# TypesType Systems

Our next topic, a large one, is
Before we start, let's spend a little time learning to use
some of Coq's more powerful automation features...
The auto tactic solves goals that are solvable by any combination of
The eauto tactic works just like auto, except that it uses
eapply instead of apply.
Using auto is always "safe" in the sense that it will never fail
and will never change the proof state: either it completely solves
the current goal, or it does nothing.
Here is a contrived example:

*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.
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 l.

at the top level, where l is a top-level lemma 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.
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.v.
If you start a proof by saying Proof with (tactic) instead of
just Proof, then writing ... instead of . after a tactic in
the body of the proof will try to solve all generated subgoals
with tactic (and fail if this doesn't work).
One common use of this facility is "Proof with auto" (or
eauto). We'll see many examples of this later in the file.
Here's another nice automation feature: it often arises that the
context contains a contradictory assumption and we want to use
inversion on it to solve the goal. We'd like to be able to say
to Coq, "find a contradictory assumption and invert it" without
giving its name explicitly.
Doing solve by inversion will find a hypothesis that can be
inverted to solve the goal, if there is one. The tactics solve
by inversion 2 and solve by inversion 3 are slightly fancier
versions which will perform two or three inversions in a row, if
necessary, to solve the goal.
(These tactics are not actually built into Coq — their
definitions are in Sflib.v.)
Caution: Overuse of solve by inversion can lead to slow proof
scripts.
If t is a tactic, then try solve [t] is a tactic that
More generally, try solve [t1 | t2 | ...] will try to solve the
goal by using t1, t2, etc. If none of them succeeds in
completely solving the goal, then try solve [t1 | t2 | ...] does
nothing.
The fact that it does nothing when it doesn't completely succeed
in solving the goal means that there is no harm in using try
solve T in situations where T might actually make no sense. In
particular, if we put it after a ; it will solve as many
subgoals as it can and leave the rest for us to solve by other
methods. It will not leave any of the subgoals in a partially
solved state.
This handy tactic replaces a goal of the form f x1 x2 ... xn = f
y1 y2 ... yn, where f is some function, with the subgoals x1 =
y1, x2 = y2,...,xn = yn. It is useful for avoiding explicit
rewriting steps, and often the generated subgoals can be quickly
cleared by auto.
When experimenting with definitions of programming languages in
Coq, we often want to see what a particular concrete term steps
to — i.e., we want to find proofs for goals of the form t ⇒*
t', where t is a completely concrete term and t' is unknown.
These proofs are simple but repetitive to do by hand. Consider for
example reducing an arithmetic expression using the small-step
relation astep defined in the previous chapter:

## 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 astep_many st := refl_step_closure (astep st).

Notation " t '/' st '⇒a*' t' " := (astep_many 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 rsc_step with (APlus (ANum 3) (ANum 12)).

apply AS_Plus2.

apply av_num.

apply AS_Mult.

apply rsc_step with (ANum 15).

apply AS_Plus.

apply rsc_refl.

Qed.

We repeatedly applied rsc_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 rsc_step. auto. simpl.

eapply rsc_step. auto. simpl.

apply rsc_refl.

Qed.

The following custom Tactic Notation definition captures this
pattern. In addition, before each rsc_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 rsc_step ;

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

apply rsc_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. *)

Qed.

Finally, this is also useful as 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.

Qed.

# Typed Arithmetic Expressions

*not*using the asnum/aslist trick that we used in ImpList.v 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

Inductive tm : Type :=

| tm_true : tm

| tm_false : tm

| tm_if : tm → tm → tm → tm

| tm_zero : tm

| tm_succ : tm → tm

| tm_pred : tm → tm

| tm_iszero : tm → tm.

Inductive bvalue : tm → Prop :=

| bv_true : bvalue tm_true

| bv_false : bvalue tm_false.

Inductive nvalue : tm → Prop :=

| nv_zero : nvalue tm_zero

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

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

Hint Constructors bvalue nvalue.

Hint Unfold value.

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

Inductive step : tm → tm → Prop :=

| ST_IfTrue : ∀ t1 t2,

(tm_if tm_true t1 t2) ⇒ t1

| ST_IfFalse : ∀ t1 t2,

(tm_if tm_false t1 t2) ⇒ t2

| ST_If : ∀ t1 t1' t2 t3,

t1 ⇒ t1' →

(tm_if t1 t2 t3) ⇒ (tm_if t1' t2 t3)

| ST_Succ : ∀ t1 t1',

t1 ⇒ t1' →

(tm_succ t1) ⇒ (tm_succ t1')

| ST_PredZero :

(tm_pred tm_zero) ⇒ tm_zero

| ST_PredSucc : ∀ t1,

nvalue t1 →

(tm_pred (tm_succ t1)) ⇒ t1

| ST_Pred : ∀ t1 t1',

t1 ⇒ t1' →

(tm_pred t1) ⇒ (tm_pred t1')

| ST_IszeroZero :

(tm_iszero tm_zero) ⇒ tm_true

| ST_IszeroSucc : ∀ t1,

nvalue t1 →

(tm_iszero (tm_succ t1)) ⇒ tm_false

| ST_Iszero : ∀ t1 t1',

t1 ⇒ t1' →

(tm_iszero t1) ⇒ (tm_iszero 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.

Notice that the step relation doesn't care about whether
expressions make global sense — it just checks that the operation
in the
The first interesting thing about the step relation in this
language is that the strong progress theorem from the Smallstep
chapter fails! That is, there are terms that are normal
forms (they can't take a step) but not values (because we have not
included them in our definition of possible "results of
evaluation"). Such terms are

*next*reduction step is being applied to the right kinds of operands. For example, the term tm_succ tm_true cannot take a step, but the almost-as-obviously-nonsensical term
tm_succ (tm_if tm_true tm_true tm_true)

can.
## 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.

☐
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.
☐

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

Using value_is_nf, we can show that the step relation is also deterministic...
☐
The next critical observation about this language is that,
although there are stuck terms, they are all "nonsensical", mixing
booleans and numbers in a way that we don't even

## 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.
The typing relation:
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 tm_true ty_Bool

| T_False :

has_type tm_false ty_Bool

| T_If : ∀ t1 t2 t3 T,

has_type t1 ty_Bool →

has_type t2 T →

has_type t3 T →

has_type (tm_if t1 t2 t3) T

| T_Zero :

has_type tm_zero ty_Nat

| T_Succ : ∀ t1,

has_type t1 ty_Nat →

has_type (tm_succ t1) ty_Nat

| T_Pred : ∀ t1,

has_type t1 ty_Nat →

has_type (tm_pred t1) ty_Nat

| T_Iszero : ∀ t1,

has_type t1 ty_Nat →

has_type (tm_iszero t1) ty_Bool.

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.

### Examples

*conservative*(or

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

Example has_type_1 :

has_type (tm_if tm_false tm_zero (tm_succ tm_zero)) ty_Nat.

Proof.

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 (tm_if tm_false tm_zero tm_true) ty_Bool.

Proof.

intros Contra. solve by inversion 2. Qed.

Example succ_hastype_nat__hastype_nat : ∀ t,

has_type (tm_succ t) ty_Nat →

has_type t ty_Nat.

Proof.

(* FILL IN HERE *) Admitted.

has_type (tm_succ t) ty_Nat →

has_type t ty_Nat.

Proof.

(* FILL IN HERE *) Admitted.

☐
The typing relation enjoys two critical properties. The first is
that well-typed normal forms are values (i.e., not stuck).
(* FILL IN HERE *)

☐

## 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'.

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. destruct IHHT1.

SCase "t1 is a value". destruct H.

SSCase "t1 is a bvalue". destruct H.

SSSCase "t1 is tm_true".

∃ t2...

SSSCase "t1 is tm_false".

∃ t3...

SSCase "t1 is an nvalue".

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

SCase "t1 can take a step".

destruct H as [t1' H1].

∃ (tm_if t1' t2 t3)...

(* FILL IN HERE *) Admitted.

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. destruct IHHT1.

SCase "t1 is a value". destruct H.

SSCase "t1 is a bvalue". destruct H.

SSSCase "t1 is tm_true".

∃ t2...

SSSCase "t1 is tm_false".

∃ t3...

SSCase "t1 is an nvalue".

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

SCase "t1 can take a step".

destruct H as [t1' H1].

∃ (tm_if t1' t2 t3)...

(* FILL IN HERE *) Admitted.

☐
This is more interesting than the strong progress theorem that we
saw in the Smallstep chapter, where
☐
The second critical property of typing is that, when a well-typed
term takes a step, the result is also a well-typed term.
This theorem is often called the
(* FILL IN HERE *)

☐

*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*. (As usual, no need to hand these in.)- 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.
- 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.

☐
Putting progress and preservation together, we can see that a
well-typed term can

## Type Soundness

*never*reach a stuck state.Definition stepmany := (refl_step_closure step).

Notation "t1 '⇒*' t2" := (stepmany 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.

Indeed, in the present — extremely simple — language,
every well-typed term can be reduced to a normal form: this is the
normalization property. (The proof is straightforward, though
somewhat tedious.) In richer languages, this property often fails,
though there are some interesting languages (such as Coq's
Fixpoint language, and the simply typed lambda-calculus, which
we'll be looking at next) where all
(* FILL IN HERE *)

☐
☐
☐
☐
☐
☐
☐
☐
☐
(* FILL IN HERE *)

☐
(* FILL IN HERE *)

☐

*well-typed*terms can be reduced to normal forms.## 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.☐

#### Exercise: 2 stars (variation1)

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

(tm_pred tm_true) ⇒ (tm_pred tm_false)

| ST_PredFalse :

(tm_pred tm_false) ⇒ (tm_pred tm_true)

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.
(tm_pred tm_true) ⇒ (tm_pred tm_false)

| ST_PredFalse :

(tm_pred tm_false) ⇒ (tm_pred tm_true)

- Determinacy of step
- Normalization of step for well-typed terms
- 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 ty_Nat →

has_type (tm_if tm_true t2 t3) ty_Nat

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

has_type (tm_if tm_true t2 t3) ty_Nat

- Determinacy of step
- Normalization of step for well-typed terms
- Progress
- Preservation

#### Exercise: 2 stars (variation3)

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

has_type t ty_Bool →

has_type (tm_succ t) ty_Bool

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

has_type (tm_succ t) ty_Bool

- Determinacy of step
- Normalization of step for well-typed terms
- Progress
- Preservation

#### Exercise: 2 stars (variation4)

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

(tm_if tm_true 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.
(tm_if tm_true t2 t3) ⇒ t3

#### Exercise: 2 stars (variation5)

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

t2 ⇒ t2' →

(tm_if t1 t2 t3) ⇒ (tm_if 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' →

(tm_if t1 t2 t3) ⇒ (tm_if t1 t2' t3)

#### Exercise: 2 stars (variation6)

Suppose instead that we add this rule:
| ST_Funny3 :

(tm_pred tm_false) ⇒ (tm_pred (tm_pred tm_false))

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

#### Exercise: 2 stars (variation7)

Suppose instead that we add this rule:
| T_Funny4 :

has_type tm_zero ty_Bool

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

#### Exercise: 2 stars (variation8)

Suppose instead that we add this rule:
| T_Funny5 :

has_type (tm_pred tm_zero) ty_Bool

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

#### 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?☐

#### 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?☐