Types.v

(** * Types: Type 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 *)

(** Before we start, let's spend a little time learning to use
    some of Coq's more powerful automation features... *)

(* ###################################################################### *)
(** ** The [auto] and [eauto] Tactics *)

(** The [auto] tactic solves goals that are solvable by any combination of 
     - [intros],
     - [apply] (with a local hypothesis, by default), and
     - [reflexivity].
       
    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: *)

Lemma auto_example_1 : forall 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 : forall 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 : forall 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
[[
      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].

    It is also sometimes necessary to add
[[
      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. *)

(** Here are some [Hint]s we will find useful. *)

Hint Constructors refl_step_closure.
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.v]. *)

(* ###################################################################### *)
(** ** The [Proof with] Tactic *)

(** 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. *)

(* ###################################################################### *)
(** ** The [solve by inversion] Tactic *)

(** 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. *)

(* ###################################################################### *)
(** ** The [try solve] Tactic *)

(** If [t] is a tactic, then [try solve [t]] is a tactic that
      - if [t] solves the goal, behaves just like [t], or
      - if [t] cannot completely solve the goal, does
        nothing.

    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. *)

(* ###################################################################### *)
(** ** The [f_equal] Tactic *)

(** 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]. *)

(* ###################################################################### *)
(** ** The [normalize] Tactic *)

(** 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: *)

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''' : exists e',
  (APlus (ANum 3) (AMult (ANum 3) (ANum 4))) / empty_state 
  ==>a* e'.
Proof.
  eapply ex_intro. normalize.
Qed.

(* ###################################################################### *)
(** * Typed Arithmetic Expressions *)

(** To motivate the discussion of type systems, let's begin as usual
    with an extremely simple toy language.  We want it to have the
    potential for programs "going wrong" because of runtime type
    errors, so we need something a tiny bit more complex than the
    language of constants and addition that we used in [Smallstep.v]:
    a single kind of data (just numbers) is too simple, but just two
    kinds (numbers and booleans) already gives us enough material to
    tell an interesting story. 

    The language definition is completely routine.  The only thing to
    notice is that we are _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 : forall t, nvalue t -> nvalue (tm_succ t).

Definition value (t:tm) := bvalue t \/ nvalue t.

Hint Constructors bvalue nvalue.
Hint Unfold value.  

(* ###################################################################### *)
(** ** Small-Step Reduction *)

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

Inductive step : tm -> tm -> Prop :=
  | ST_IfTrue : forall t1 t2,
      (tm_if tm_true t1 t2) ==> t1
  | ST_IfFalse : forall t1 t2,
      (tm_if tm_false t1 t2) ==> t2
  | ST_If : forall t1 t1' t2 t3,
      t1 ==> t1' ->
      (tm_if t1 t2 t3) ==> (tm_if t1' t2 t3)
  | ST_Succ : forall t1 t1',
      t1 ==> t1' ->
      (tm_succ t1) ==> (tm_succ t1')
  | ST_PredZero :
      (tm_pred tm_zero) ==> tm_zero
  | ST_PredSucc : forall t1,
      nvalue t1 ->
      (tm_pred (tm_succ t1)) ==> t1
  | ST_Pred : forall t1 t1',
      t1 ==> t1' ->
      (tm_pred t1) ==> (tm_pred t1')
  | ST_IszeroZero :
      (tm_iszero tm_zero) ==> tm_true
  | ST_IszeroSucc : forall t1,
       nvalue t1 ->
      (tm_iszero (tm_succ t1)) ==> tm_false
  | ST_Iszero : forall 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 _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 *)

(** 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 _stuck_. *)

Notation step_normal_form := (normal_form step).

Definition stuck (t:tm) : Prop :=
  step_normal_form t /\ ~ value t.

Hint Unfold stuck.

(** **** Exercise: 2 stars, optional (some_tm_is_stuck) *)
Example some_tm_is_stuck :
  exists t, stuck t.
Proof.
  (* 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 : forall t,
  value t -> step_normal_form t.
Proof.
  (* 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:
  partial_function step.
Proof with eauto.
  (* FILL IN HERE *) Admitted.
(** [] *)

(* ###################################################################### *)
(** ** Typing *)

(** 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 _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 := 
  | ty_Bool : ty
  | ty_Nat : ty.

(** 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 : forall 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 : forall t1,
       has_type t1 ty_Nat ->
       has_type (tm_succ t1) ty_Nat
  | T_Pred : forall t1,
       has_type t1 ty_Nat ->
       has_type (tm_pred t1) ty_Nat
  | T_Iszero : forall 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 *)

(** It's important to realize that the typing relation is a
    _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.

(** **** Exercise: 1 star (succ_hastype_nat__hastype_nat) *)
Example succ_hastype_nat__hastype_nat : forall t,
  has_type (tm_succ t) ty_Nat ->
  has_type t ty_Nat.  
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

(* ###################################################################### *)
(** ** Progress *)

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

(** **** 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].

    (* FILL IN HERE *)
[]
*)

(** **** Exercise: 3 stars (finish_progress) *)
Theorem progress : forall t T,
  has_type t T ->
  value t \/ exists 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".
          exists t2...
        SSSCase "t1 is tm_false". 
          exists 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].
      exists (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 _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 *)

(** 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 _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].  

        Inspecting the rules for the small-step reduction relation and
        remembering that [t] has the form [if ...], we see that the
        only ones that could have been used to prove [t ==> t'] are
        [ST_IfTrue], [ST_IfFalse], or [ST_If].

           - 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.

    (* FILL IN HERE *)
[]
*)

(** **** Exercise: 2 stars (finish_preservation) *)
Theorem preservation : forall 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.
(** [] *)

(** **** 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' : forall t t' T,
  has_type t T ->
  t ==> t' ->
  has_type t' T.
Proof with eauto.
  (* FILL IN HERE *) Admitted.
(** [] *)

(* ###################################################################### *)
(** ** Type Soundness *)

(** Putting progress and preservation together, we can see that a
    well-typed term can _never_ reach a stuck state.  *)

Definition stepmany := (refl_step_closure step).
Notation "t1 '==>*' t2" := (stepmany t1 t2) (at level 40).

Corollary soundness : forall 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 _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.

    (* FILL IN HERE *)
[]
*)

(** **** 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.
      - 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 : forall 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.)
      - 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 : forall 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.)
      - 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 : forall 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.

[]
*)

(** **** Exercise: 2 stars (variation5) *)
(** Suppose instead that we add this rule:
[[
      | ST_Funny2 : forall 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.

[]
*)

(** **** 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.

[]
*)

(** **** 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.

[]
*)

(** **** 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.

[]
*)

(** **** 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]?

(* FILL IN HERE *)
[]
*)

(** **** 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?

(* FILL IN HERE *)
[]
*)