))

☐

#### Exercise: 2 stars, standard, optional (variation5)

Suppose instead that we add this rule:
| T_Funny4 :

⊢ zro ∈ Bool

Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
(* FILL IN HERE *)⊢ zro ∈ Bool

☐

#### Exercise: 2 stars, standard, optional (variation6)

Suppose instead that we add this rule:
| T_Funny5 :

⊢ prd zro ∈ Bool

Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
(* FILL IN HERE *)⊢ prd zro ∈ 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.

(* Mon Mar 25 14:39:38 EDT 2019 *)