# 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,-deprecated-hint-without-locality".

From Coq Require Import Arith.Arith.

From PLF Require Import Maps.

From PLF Require Import Smallstep.

Set Default Goal Selector "!".

Hint Constructors multi : core.

From Coq Require Import Arith.Arith.

From PLF Require Import Maps.

From PLF Require Import Smallstep.

Set Default Goal Selector "!".

Hint Constructors multi : core.

# Typed Arithmetic Expressions

## Syntax

t ::= true

| false

| if t then t else t

| 0

| succ t

| pred t

| iszero t And here it is formally:

Module TM.

Inductive tm : Type :=

| tru : tm

| fls : tm

| ite : tm → tm → tm → tm

| zro : tm

| scc : tm → tm

| prd : tm → tm

| iszro : tm → tm.

Declare Custom Entry tm.

Declare Scope tm_scope.

Notation "'true'" := true (at level 1): tm_scope.

Notation "'true'" := (tru) (in custom tm at level 0): tm_scope.

Notation "'false'" := false (at level 1): tm_scope.

Notation "'false'" := (fls) (in custom tm at level 0): tm_scope.

Notation "<{ e }>" := e (e custom tm at level 99): tm_scope.

Notation "( x )" := x (in custom tm, x at level 99): tm_scope.

Notation "x" := x (in custom tm at level 0, x constr at level 0): tm_scope.

Notation "'0'" := (zro) (in custom tm at level 0): tm_scope.

Notation "'0'" := 0 (at level 1): tm_scope.

Notation "'succ' x" := (scc x) (in custom tm at level 90, x custom tm at level 80): tm_scope.

Notation "'pred' x" := (prd x) (in custom tm at level 90, x custom tm at level 80): tm_scope.

Notation "'iszero' x" := (iszro x) (in custom tm at level 80, x custom tm at level 70): tm_scope.

Notation "'if' c 'then' t 'else' e" := (ite c t e)

(in custom tm at level 90, c custom tm at level 80,

t custom tm at level 80, e custom tm at level 80): tm_scope.

Local Open Scope tm_scope.

Inductive tm : Type :=

| tru : tm

| fls : tm

| ite : tm → tm → tm → tm

| zro : tm

| scc : tm → tm

| prd : tm → tm

| iszro : tm → tm.

Declare Custom Entry tm.

Declare Scope tm_scope.

Notation "'true'" := true (at level 1): tm_scope.

Notation "'true'" := (tru) (in custom tm at level 0): tm_scope.

Notation "'false'" := false (at level 1): tm_scope.

Notation "'false'" := (fls) (in custom tm at level 0): tm_scope.

Notation "<{ e }>" := e (e custom tm at level 99): tm_scope.

Notation "( x )" := x (in custom tm, x at level 99): tm_scope.

Notation "x" := x (in custom tm at level 0, x constr at level 0): tm_scope.

Notation "'0'" := (zro) (in custom tm at level 0): tm_scope.

Notation "'0'" := 0 (at level 1): tm_scope.

Notation "'succ' x" := (scc x) (in custom tm at level 90, x custom tm at level 80): tm_scope.

Notation "'pred' x" := (prd x) (in custom tm at level 90, x custom tm at level 80): tm_scope.

Notation "'iszero' x" := (iszro x) (in custom tm at level 80, x custom tm at level 70): tm_scope.

Notation "'if' c 'then' t 'else' e" := (ite c t e)

(in custom tm at level 90, c custom tm at level 80,

t custom tm at level 80, e custom tm at level 80): tm_scope.

Local Open Scope tm_scope.

*Values*are true, false, and numeric values...

Inductive bvalue : tm → Prop :=

| bv_true : bvalue <{ true }>

| bv_false : bvalue <{ false }>.

Inductive nvalue : tm → Prop :=

| nv_0 : nvalue <{ 0 }>

| nv_succ : ∀ t, nvalue t → nvalue <{ succ t }>.

Definition value (t : tm) := bvalue t ∨ nvalue t.

Hint Constructors bvalue nvalue : core.

Hint Unfold value : core.

| bv_true : bvalue <{ true }>

| bv_false : bvalue <{ false }>.

Inductive nvalue : tm → Prop :=

| nv_0 : nvalue <{ 0 }>

| nv_succ : ∀ t, nvalue t → nvalue <{ succ t }>.

Definition value (t : tm) := bvalue t ∨ nvalue t.

Hint Constructors bvalue nvalue : core.

Hint Unfold value : core.

## Operational Semantics

(ST_IfTrue) | |

if true then t_{1} else t_{2} --> t_{1} |

(ST_IfFalse) | |

if false then t_{1} else t_{2} --> t_{2} |

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

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

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

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

(ST_Pred0) | |

pred 0 --> 0 |

numeric value v | (ST_PredSucc) |

pred (succ v) --> v |

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

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

(ST_IsZero0) | |

iszero 0 --> true |

numeric value v | (ST_IszeroSucc) |

iszero (succ v) --> false |

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

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

Inductive step : tm → tm → Prop :=

| ST_IfTrue : ∀ t

_{1}t

_{2},

<{ if true then t

_{1}else t

_{2}}> --> t

_{1}

| ST_IfFalse : ∀ t

_{1}t

_{2},

<{ if false then t

_{1}else t

_{2}}> --> t

_{2}

| ST_If : ∀ c c' t

_{2}t

_{3},

c --> c' →

<{ if c then t

_{2}else t

_{3}}> --> <{ if c' then t

_{2}else t

_{3}}>

| ST_Succ : ∀ t

_{1}t

_{1}',

t

_{1}--> t

_{1}' →

<{ succ t

_{1}}> --> <{ succ t

_{1}' }>

| ST_Pred0 :

<{ pred 0 }> --> <{ 0 }>

| ST_PredSucc : ∀ v,

nvalue v →

<{ pred (succ v) }> --> v

| ST_Pred : ∀ t

_{1}t

_{1}',

t

_{1}--> t

_{1}' →

<{ pred t

_{1}}> --> <{ pred t

_{1}' }>

| ST_Iszero0 :

<{ iszero 0 }> --> <{ true }>

| ST_IszeroSucc : ∀ v,

nvalue v →

<{ iszero (succ v) }> --> <{ false }>

| ST_Iszero : ∀ t

_{1}t

_{1}',

t

_{1}--> t

_{1}' →

<{ iszero t

_{1}}> --> <{ iszero t

_{1}' }>

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

succ (if true then true else true) can take a step (once, before becoming stuck).

*next*reduction step is being applied to the right kinds of operands. For example, the term succ true cannot take a step, but the almost as obviously nonsensical termsucc (if true then true else true) 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 : core.

Definition stuck (t : tm) : Prop :=

step_normal_form t ∧ ¬ value t.

Hint Unfold stuck : core.

*not*the same in this language, the set of values is a subset of the set of normal forms.

#### 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

*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.

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_True) | |

|-- true ∈ Bool |

(T_False) | |

|-- false ∈ Bool |

|-- t_{1} ∈ Bool |-- t_{2} ∈ T |-- t_{3} ∈ T |
(T_If) |

|-- if t_{1} then t_{2} else t_{3} ∈ T |

(T_0) | |

|-- 0 ∈ Nat |

|-- t_{1} ∈ Nat |
(T_Succ) |

|-- succ t_{1} ∈ Nat |

|-- t_{1} ∈ Nat |
(T_Pred) |

|-- pred t_{1} ∈ Nat |

|-- t_{1} ∈ Nat |
(T_Iszero) |

|-- iszero t_{1} ∈ Bool |

Inductive has_type : tm → ty → Prop :=

| T_True :

|-- <{ true }> \in Bool

| T_False :

|-- <{ false }> \in Bool

| T_If : ∀ t

_{1}t

_{2}t

_{3}T,

|-- t

_{1}\in Bool →

|-- t

_{2}\in T →

|-- t

_{3}\in T →

|-- <{ if t

_{1}then t

_{2}else t

_{3}}> \in T

| T_0 :

|-- <{ 0 }> \in Nat

| T_Succ : ∀ t

_{1},

|-- t

_{1}\in Nat →

|-- <{ succ t

_{1}}> \in Nat

| T_Pred : ∀ t

_{1},

|-- t

_{1}\in Nat →

|-- <{ pred t

_{1}}> \in Nat

| T_Iszero : ∀ t

_{1},

|-- t

_{1}\in Nat →

|-- <{ iszero t

_{1}}> \in Bool

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

Hint Constructors has_type : core.

Example has_type_1 :

|-- <{ if false then 0 else (succ 0) }> \in Nat.

Proof.

apply T_If.

- apply T_False.

- apply T_0.

- apply T_Succ. apply T_0.

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 :

¬ ( |-- <{ if false then 0 else true}> \in Bool ).

¬ ( |-- <{ if false then 0 else true}> \in Bool ).

Proof.

intros Contra. solve_by_inverts 2. Qed.

intros Contra. solve_by_inverts 2. Qed.

Example succ_hastype_nat__hastype_nat : ∀ t,

|-- <{succ t}> \in Nat →

|-- t \in Nat.

Proof.

(* FILL IN HERE *) Admitted.

☐

|-- <{succ t}> \in Nat →

|-- t \in Nat.

Proof.

(* FILL IN HERE *) Admitted.

☐

### Canonical forms

Lemma bool_canonical : ∀ t,

|-- t \in Bool → value t → bvalue t.

Lemma nat_canonical : ∀ t,

|-- t \in Nat → value t → nvalue 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

*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_True and

T_False, are eliminated immediately by auto *)

- (* T_If *)

right. destruct IHHT1.

+ (* t

apply (bool_canonical t

destruct H.

× ∃ t

× ∃ t

+ (* t

destruct H as [t

∃ (<{ if t

(* FILL IN HERE *) Admitted.

intros t T HT.

induction HT; auto.

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

T_False, are eliminated immediately by auto *)

- (* T_If *)

right. destruct IHHT1.

+ (* t

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

_{1}HT_{1}) in H.destruct H.

× ∃ t

_{2}. auto.× ∃ t

_{3}. auto.+ (* t

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

_{1}' H_{1}].∃ (<{ if t

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

☐

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

Complete the corresponding informal proof:*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.- If the last rule in the derivation is T_If, then t = if t
_{1}then t_{2}else t_{3}, with |-- t_{1}\in Bool, |-- t_{2}\in T and |-- t_{3}\in 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}\in Bool we have that t_{1}is a bvalue -- i.e., it is either true or false. If t_{1}= true, then t steps to t_{2}by ST_IfTrue, while if t_{1}= false, then t steps to t_{3}by ST_IfFalse. Either way, t can step, which is what we wanted to show. - If t
_{1}itself can take a step, then, by ST_If, 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)

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_If *) inversion HE; subst; clear HE.

+ (* ST_IFTrue *) assumption.

+ (* ST_IfFalse *) assumption.

+ (* ST_If *) apply T_If; 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_If *) inversion HE; subst; clear HE.

+ (* ST_IFTrue *) assumption.

+ (* ST_IfFalse *) assumption.

+ (* ST_If *) apply T_If; try assumption.

apply IHHT1; assumption.

(* FILL IN HERE *) Admitted.

☐

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

Complete the following informal proof:*Theorem*: If |-- t \in T and t --> t', then |-- t' \in T.*Proof*: By induction on a derivation of |-- t \in T.- If the last rule in the derivation is T_If, then t = if t
_{1}then t_{2}else t_{3}, with |-- t_{1}\in Bool, |-- t_{2}\in T and |-- t_{3}\in T.- If the last rule was ST_IfTrue, then t' = t
_{2}. But we know that |-- t_{2}\in T, so we are done. - If the last rule was ST_IfFalse, then t' = t
_{3}. But we know that |-- t_{3}\in T, so we are done. - If the last rule was ST_If, then t' = if t
_{1}' then t_{2}else t_{3}, where t_{1}--> t_{1}'. We know |-- t_{1}\in Bool so, by the IH, |-- t_{1}' \in Bool. The T_If rule then gives us |-- if t_{1}' then t_{2}else t_{3}\in T, as required.

- If the last rule was ST_IfTrue, 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 \in T →

t --> t' →

|-- t' \in T.

Proof with eauto.

(* FILL IN HERE *) Admitted.

☐

|-- t \in T →

t --> t' →

|-- t' \in 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 \in T →

t -->* t' →

~(stuck t').

Notation "t

_{1}'-->*' t_{2}" := (multistep t_{1}t_{2}) (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: 3 stars, standard, especially useful (subject_expansion)

Having seen the subject reduction property, one might wonder whether the opposite 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.

Theorem subject_expansion:

(∀ t t' T, t --> t' ∧ |-- t' \in T → |-- t \in T)

∨

¬ (∀ t t' T, t --> t' ∧ |-- t' \in T → |-- t \in T).

Proof.

(* FILL IN HERE *) Admitted.

☐

(∀ t t' T, t --> t' ∧ |-- t' \in T → |-- t \in T)

∨

¬ (∀ t t' T, t --> t' ∧ |-- t' \in T → |-- t \in T).

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 2 stars, standard (variation1)

Suppose that we add this new rule to the typing relation:| T_SuccBool : ∀ t,

|-- t \in Bool →

|-- <{ succ t }> \in 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.

- 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

_{2}t

_{3},

(<{ if true then t

_{2}else t

_{3}}>) --> t

_{3}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 *)

(* 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

_{1}t

_{2}t

_{2}' t

_{3},

t

_{2}--> t

_{2}' →

(<{ if t

_{1}then t

_{2}else t

_{3}}>) --> (<{ if t

_{1}then t

_{2}' else t

_{3}}>) 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 *)

☐

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

Suppose instead that we add this rule:| ST_Funny3 :

(<{pred false}>) --> (<{ pred (pred false)}>) 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 *)

☐

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

Suppose instead that we add this rule:| T_Funny4 :

|-- <{ 0 }> \in 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 *)

☐

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

Suppose instead that we add this rule:| T_Funny5 :

|-- <{ pred 0 }> \in 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 *)

☐

#### 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_pred0)

The reduction rule ST_Pred0 is a bit counter-intuitive: we might feel that it makes more sense for the predecessor of 0 to be undefined, rather than being defined to be 0. 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_pred0 : option (nat×string) := None.

☐

Definition manual_grade_for_remove_pred0 : option (nat×string) := None.

☐