# 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".

Require Import Coq.Arith.Arith.

From PLF Require Import Maps.

From PLF Require Import Imp.

From PLF Require Import Smallstep.

Hint Constructors multi.

Require Import Coq.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 ::= true

| false

| if t then t else t

| 0

| succ t

| pred t

| iszero t

And here it is 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.

| 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 update.

| 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 update.

## 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_PredZero) | |

pred 0 ==> 0 |

numeric value v_{1} |
(ST_PredSucc) |

pred (succ v_{1}) ==> v_{1} |

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

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

(ST_IszeroZero) | |

iszero 0 ==> true |

numeric value v_{1} |
(ST_IszeroSucc) |

iszero (succ v_{1}) ==> false |

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

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

Reserved Notation "t

Inductive step : tm → tm → Prop :=

| ST_IfTrue : ∀ t

(tif ttrue t

| ST_IfFalse : ∀ t

(tif tfalse t

| ST_If : ∀ t

t

(tif t

| ST_Succ : ∀ t

t

(tsucc t

| ST_PredZero :

(tpred tzero) ==> tzero

| ST_PredSucc : ∀ t

nvalue t

(tpred (tsucc t

| ST_Pred : ∀ t

t

(tpred t

| ST_IszeroZero :

(tiszero tzero) ==> ttrue

| ST_IszeroSucc : ∀ t

nvalue t

(tiszero (tsucc t

| ST_Iszero : ∀ t

t

(tiszero t

where "t

Hint Constructors step.

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

| ST_IfTrue : ∀ t

_{1}t_{2},(tif ttrue t

_{1}t_{2}) ==> t_{1}| ST_IfFalse : ∀ t

_{1}t_{2},(tif tfalse t

_{1}t_{2}) ==> t_{2}| ST_If : ∀ t

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

_{1}==> t_{1}' →(tif t

_{1}t_{2}t_{3}) ==> (tif t_{1}' t_{2}t_{3})| ST_Succ : ∀ t

_{1}t_{1}',t

_{1}==> t_{1}' →(tsucc t

_{1}) ==> (tsucc t_{1}')| ST_PredZero :

(tpred tzero) ==> tzero

| ST_PredSucc : ∀ t

_{1},nvalue t

_{1}→(tpred (tsucc t

_{1})) ==> t_{1}| ST_Pred : ∀ t

_{1}t_{1}',t

_{1}==> t_{1}' →(tpred t

_{1}) ==> (tpred t_{1}')| ST_IszeroZero :

(tiszero tzero) ==> ttrue

| ST_IszeroSucc : ∀ t

_{1},nvalue t

_{1}→(tiszero (tsucc t

_{1})) ==> tfalse| ST_Iszero : ∀ t

_{1}t_{1}',t

_{1}==> t_{1}' →(tiszero t

_{1}) ==> (tiszero 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
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 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 included in 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 (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, 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 :=

| TBool : ty

| TNat : ty.

| TBool : ty

| TNat : 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_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_Zero) | |

|- 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 |

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

Inductive has_type : tm → ty → Prop :=

| T_True :

|- ttrue ∈ TBool

| T_False :

|- tfalse ∈ TBool

| T_If : ∀ t

|- t

|- t

|- t

|- tif t

| T_Zero :

|- tzero ∈ TNat

| T_Succ : ∀ t

|- t

|- tsucc t

| T_Pred : ∀ t

|- t

|- tpred t

| T_Iszero : ∀ t

|- t

|- tiszero t

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

Hint Constructors has_type.

Example has_type_1 :

|- tif tfalse tzero (tsucc tzero) ∈ TNat.

Proof.

apply T_If.

- apply T_False.

- apply T_Zero.

- apply T_Succ.

+ apply T_Zero.

Qed.

Inductive has_type : tm → ty → Prop :=

| T_True :

|- ttrue ∈ TBool

| T_False :

|- tfalse ∈ TBool

| T_If : ∀ t

_{1}t_{2}t_{3}T,|- t

_{1}∈ TBool →|- t

_{2}∈ T →|- t

_{3}∈ T →|- tif t

_{1}t_{2}t_{3}∈ T| T_Zero :

|- tzero ∈ TNat

| T_Succ : ∀ t

_{1},|- t

_{1}∈ TNat →|- tsucc t

_{1}∈ TNat| T_Pred : ∀ t

_{1},|- t

_{1}∈ TNat →|- tpred t

_{1}∈ TNat| T_Iszero : ∀ t

_{1},|- t

_{1}∈ TNat →|- tiszero t

_{1}∈ TBoolwhere "'|-' t '∈' T" := (has_type t T).

Hint Constructors has_type.

Example has_type_1 :

|- tif tfalse tzero (tsucc tzero) ∈ TNat.

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.)
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 :

¬ (|- tif tfalse tzero ttrue ∈ TBool).

¬ (|- tif tfalse tzero ttrue ∈ TBool).

Proof.

intros Contra. solve_by_inverts 2. Qed.

intros Contra. solve_by_inverts 2. Qed.

Example succ_hastype_nat__hastype_nat : ∀ t,

|- tsucc t ∈ TNat →

|- t ∈ TNat.

Proof.

(* FILL IN HERE *) Admitted.

☐
|- tsucc t ∈ TNat →

|- t ∈ TNat.

Proof.

(* FILL IN HERE *) Admitted.

### Canonical forms

Lemma bool_canonical : ∀ t,

|- t ∈ TBool → value t → bvalue t.

Lemma nat_canonical : ∀ t,

|- t ∈ TNat → value t → nvalue t.

|- t ∈ TBool → value t → bvalue t.

Proof.

intros t HT HV.

inversion HV; auto.

induction H; inversion HT; auto.

Qed.

intros t HT HV.

inversion HV; auto.

induction H; inversion HT; auto.

Qed.

Lemma nat_canonical : ∀ t,

|- t ∈ TNat → value t → nvalue t.

Proof.

intros t HT HV.

inversion HV.

inversion H; subst; inversion HT.

auto.

Qed.

intros t HT HV.

inversion HV.

inversion H; subst; inversion HT.

auto.

Qed.

## Progress

*progress*.

#### Exercise: 3 stars (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_True and

T_False, were eliminated immediately by auto *)

- (* T_If *)

right. inversion IHHT1; clear IHHT1.

+ (* t

apply (bool_canonical t

inversion H; subst; clear H.

∃ t

∃ t

+ (* t

inversion H as [t

∃ (tif t

(* FILL IN HERE *) Admitted.

intros t T HT.

induction HT...

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

T_False, were eliminated immediately by auto *)

- (* T_If *)

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}].∃ (tif 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_If, then t = if 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 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 (prod nat string) := None.

☐
Definition manual_grade_for_finish_progress_informal : option (prod 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 (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_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 ∈ 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 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_IfTrue, then t' = t
_{2}. But we know that |- t_{2}∈ T, so we are done. - If the last rule was ST_IfFalse, then t' = t
_{3}. But we know that |- t_{3}∈ 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}∈ Bool so, by the IH, |- t_{1}' ∈ Bool. The T_If rule then gives us |- if t_{1}' then t_{2}else t_{3}∈ 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 (prod nat string) := None.

☐
Definition manual_grade_for_finish_preservation_informal : option (prod nat string) := None.

#### 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 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.

# Aside: the normalize Tactic

Module NormalizePlayground.

Import Smallstep.

Example step_example1 :

(P (C 3) (P (C 3) (C 4)))

==>* (C 10).

Proof.

apply multi_step with (P (C 3) (C 7)).

apply ST_Plus2.

apply v_const.

apply ST_PlusConstConst.

apply multi_step with (C 10).

apply ST_PlusConstConst.

apply multi_refl.

Qed.

Import Smallstep.

Example step_example1 :

(P (C 3) (P (C 3) (C 4)))

==>* (C 10).

Proof.

apply multi_step with (P (C 3) (C 7)).

apply ST_Plus2.

apply v_const.

apply ST_PlusConstConst.

apply multi_step with (C 10).

apply ST_PlusConstConst.

apply multi_refl.

Qed.

The proof repeatedly applies multi_step until the term reaches a
normal form. Fortunately The sub-proofs for the intermediate
steps are simple enough that auto, with appropriate hints, can
solve them.

Hint Constructors step value.

Example step_example1' :

(P (C 3) (P (C 3) (C 4)))

==>* (C 10).

Proof.

eapply multi_step. auto. simpl.

eapply multi_step. auto. simpl.

apply multi_refl.

Qed.

Example step_example1' :

(P (C 3) (P (C 3) (C 4)))

==>* (C 10).

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 step, we print out the current
goal, so that we can follow how the term is being reduced.

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 step_example1'' :

(P (C 3) (P (C 3) (C 4)))

==>* (C 10).

Proof.

normalize.

(* The print_goal in the normalize tactic shows

a trace of how the expression reduced...

(P (C 3) (P (C 3) (C 4)) ==>* C 10)

(P (C 3) (C 7) ==>* C 10)

(C 10 ==>* C 10)

*)

Qed.

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 step_example1'' :

(P (C 3) (P (C 3) (C 4)))

==>* (C 10).

Proof.

normalize.

(* The print_goal in the normalize tactic shows

a trace of how the expression reduced...

(P (C 3) (P (C 3) (C 4)) ==>* C 10)

(P (C 3) (C 7) ==>* C 10)

(C 10 ==>* C 10)

*)

Qed.

The normalize tactic also provides a simple way to calculate the
normal form of a term, by starting with a goal with an existentially
bound variable.

Example step_example1''' : ∃ e',

(P (C 3) (P (C 3) (C 4)))

==>* e'.

Proof.

eapply ex_intro. normalize.

(* This time, the trace is:

(P (C 3) (P (C 3) (C 4)) ==>* ?e')

(P (C 3) (C 7) ==>* ?e')

(C 10 ==>* ?e')

where ?e' is the variable ``guessed'' by eapply. *)

Qed.

(P (C 3) (P (C 3) (C 4)))

==>* e'.

Proof.

eapply ex_intro. normalize.

(* This time, the trace is:

(P (C 3) (P (C 3) (C 4)) ==>* ?e')

(P (C 3) (C 7) ==>* ?e')

(C 10 ==>* ?e')

where ?e' is the variable ``guessed'' by eapply. *)

Qed.

Theorem normalize_ex : ∃ e',

(P (C 3) (P (C 2) (C 1)))

==>* e'.

Proof.

(* FILL IN HERE *) Admitted.

☐
(P (C 3) (P (C 2) (C 1)))

==>* e'.

Proof.

(* FILL IN HERE *) Admitted.

Theorem normalize_ex' : ∃ e',

(P (C 3) (P (C 2) (C 1)))

==>* e'.

Proof.

(* FILL IN HERE *) Admitted.

☐
(P (C 3) (P (C 2) (C 1)))

==>* e'.

Proof.

(* FILL IN HERE *) Admitted.

End NormalizePlayground.

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.

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.

## Additional Exercises

#### Exercise: 2 stars, 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 (prod nat string) := None.

☐
Definition manual_grade_for_subject_expansion : option (prod nat string) := None.

#### Exercise: 2 stars (variation1)

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

|- t ∈ TBool →

|- tsucc t ∈ TBool

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 ∈ TBool →

|- tsucc t ∈ TBool

- Determinism of step
- Progress
- Preservation

(* Do not modify the following line: *)

Definition manual_grade_for_variation1 : option (prod nat string) := None.

☐
Definition manual_grade_for_variation1 : option (prod nat string) := None.

#### Exercise: 2 stars (variation2)

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

(tif ttrue t

Which of the above properties become false in the presence of
this rule? For each one that does, give a counter-example.
_{2}t_{3},(tif ttrue t

_{2}t_{3}) ==> t_{3}
(* Do not modify the following line: *)

Definition manual_grade_for_variation2 : option (prod nat string) := None.

☐
Definition manual_grade_for_variation2 : option (prod nat string) := None.

#### Exercise: 2 stars, optional (variation3)

Suppose instead that we add this rule:
| ST_Funny2 : ∀ t

t

(tif t

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

_{2}==> t_{2}' →(tif t

_{1}t_{2}t_{3}) ==> (tif t_{1}t_{2}' t_{3})#### Exercise: 2 stars, optional (variation4)

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, optional (variation5)

Suppose instead that we add this rule:
| T_Funny4 :

|- tzero ∈ TBool

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

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

Suppose instead that we add this rule:
| T_Funny5 :

|- 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.
|- 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 reduction rule ST_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?
(* Do not modify the following line: *)

Definition manual_grade_for_remove_predzero : option (prod nat string) := None.

☐
Definition manual_grade_for_remove_predzero : option (prod 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 (prod nat string) := None.

☐
Definition manual_grade_for_prog_pres_bigstep : option (prod nat string) := None.