# TypesType Systems

(* $Date: 2011-06-03 13:58:55 -0400 (Fri, 03 Jun 2011) $ *)

Require Export Smallstep.

Our next topic, a large one, is

*type systems*— static program analyses that classify expressions according to the "shapes" of their results. We'll begin with a typed version of a very simple language with just booleans and numbers, to introduce the basic ideas of types, typing rules, and the fundamental theorems about type systems:*type preservation*and*progress*. Then we'll move on to the*simply typed lambda-calculus*, which lives at the core of every modern functional programming language (including Coq).# More Automation

## The auto and eauto Tactics

- intros,
- apply (with a local hypothesis, by default), and
- reflexivity.

Lemma auto_example_1 : ∀P Q R S T U : Prop,

(P → Q) →

(P → R) →

(T → R) →

(S → T → U) →

((P→Q) → (P→S)) →

T →

P →

U.

Proof. auto. Qed.

When searching for potential proofs of the current goal, auto
and eauto consider the hypotheses in the current context
together with a

*hint database*of other lemmas and constructors. Some of the lemmas and constructors we've already seen — e.g., conj, or_introl, and or_intror — are installed in this hint database by default.Lemma auto_example_2 : ∀P Q R : Prop,

Q →

(Q → R) →

P ∨ (Q ∧ R).

Proof.

auto. Qed.

We can extend the hint database just for the purposes of one
application of auto or eauto by writing auto using ....
E.g., if conj, or_introl, and or_intror had

*not*already been in the hint database, we could have done this instead:Lemma auto_example_2a : ∀P Q R : Prop,

Q →

(Q → R) →

P ∨ (Q ∧ R).

Proof.

auto using conj, or_introl, or_intror. Qed.

Of course, in any given development there will also be some of our
own specific constructors and lemmas that are used very often in
proofs. We can add these to the global hint database by writing
It is also sometimes necessary to add
Here are some Hints we will find useful.

Hint Resolve T.

at the top level, where T is a top-level theorem or a
constructor of an inductively defined proposition (i.e., anything
whose type is an implication). As a shorthand, we can write
Hint Constructors c.

to tell Coq to do a Hint Resolve for *all*of the constructors from the inductive definition of c.
Hint Unfold d.

where d is a defined symbol, so that auto knows to expand
uses of d and enable further possibilities for applying
lemmas that it knows about.
Hint Constructors multi.

Hint Resolve beq_id_eq beq_id_false_not_eq.

Warning: Just as with Coq's other automation facilities, it is
easy to overuse auto and eauto and wind up with proofs that
are impossible to understand later!
Also, overuse of eauto can make proof scripts very slow. Get in
the habit of using auto most of the time and eauto only when
necessary.
For much more detailed information about using auto and eauto,
see the chapter UseAuto.

## The Proof with Tactic

## The solve by inversion Tactic

## The try solve Tactic

- if t solves the goal, behaves just like t, or
- if t cannot completely solve the goal, does nothing.

## The f_equal Tactic

## The normalize Tactic

Definition amultistep st := multi (astep st).

Notation " t '/' st '⇒a*' t' " := (amultistep st t t')

(at level 40, st at level 39).

Notation " t '/' st '⇒a*' t' " := (amultistep st t t')

(at level 40, st at level 39).

Example astep_example1 :

(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state

⇒a* (ANum 15).

Proof.

apply multi_step with (APlus (ANum 3) (ANum 12)).

apply AS_Plus2.

apply av_num.

apply AS_Mult.

apply multi_step with (ANum 15).

apply AS_Plus.

apply multi_refl.

Qed.

We repeatedly applied multi_step until we got to a normal
form. The proofs that the intermediate steps are possible are
simple enough that auto, with appropriate hints, can solve
them.

Hint Constructors astep aval.

Example astep_example1' :

(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state

⇒a* (ANum 15).

Proof.

eapply multi_step. auto. simpl.

eapply multi_step. auto. simpl.

apply multi_refl.

Qed.

The following custom Tactic Notation definition captures this
pattern. In addition, before each multi_step we print out the
current goal, so that the user can follow how the term is being
evaluated.

Tactic Notation "print_goal" := match goal with ⊢ ?x => idtac x end.

Tactic Notation "normalize" :=

repeat (print_goal; eapply multi_step ;

[ (eauto 10; fail) | (instantiate; simpl)]);

apply multi_refl.

Example astep_example1'' :

(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state

⇒a* (ANum 15).

Proof.

normalize.

(* At this point in the proof script, the Coq response shows

a trace of how the expression evaluated.

(APlus (ANum 3) (AMult (ANum 3) (ANum 4)) / empty_state ==>a* ANum 15)

(multi (astep empty_state) (APlus (ANum 3) (ANum 12)) (ANum 15))

(multi (astep empty_state) (ANum 15) (ANum 15))

*)

Qed.

The normalize tactic also provides a simple way to calculate
what the normal form of a term is, by proving a goal with an
existential variable in it.

Example astep_example1''' : ∃e',

(APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state

⇒a* e'.

Proof.

eapply ex_intro. normalize.

(* This time the trace will be:

(APlus (ANum 3) (AMult (ANum 3) (ANum 4)) / empty_state ==>a* ??)

(multi (astep empty_state) (APlus (ANum 3) (ANum 12)) ??)

(multi (astep empty_state) (ANum 15) ??)

where ?? is the variable ``guessed'' by eapply.

*)

Qed.

Theorem normalize_ex : ∃e',

(AMult (ANum 3) (AMult (ANum 2) (ANum 1))) / empty_state

⇒a* e'.

Proof.

(* FILL IN HERE *) Admitted.

(AMult (ANum 3) (AMult (ANum 2) (ANum 1))) / empty_state

⇒a* e'.

Proof.

(* FILL IN HERE *) Admitted.

Theorem normalize_ex' : ∃e',

(AMult (ANum 3) (AMult (ANum 2) (ANum 1))) / empty_state

⇒a* e'.

Proof.

(* FILL IN HERE *) Admitted.

☐

# Typed Arithmetic Expressions

*not*using the asnum/aslist trick that we used in chapter HoareList to make all the operations total by forcibly coercing the arguments to + (for example) into numbers. Instead, we simply let terms get stuck if they try to use an operator with the wrong kind of operands: the step relation doesn't relate them to anything.

## Syntax

t ::= true

| false

| if t then t else t

| 0

| succ t

| pred t

| iszero t

Formally:
| false

| if t then t else t

| 0

| succ t

| pred t

| iszero t

Inductive tm : Type :=

| ttrue : tm

| tfalse : tm

| tif : tm → tm → tm → tm

| tzero : tm

| tsucc : tm → tm

| tpred : tm → tm

| tiszero : tm → tm.

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

Inductive bvalue : tm → Prop :=

| bv_true : bvalue ttrue

| bv_false : bvalue tfalse.

Inductive nvalue : tm → Prop :=

| nv_zero : nvalue tzero

| nv_succ : ∀t, nvalue t → nvalue (tsucc t).

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

Hint Constructors bvalue nvalue.

Hint Unfold value.

Hint Unfold value.

## Operational Semantics

(ST_IfTrue) | |

if true then t1 else t2 ⇒ t1 |

(ST_IfFalse) | |

if false then t1 else t2 ⇒ t2 |

t1 ⇒ t1' | (ST_If) |

if t1 then t2 else t3 ⇒ | |

if t1' then t2 else t3 |

t1 ⇒ t1' | (ST_Succ) |

succ t1 ⇒ succ t1' |

(ST_PredZero) | |

pred 0 ⇒ 0 |

numeric value v1 | (ST_PredSucc) |

pred (succ v1) ⇒ v1 |

t1 ⇒ t1' | (ST_Pred) |

pred t1 ⇒ pred t1' |

(ST_IszeroZero) | |

iszero 0 ⇒ true |

numeric value v1 | (ST_IszeroSucc) |

iszero (succ v1) ⇒ false |

t1 ⇒ t1' | (ST_Iszero) |

iszero t1 ⇒ iszero t1' |

Reserved Notation "t1 '⇒' t2" (at level 40).

Inductive step : tm → tm → Prop :=

| ST_IfTrue : ∀t1 t2,

(tif ttrue t1 t2) ⇒ t1

| ST_IfFalse : ∀t1 t2,

(tif tfalse t1 t2) ⇒ t2

| ST_If : ∀t1 t1' t2 t3,

t1 ⇒ t1' →

(tif t1 t2 t3) ⇒ (tif t1' t2 t3)

| ST_Succ : ∀t1 t1',

t1 ⇒ t1' →

(tsucc t1) ⇒ (tsucc t1')

| ST_PredZero :

(tpred tzero) ⇒ tzero

| ST_PredSucc : ∀t1,

nvalue t1 →

(tpred (tsucc t1)) ⇒ t1

| ST_Pred : ∀t1 t1',

t1 ⇒ t1' →

(tpred t1) ⇒ (tpred t1')

| ST_IszeroZero :

(tiszero tzero) ⇒ ttrue

| ST_IszeroSucc : ∀t1,

nvalue t1 →

(tiszero (tsucc t1)) ⇒ tfalse

| ST_Iszero : ∀t1 t1',

t1 ⇒ t1' →

(tiszero t1) ⇒ (tiszero t1')

where "t1 '⇒' t2" := (step t1 t2).

Tactic Notation "step_cases" tactic(first) ident(c) :=

first;

[ Case_aux c "ST_IfTrue" | Case_aux c "ST_IfFalse" | Case_aux c "ST_If"

| Case_aux c "ST_Succ" | Case_aux c "ST_PredZero"

| Case_aux c "ST_PredSucc" | Case_aux c "ST_Pred"

| Case_aux c "ST_IszeroZero" | Case_aux c "ST_IszeroSucc"

| Case_aux c "ST_Iszero" ].

Hint Constructors step.

first;

[ Case_aux c "ST_IfTrue" | Case_aux c "ST_IfFalse" | Case_aux c "ST_If"

| Case_aux c "ST_Succ" | Case_aux c "ST_PredZero"

| Case_aux c "ST_PredSucc" | Case_aux c "ST_Pred"

| Case_aux c "ST_IszeroZero" | Case_aux c "ST_IszeroSucc"

| Case_aux c "ST_Iszero" ].

Hint Constructors step.

Notice that the step relation doesn't care about whether
expressions make 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 succ true (i.e., tsucc ttrue in the formal syntax) cannot take a step, but the almost-as-obviously-nonsensical term
succ (if true then true else true)

can take *one*step.## 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.

Example some_term_is_stuck :

∃t, stuck t.

∃t, stuck t.

Proof.

(* FILL IN HERE *) Admitted.

(* FILL IN HERE *) Admitted.

☐
However, although values and normal forms are not the same in this
language, the former set is included in the latter. This is
important because it shows we did not accidentally define things
so that some value could still take a step.

#### Exercise: 3 stars, optional (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.Lemma value_is_nf : ∀t,

value t → step_normal_form t.

Proof.

(* FILL IN HERE *) Admitted.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 3 stars, optional (step_deterministic)

Using value_is_nf, we can show that the step relation is also deterministic...Theorem step_deterministic:

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 :=

| TBool : ty

| TNat : ty.

In informal notation, the typing relation is often written
⊢ t : T, pronounced "t has type T." The ⊢ symbol is
called a "turnstile". (Below, we're going to see richer typing
relations where an additional "context" argument is written to the
left of the turnstile. Here, the context is always empty.)

(T_True) | |

⊢ true : Bool |

(T_False) | |

⊢ false : Bool |

⊢ t1 : Bool ⊢ t2 : T ⊢ t3 : T | (T_If) |

⊢ if t1 then t2 else t3 : T |

(T_Zero) | |

⊢ 0 : Nat |

⊢ t1 : Nat | (T_Succ) |

⊢ succ t1 : Nat |

⊢ t1 : Nat | (T_Pred) |

⊢ pred t1 : Nat |

⊢ t1 : Nat | (T_IsZero) |

⊢ iszero t1 : Bool |

Inductive has_type : tm → ty → Prop :=

| T_True :

has_type ttrue TBool

| T_False :

has_type tfalse TBool

| T_If : ∀t1 t2 t3 T,

has_type t1 TBool →

has_type t2 T →

has_type t3 T →

has_type (tif t1 t2 t3) T

| T_Zero :

has_type tzero TNat

| T_Succ : ∀t1,

has_type t1 TNat →

has_type (tsucc t1) TNat

| T_Pred : ∀t1,

has_type t1 TNat →

has_type (tpred t1) TNat

| T_Iszero : ∀t1,

has_type t1 TNat →

has_type (tiszero t1) TBool.

Tactic Notation "has_type_cases" tactic(first) ident(c) :=

first;

[ Case_aux c "T_True" | Case_aux c "T_False" | Case_aux c "T_If"

| Case_aux c "T_Zero" | Case_aux c "T_Succ" | Case_aux c "T_Pred"

| Case_aux c "T_Iszero" ].

Hint Constructors has_type.

first;

[ Case_aux c "T_True" | Case_aux c "T_False" | Case_aux c "T_If"

| Case_aux c "T_Zero" | Case_aux c "T_Succ" | Case_aux c "T_Pred"

| Case_aux c "T_Iszero" ].

Hint Constructors has_type.

### Examples

*conservative*(or

*static*) approximation: it does not calculate the type of the normal form of a term.

Example has_type_1 :

has_type (tif tfalse tzero (tsucc tzero)) TNat.

Proof.

apply T_If.

apply T_False.

apply T_Zero.

apply T_Succ.

apply T_Zero.

Qed.

apply T_If.

apply T_False.

apply T_Zero.

apply T_Succ.

apply T_Zero.

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.)

Example has_type_not :

~ has_type (tif tfalse tzero ttrue) TBool.

Proof.

intros Contra. solve by inversion 2. Qed.

intros Contra. solve by inversion 2. Qed.

Example succ_hastype_nat__hastype_nat : ∀t,

has_type (tsucc t) TNat →

has_type t TNat.

Proof.

(* FILL IN HERE *) Admitted.

has_type (tsucc t) TNat →

has_type t TNat.

Proof.

(* FILL IN HERE *) Admitted.

☐

## Progress

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

Complete the following 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_If, then t = if t1
then t2 else t3, with ⊢ t1 : Bool, ⊢ t2 : T and ⊢ t3
: T. By the IH, either t1 is a value or else t1 can step
to some t1'.
- If t1 is a value, then it is either an nvalue or a
bvalue. But it cannot be an nvalue, because we know
⊢ t1 : Bool and there are no rules assigning type
Bool to any term that could be an nvalue. So t1
is a bvalue — i.e., it is either true or false.
If t1 = true, then t steps to t2 by ST_IfTrue,
while if t1 = false, then t steps to t3 by
ST_IfFalse. Either way, t can step, which is what
we wanted to show.
- If t1 itself can take a step, then, by ST_If, so can t.

- If t1 is a value, then it is either an nvalue or a
bvalue. But it cannot be an nvalue, because we know
⊢ t1 : Bool and there are no rules assigning type
Bool to any term that could be an nvalue. So t1
is a bvalue — i.e., it is either true or false.
If t1 = true, then t steps to t2 by ST_IfTrue,
while if t1 = false, then t steps to t3 by
ST_IfFalse. Either way, t can step, which is what
we wanted to show.

☐

#### Exercise: 3 stars (finish_progress)

Theorem progress : ∀t T,

has_type t T →

value t ∨ ∃t', t ⇒ t'.

has_type t T →

value t ∨ ∃t', t ⇒ t'.

Proof with auto.

intros t T HT.

has_type_cases (induction HT) Case...

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

T_False, were eliminated immediately by auto *)

Case "T_If".

right. inversion IHHT1; clear IHHT1.

SCase "t1 is a value". inversion H; clear H.

SSCase "t1 is a bvalue". inversion H0; clear H0.

SSSCase "t1 is ttrue".

∃t2...

SSSCase "t1 is tfalse".

∃t3...

SSCase "t1 is an nvalue".

solve by inversion 2. (* on H and HT1 *)

SCase "t1 can take a step".

inversion H as [t1' H1].

∃(tif t1' t2 t3)...

(* FILL IN HERE *) Admitted.

intros t T HT.

has_type_cases (induction HT) Case...

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

T_False, were eliminated immediately by auto *)

Case "T_If".

right. inversion IHHT1; clear IHHT1.

SCase "t1 is a value". inversion H; clear H.

SSCase "t1 is a bvalue". inversion H0; clear H0.

SSSCase "t1 is ttrue".

∃t2...

SSSCase "t1 is tfalse".

∃t3...

SSCase "t1 is an nvalue".

solve by inversion 2. (* on H and HT1 *)

SCase "t1 can take a step".

inversion H as [t1' H1].

∃(tif t1' t2 t3)...

(* FILL IN HERE *) Admitted.

☐

This is more interesting than the strong progress theorem that we
saw in the Smallstep chapter, where
☐

*all*normal forms were values. Here, a term can be stuck, but only if it is ill typed.#### Exercise: 1 star (step_review)

Quick review. Answer*true*or*false*. In this language...- Every well-typed normal form is a value.
- Every value is a normal form.
- The single-step evaluation relation is
a partial function (i.e., it is deterministic).
- The single-step evaluation relation is a
*total*function.

## Type Preservation

*subject reduction*property, 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.

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

Complete the following 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_If, then t = if t1
then t2 else t3, with ⊢ t1 : Bool, ⊢ t2 : T and ⊢ t3
: T.
- If the last rule was ST_IfTrue, then t' = t2. But we
know that ⊢ t2 : T, so we are done.
- If the last rule was ST_IfFalse, then t' = t3. But we
know that ⊢ t3 : T, so we are done.
- If the last rule was ST_If, then t' = if t1' then t2 else t3, where t1 ⇒ t1'. We know ⊢ t1 : Bool so, by the IH, ⊢ t1' : Bool. The T_If rule then gives us ⊢ if t1' then t2 else t3 : T, as required.

- If the last rule was ST_IfTrue, then t' = t2. But we
know that ⊢ t2 : T, so we are done.

☐

#### Exercise: 2 stars (finish_preservation)

Theorem preservation : ∀t t' T,

has_type t T →

t ⇒ t' →

has_type t' T.

Proof with auto.

intros t t' T HT HE.

generalize dependent t'.

has_type_cases (induction HT) Case;

(* 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 inversion).

Case "T_If". inversion HE; subst.

SCase "ST_IFTrue". assumption.

SCase "ST_IfFalse". assumption.

SCase "ST_If". apply T_If; try assumption.

apply IHHT1; assumption.

(* FILL IN HERE *) Admitted.

has_type t T →

t ⇒ t' →

has_type t' T.

Proof with auto.

intros t t' T HT HE.

generalize dependent t'.

has_type_cases (induction HT) Case;

(* 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 inversion).

Case "T_If". inversion HE; subst.

SCase "ST_IFTrue". assumption.

SCase "ST_IfFalse". assumption.

SCase "ST_If". apply T_If; try assumption.

apply IHHT1; assumption.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 3 stars (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 proof 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,

has_type t T →

t ⇒ t' →

has_type t' T.

Proof with eauto.

(* FILL IN HERE *) Admitted.

☐

## Type Soundness

*never*reach a stuck state.

Definition multistep := (multi step).

Notation "t1 '⇒*' t2" := (multistep t1 t2) (at level 40).

Notation "t1 '⇒*' t2" := (multistep t1 t2) (at level 40).

Corollary soundness : ∀t t' T,

has_type 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, recommended (subject_expansion)

Having seen the subject reduction property, it is reasonable to wonder whether the opposity property — subject*expansion*— also holds. That is, is it always the case that, if t ⇒ t' and has_type t' T, then has_type 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 if you like.)

☐

#### Exercise: 2 stars (variation1)

Suppose we add the following two new rules to the reduction relation:
| ST_PredTrue :

(tpred ttrue) ⇒ (tpred tfalse)

| ST_PredFalse :

(tpred tfalse) ⇒ (tpred ttrue)

Which of the following properties remain true in the presence
of these rules? For each one, write either "remains true" or
else "becomes false." If a property becomes false, give a
counterexample.
(tpred ttrue) ⇒ (tpred tfalse)

| ST_PredFalse :

(tpred tfalse) ⇒ (tpred ttrue)

- Determinism of step
- Progress
- Preservation

#### Exercise: 2 stars (variation2)

Suppose, instead, that we add this new rule to the typing relation:
| T_IfFunny : ∀ t2 t3,

has_type t2 TNat →

has_type (tif ttrue t2 t3) TNat

Which of the following properties remain true in the presence of
this rule? (Answer in the same style as above.)
has_type t2 TNat →

has_type (tif ttrue t2 t3) TNat

- Determinism of step
- Progress
- Preservation

#### Exercise: 2 stars (variation3)

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

has_type t TBool →

has_type (tsucc t) TBool

Which of the following properties remain true in the presence of
this rule? (Answer in the same style as above.)
has_type t TBool →

has_type (tsucc t) TBool

- Determinism of step
- Progress
- Preservation

#### Exercise: 2 stars (variation4)

Suppose, instead, that we add this new rule to the step relation:
| ST_Funny1 : ∀ t2 t3,

(tif ttrue t2 t3) ⇒ t3

Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
(tif ttrue t2 t3) ⇒ t3

#### Exercise: 2 stars (variation5)

Suppose instead that we add this rule:
| ST_Funny2 : ∀ t1 t2 t2' t3,

t2 ⇒ t2' →

(tif t1 t2 t3) ⇒ (tif t1 t2' t3)

Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
t2 ⇒ t2' →

(tif t1 t2 t3) ⇒ (tif t1 t2' t3)

#### Exercise: 2 stars (variation6)

Suppose instead that we add this rule:
| ST_Funny3 :

(tpred tfalse) ⇒ (tpred (tpred tfalse))

Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
(tpred tfalse) ⇒ (tpred (tpred tfalse))

#### Exercise: 2 stars (variation7)

Suppose instead that we add this rule:
| T_Funny4 :

has_type tzero TBool

Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
has_type tzero TBool

#### Exercise: 2 stars (variation8)

Suppose instead that we add this rule:
| T_Funny5 :

has_type (tpred tzero) TBool

Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
has_type (tpred tzero) TBool

#### Exercise: 3 stars, 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. ☐#### Exercise: 1 star (remove_predzero)

The evaluation rule E_PredZero is a bit counter-intuitive: we might feel that it makes more sense for the predecessor of zero to be undefined, rather than being defined to be zero. Can we achieve this simply by removing the rule from the definition of step? Would doing so create any problems elsewhere?☐

#### Exercise: 4 stars, optional (prog_pres_bigstep)

Suppose our evaluation relation is defined in the big-step style. What are the appropriate analogs of the progress and preservation properties?☐