EquivProgram Equivalence


Require Import Coq.Bool.Bool.
Require Import Coq.Arith.Arith.
Require Import Coq.Arith.EqNat.
Require Import Coq.omega.Omega.
Require Import Coq.Lists.List.
Import ListNotations.
Require Import Coq.Logic.FunctionalExtensionality.

Require Import SfLib.
Require Import Maps.
Require Import Imp.

Some general advice for working on exercises:

  • Most of the Coq proofs we ask you to do are similar to proofs that we've provided. Before starting to work on the homework problems, take the time to work through our proofs (both informally, on paper, and in Coq) and make sure you understand them in detail. This will save you a lot of time.
  • The Coq proofs we're doing now are sufficiently complicated that it is more or less impossible to complete them simply by random experimentation or "following your nose." You need to start with an idea about why the property is true and how the proof is going to go. The best way to do this is to write out at least a sketch of an informal proof on paper — one that intuitively convinces you of the truth of the theorem — before starting to work on the formal one. Alternately, grab a friend and try to convince them that the theorem is true; then try to formalize your explanation.
  • Use automation to save work! Some of the proofs in this chapter's exercises are pretty long if you try to write out all the cases explicitly.

Behavioral Equivalence

In the last chapter, we investigated the correctness of a very simple program transformation: the optimize_0plus function. The programming language we were considering was the first version of the language of arithmetic expressions — with no variables — so in that setting it was very easy to define what it means for a program transformation to be correct: it should always yield a program that evaluates to the same number as the original.
To go further and talk about the correctness of program transformations in the full Imp language, we need to consider the role of variables and state.

Definitions

For aexps and bexps with variables, the definition we want is clear. We say that two aexps or bexps are behaviorally equivalent if they evaluate to the same result in every state.

Definition aequiv (a1 a2 : aexp) : Prop :=
  (st:state),
    aeval st a1 = aeval st a2.

Definition bequiv (b1 b2 : bexp) : Prop :=
  (st:state),
    beval st b1 = beval st b2.

For commands, the situation is a little more subtle. We can't simply say "two commands are behaviorally equivalent if they evaluate to the same ending state whenever they are started in the same initial state," because some commands (in some starting states) don't terminate in any final state at all! What we need instead is this: two commands are behaviorally equivalent if, for any given starting state, they either both diverge or both terminate in the same final state. A compact way to express this is "if the first one terminates in a particular state then so does the second, and vice versa."

Definition cequiv (c1 c2 : com) : Prop :=
  (st st' : state),
    (c1 / st st') (c2 / st st').

Exercise: 2 stars (equiv_classes)

Given the following programs, group together those that are equivalent in Imp. Your answer should be given as a list of lists, where each sub-list represents a group of equivalent programs. For example, if you think programs (a) through (h) are all equivalent to each other, but not to (i), your answer should look like this:
    [ [prog_a;prog_b;prog_c;prog_d;prog_e;prog_f;prog_g;prog_h] ;
      [prog_i] ]
Write down your answer below in the definition of equiv_classes.

Definition prog_a : com :=
  WHILE BNot (BLe (AId X) (ANum 0)) DO
    X ::= APlus (AId X) (ANum 1)
  END.

Definition prog_b : com :=
  IFB BEq (AId X) (ANum 0) THEN
    X ::= APlus (AId X) (ANum 1);;
    Y ::= ANum 1
  ELSE
    Y ::= ANum 0
  FI;;
  X ::= AMinus (AId X) (AId Y);;
  Y ::= ANum 0.

Definition prog_c : com :=
  SKIP.

Definition prog_d : com :=
  WHILE BNot (BEq (AId X) (ANum 0)) DO
    X ::= APlus (AMult (AId X) (AId Y)) (ANum 1)
  END.

Definition prog_e : com :=
  Y ::= ANum 0.

Definition prog_f : com :=
  Y ::= APlus (AId X) (ANum 1);;
  WHILE BNot (BEq (AId X) (AId Y)) DO
    Y ::= APlus (AId X) (ANum 1)
  END.

Definition prog_g : com :=
  WHILE BTrue DO
    SKIP
  END.

Definition prog_h : com :=
  WHILE BNot (BEq (AId X) (AId X)) DO
    X ::= APlus (AId X) (ANum 1)
  END.

Definition prog_i : com :=
  WHILE BNot (BEq (AId X) (AId Y)) DO
    X ::= APlus (AId Y) (ANum 1)
  END.

Definition equiv_classes : list (list com) :=
(* FILL IN HERE *) admit.

Examples

Here are some simple examples of equivalences of arithmetic and boolean expressions.

Theorem aequiv_example:
  aequiv (AMinus (AId X) (AId X)) (ANum 0).
Proof.
  intros st. simpl. omega.
Qed.

Theorem bequiv_example:
  bequiv (BEq (AMinus (AId X) (AId X)) (ANum 0)) BTrue.
Proof.
  intros st. unfold beval.
  rewrite aequiv_example. reflexivity.
Qed.

For examples of command equivalence, let's start by looking at some trivial program transformations involving SKIP:

Theorem skip_left: c,
  cequiv
     (SKIP;; c)
     c.
Proof.
  (* WORKED IN CLASS *)
  intros c st st'.
  split; intros H.
  - (* -> *)
    inversion H. subst.
    inversion H2. subst.
    assumption.
  - (* <- *)
    apply E_Seq with st.
    apply E_Skip.
    assumption.
Qed.

Exercise: 2 stars (skip_right)

Prove that adding a SKIP after a command results in an equivalent program

Theorem skip_right: c,
  cequiv
    (c;; SKIP)
    c.
Proof.
  (* FILL IN HERE *) Admitted.
Similarly, here is a simple transformations that simplifies IFB commands:

Theorem IFB_true_simple: c1 c2,
  cequiv
    (IFB BTrue THEN c1 ELSE c2 FI)
    c1.
Proof.
  intros c1 c2.
  split; intros H.
  - (* -> *)
    inversion H; subst. assumption. inversion H5.
  - (* <- *)
    apply E_IfTrue. reflexivity. assumption. Qed.

Of course, few programmers would be tempted to write a conditional whose guard is literally BTrue. A more interesting case is when the guard is equivalent to true:
Theorem: If b is equivalent to BTrue, then IFB b THEN c1 ELSE c2 FI is equivalent to c1.
Proof:
  • () We must show, for all st and st', that if IFB b THEN c1 ELSE c2 FI / st st' then c1 / st st'.
    Proceed by cases on the rules that could possibly have been used to show IFB b THEN c1 ELSE c2 FI / st st', namely E_IfTrue and E_IfFalse.
    • Suppose the final rule rule in the derivation of IFB b THEN c1 ELSE c2 FI / st st' was E_IfTrue. We then have, by the premises of E_IfTrue, that c1 / st st'. This is exactly what we set out to prove.
    • On the other hand, suppose the final rule in the derivation of IFB b THEN c1 ELSE c2 FI / st st' was E_IfFalse. We then know that beval st b = false and c2 / st st'.
      Recall that b is equivalent to BTrue, i.e. forall st, beval st b = beval st BTrue. In particular, this means that beval st b = true, since beval st BTrue = true. But this is a contradiction, since E_IfFalse requires that beval st b = false. Thus, the final rule could not have been E_IfFalse.
  • () We must show, for all st and st', that if c1 / st st' then IFB b THEN c1 ELSE c2 FI / st st'.
    Since b is equivalent to BTrue, we know that beval st b = beval st BTrue = true. Together with the assumption that c1 / st st', we can apply E_IfTrue to derive IFB b THEN c1 ELSE c2 FI / st st'.
Here is the formal version of this proof:

Theorem IFB_true: b c1 c2,
     bequiv b BTrue
     cequiv
       (IFB b THEN c1 ELSE c2 FI)
       c1.
Proof.
  intros b c1 c2 Hb.
  split; intros H.
  - (* -> *)
    inversion H; subst.
    + (* b evaluates to true *)
      assumption.
    + (* b evaluates to false (contradiction) *)
      unfold bequiv in Hb. simpl in Hb.
      rewrite Hb in H5.
      inversion H5.
  - (* <- *)
    apply E_IfTrue; try assumption.
    unfold bequiv in Hb. simpl in Hb.
    rewrite Hb. reflexivity. Qed.

Exercise: 2 stars, recommended (IFB_false)

Theorem IFB_false: b c1 c2,
  bequiv b BFalse
  cequiv
    (IFB b THEN c1 ELSE c2 FI)
    c2.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars (swap_if_branches)

Show that we can swap the branches of an IF by negating its condition

Theorem swap_if_branches: b e1 e2,
  cequiv
    (IFB b THEN e1 ELSE e2 FI)
    (IFB BNot b THEN e2 ELSE e1 FI).
Proof.
  (* FILL IN HERE *) Admitted.
For WHILE loops, we can give a similar pair of theorems. A loop whose guard is equivalent to BFalse is equivalent to SKIP, while a loop whose guard is equivalent to BTrue is equivalent to WHILE BTrue DO SKIP END (or any other non-terminating program). The first of these facts is easy.

Theorem WHILE_false : b c,
     bequiv b BFalse
     cequiv
       (WHILE b DO c END)
       SKIP.
Proof.
  intros b c Hb. split; intros H.
  - (* -> *)
    inversion H; subst.
    + (* E_WhileEnd *)
      apply E_Skip.
    + (* E_WhileLoop *)
      rewrite Hb in H2. inversion H2.
  - (* <- *)
    inversion H; subst.
    apply E_WhileEnd.
    rewrite Hb.
    reflexivity. Qed.

Exercise: 2 stars, advanced, optional (WHILE_false_informal)

Write an informal proof of WHILE_false.
(* FILL IN HERE *)
To prove the second fact, we need an auxiliary lemma stating that WHILE loops whose guards are equivalent to BTrue never terminate:
Lemma: If b is equivalent to BTrue, then it cannot be the case that (WHILE b DO c END) / st st'.
Proof: Suppose that (WHILE b DO c END) / st st'. We show, by induction on a derivation of (WHILE b DO c END) / st st', that this assumption leads to a contradiction.
  • Suppose (WHILE b DO c END) / st st' is proved using rule E_WhileEnd. Then by assumption beval st b = false. But this contradicts the assumption that b is equivalent to BTrue.
  • Suppose (WHILE b DO c END) / st st' is proved using rule E_WhileLoop. Then we are given the induction hypothesis that (WHILE b DO c END) / st st' is contradictory, which is exactly what we are trying to prove!
  • Since these are the only rules that could have been used to prove (WHILE b DO c END) / st st', the other cases of the induction are immediately contradictory.

Lemma WHILE_true_nonterm : b c st st',
     bequiv b BTrue
     ~( (WHILE b DO c END) / st st' ).
Proof.
  (* WORKED IN CLASS *)
  intros b c st st' Hb.
  intros H.
  remember (WHILE b DO c END) as cw eqn:Heqcw.
  induction H;
    (* Most rules don't apply, and we can rule them out
       by inversion *)

    inversion Heqcw; subst; clear Heqcw.
  (* The two interesting cases are the ones for WHILE loops: *)
  - (* E_WhileEnd *) (* contradictory -- b is always true! *)
    unfold bequiv in Hb.
    (* rewrite is able to instantiate the quantifier in st *)
    rewrite Hb in H. inversion H.
  - (* E_WhileLoop *) (* immediate from the IH *)
    apply IHceval2. reflexivity. Qed.

Exercise: 2 stars, optional (WHILE_true_nonterm_informal)

Explain what the lemma WHILE_true_nonterm means in English.
(* FILL IN HERE *)

Exercise: 2 stars, recommended (WHILE_true)

Prove the following theorem. Hint: You'll want to use WHILE_true_nonterm here.

Theorem WHILE_true: b c,
     bequiv b BTrue
     cequiv
       (WHILE b DO c END)
       (WHILE BTrue DO SKIP END).
Proof.
  (* FILL IN HERE *) Admitted.

Theorem loop_unrolling: b c,
  cequiv
    (WHILE b DO c END)
    (IFB b THEN (c;; WHILE b DO c END) ELSE SKIP FI).
Proof.
  (* WORKED IN CLASS *)
  intros b c st st'.
  split; intros Hce.
  - (* -> *)
    inversion Hce; subst.
    + (* loop doesn't run *)
      apply E_IfFalse. assumption. apply E_Skip.
    + (* loop runs *)
      apply E_IfTrue. assumption.
      apply E_Seq with (st' := st'0). assumption. assumption.
  - (* <- *)
    inversion Hce; subst.
    + (* loop runs *)
      inversion H5; subst.
      apply E_WhileLoop with (st' := st'0).
      assumption. assumption. assumption.
    + (* loop doesn't run *)
      inversion H5; subst. apply E_WhileEnd. assumption. Qed.

Exercise: 2 stars, optional (seq_assoc)

Theorem seq_assoc : c1 c2 c3,
  cequiv ((c1;;c2);;c3) (c1;;(c2;;c3)).
Proof.
  (* FILL IN HERE *) Admitted.
Proving program properties involving assignments is one place where the functional-extensionality axiom introduced in the Logic chapter often comes in handy. Here are an example and an exercise.

Theorem identity_assignment : (X:id),
  cequiv
    (X ::= AId X)
    SKIP.
Proof.
   intros. split; intro H.
     - (* -> *)
       inversion H; subst. simpl.
       replace (t_update st X (st X)) with st.
       constructor.
       apply functional_extensionality. intro.
       rewrite t_update_same; reflexivity.
     - (* <- *)
       inversion H; subst.
       assert (st' = (t_update st' X (st' X))).
          apply functional_extensionality. intro.
          rewrite t_update_same; reflexivity.
       rewrite H0 at 2.
       constructor. reflexivity.
Qed.

Exercise: 2 stars, recommended (assign_aequiv)

Theorem assign_aequiv : X e,
  aequiv (AId X) e
  cequiv SKIP (X ::= e).
Proof.
  (* FILL IN HERE *) Admitted.

Properties of Behavioral Equivalence

We now turn to developing some of the properties of the program equivalences we have defined.

Behavioral Equivalence is an Equivalence

First, we verify that the equivalences on aexps, bexps, and coms really are equivalences — i.e., that they are reflexive, symmetric, and transitive. The proofs are all easy.

Lemma refl_aequiv : (a : aexp), aequiv a a.
Proof.
  intros a st. reflexivity. Qed.

Lemma sym_aequiv : (a1 a2 : aexp),
  aequiv a1 a2 aequiv a2 a1.
Proof.
  intros a1 a2 H. intros st. symmetry. apply H. Qed.

Lemma trans_aequiv : (a1 a2 a3 : aexp),
  aequiv a1 a2 aequiv a2 a3 aequiv a1 a3.
Proof.
  unfold aequiv. intros a1 a2 a3 H12 H23 st.
  rewrite (H12 st). rewrite (H23 st). reflexivity. Qed.

Lemma refl_bequiv : (b : bexp), bequiv b b.
Proof.
  unfold bequiv. intros b st. reflexivity. Qed.

Lemma sym_bequiv : (b1 b2 : bexp),
  bequiv b1 b2 bequiv b2 b1.
Proof.
  unfold bequiv. intros b1 b2 H. intros st. symmetry. apply H. Qed.

Lemma trans_bequiv : (b1 b2 b3 : bexp),
  bequiv b1 b2 bequiv b2 b3 bequiv b1 b3.
Proof.
  unfold bequiv. intros b1 b2 b3 H12 H23 st.
  rewrite (H12 st). rewrite (H23 st). reflexivity. Qed.

Lemma refl_cequiv : (c : com), cequiv c c.
Proof.
  unfold cequiv. intros c st st'. apply iff_refl. Qed.

Lemma sym_cequiv : (c1 c2 : com),
  cequiv c1 c2 cequiv c2 c1.
Proof.
  unfold cequiv. intros c1 c2 H st st'.
  assert (c1 / st st' c2 / st st') as H'.
  { (* Proof of assertion *) apply H. }
  apply iff_sym. assumption.
Qed.

Lemma iff_trans : (P1 P2 P3 : Prop),
  (P1 P2) (P2 P3) (P1 P3).
Proof.
  intros P1 P2 P3 H12 H23.
  inversion H12. inversion H23.
  split; intros A.
    apply H1. apply H. apply A.
    apply H0. apply H2. apply A. Qed.

Lemma trans_cequiv : (c1 c2 c3 : com),
  cequiv c1 c2 cequiv c2 c3 cequiv c1 c3.
Proof.
  unfold cequiv. intros c1 c2 c3 H12 H23 st st'.
  apply iff_trans with (c2 / st st'). apply H12. apply H23. Qed.

Behavioral Equivalence is a Congruence

Less obviously, behavioral equivalence is also a congruence. That is, the equivalence of two subprograms implies the equivalence of the larger programs in which they are embedded:
aequiv a1 a1'  

cequiv (i ::= a1) (i ::= a1')
cequiv c1 c1'
cequiv c2 c2'  

cequiv (c1;;c2) (c1';;c2')
...and so on.
(Note that we are using the inference rule notation here not as part of a definition, but simply to write down some valid implications in a readable format. We prove these implications below.)
We will see a concrete example of why these congruence properties are important in the following section (in the proof of fold_constants_com_sound), but the main idea is that they allow us to replace a small part of a large program with an equivalent small part and know that the whole large programs are equivalent without doing an explicit proof about the non-varying parts — i.e., the "proof burden" of a small change to a large program is proportional to the size of the change, not the program.

Theorem CAss_congruence : i a1 a1',
  aequiv a1 a1'
  cequiv (CAss i a1) (CAss i a1').
Proof.
  intros i a1 a2 Heqv st st'.
  split; intros Hceval.
  - (* -> *)
    inversion Hceval. subst. apply E_Ass.
    rewrite Heqv. reflexivity.
  - (* <- *)
    inversion Hceval. subst. apply E_Ass.
    rewrite Heqv. reflexivity. Qed.

The congruence property for loops is a little more interesting, since it requires induction.
Theorem: Equivalence is a congruence for WHILE — that is, if b1 is equivalent to b1' and c1 is equivalent to c1', then WHILE b1 DO c1 END is equivalent to WHILE b1' DO c1' END.
Proof: Suppose b1 is equivalent to b1' and c1 is equivalent to c1'. We must show, for every st and st', that WHILE b1 DO c1 END / st st' iff WHILE b1' DO c1' END / st st'. We consider the two directions separately.
  • () We show that WHILE b1 DO c1 END / st st' implies WHILE b1' DO c1' END / st st', by induction on a derivation of WHILE b1 DO c1 END / st st'. The only nontrivial cases are when the final rule in the derivation is E_WhileEnd or E_WhileLoop.
    • E_WhileEnd: In this case, the form of the rule gives us beval st b1 = false and st = st'. But then, since b1 and b1' are equivalent, we have beval st b1' = false, and E-WhileEnd applies, giving us WHILE b1' DO c1' END / st st', as required.
    • E_WhileLoop: The form of the rule now gives us beval st b1 = true, with c1 / st st'0 and WHILE b1 DO c1 END / st'0 st' for some state st'0, with the induction hypothesis WHILE b1' DO c1' END / st'0 st'.
      Since c1 and c1' are equivalent, we know that c1' / st st'0. And since b1 and b1' are equivalent, we have beval st b1' = true. Now E-WhileLoop applies, giving us WHILE b1' DO c1' END / st st', as required.
  • () Similar.

Theorem CWhile_congruence : b1 b1' c1 c1',
  bequiv b1 b1' cequiv c1 c1'
  cequiv (WHILE b1 DO c1 END) (WHILE b1' DO c1' END).
Proof.
  (* WORKED IN CLASS *)
  unfold bequiv,cequiv.
  intros b1 b1' c1 c1' Hb1e Hc1e st st'.
  split; intros Hce.
  - (* -> *)
    remember (WHILE b1 DO c1 END) as cwhile eqn:Heqcwhile.
    induction Hce; inversion Heqcwhile; subst.
    + (* E_WhileEnd *)
      apply E_WhileEnd. rewrite Hb1e. apply H.
    + (* E_WhileLoop *)
      apply E_WhileLoop with (st' := st').
      * (* show loop runs *) rewrite Hb1e. apply H.
      * (* body execution *)
        apply (Hc1e st st'). apply Hce1.
      * (* subsequent loop execution *)
        apply IHHce2. reflexivity.
  - (* <- *)
    remember (WHILE b1' DO c1' END) as c'while eqn:Heqc'while.
    induction Hce; inversion Heqc'while; subst.
    + (* E_WhileEnd *)
      apply E_WhileEnd. rewrite Hb1e. apply H.
    + (* E_WhileLoop *)
      apply E_WhileLoop with (st' := st').
      * (* show loop runs *) rewrite Hb1e. apply H.
      * (* body execution *)
        apply (Hc1e st st'). apply Hce1.
      * (* subsequent loop execution *)
        apply IHHce2. reflexivity. Qed.

Exercise: 3 stars, optional (CSeq_congruence)

Theorem CSeq_congruence : c1 c1' c2 c2',
  cequiv c1 c1' cequiv c2 c2'
  cequiv (c1;;c2) (c1';;c2').
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars (CIf_congruence)

Theorem CIf_congruence : b b' c1 c1' c2 c2',
  bequiv b b' cequiv c1 c1' cequiv c2 c2'
  cequiv (IFB b THEN c1 ELSE c2 FI) (IFB b' THEN c1' ELSE c2' FI).
Proof.
  (* FILL IN HERE *) Admitted.
For example, here are two equivalent programs and a proof of their equivalence...

Example congruence_example:
  cequiv
    (* Program 1: *)
    (X ::= ANum 0;;
     IFB (BEq (AId X) (ANum 0))
     THEN
       Y ::= ANum 0
     ELSE
       Y ::= ANum 42
     FI)
    (* Program 2: *)
    (X ::= ANum 0;;
     IFB (BEq (AId X) (ANum 0))
     THEN
       Y ::= AMinus (AId X) (AId X) (* <--- changed here *)
     ELSE
       Y ::= ANum 42
     FI).
Proof.
  apply CSeq_congruence.
    apply refl_cequiv.
    apply CIf_congruence.
      apply refl_bequiv.
      apply CAss_congruence. unfold aequiv. simpl.
        symmetry. apply minus_diag.
      apply refl_cequiv.
Qed.

Program Transformations

A program transformation is a function that takes a program as input and produces some variant of the program as its output. Compiler optimizations such as constant folding are a canonical example, but there are many others.
A program transformation is sound if it preserves the behavior of the original program.
We can define a notion of soundness for translations of aexps, bexps, and coms.

Definition atrans_sound (atrans : aexp aexp) : Prop :=
  (a : aexp),
    aequiv a (atrans a).

Definition btrans_sound (btrans : bexp bexp) : Prop :=
  (b : bexp),
    bequiv b (btrans b).

Definition ctrans_sound (ctrans : com com) : Prop :=
  (c : com),
    cequiv c (ctrans c).

The Constant-Folding Transformation

An expression is constant when it contains no variable references.
Constant folding is an optimization that finds constant expressions and replaces them by their values.

Fixpoint fold_constants_aexp (a : aexp) : aexp :=
  match a with
  | ANum nANum n
  | AId iAId i
  | APlus a1 a2
      match (fold_constants_aexp a1, fold_constants_aexp a2) with
      | (ANum n1, ANum n2) ⇒ ANum (n1 + n2)
      | (a1', a2') ⇒ APlus a1' a2'
      end
  | AMinus a1 a2
      match (fold_constants_aexp a1, fold_constants_aexp a2) with
      | (ANum n1, ANum n2) ⇒ ANum (n1 - n2)
      | (a1', a2') ⇒ AMinus a1' a2'
      end
  | AMult a1 a2
      match (fold_constants_aexp a1, fold_constants_aexp a2) with
      | (ANum n1, ANum n2) ⇒ ANum (n1 * n2)
      | (a1', a2') ⇒ AMult a1' a2'
      end
  end.

Example fold_aexp_ex1 :
    fold_constants_aexp
      (AMult (APlus (ANum 1) (ANum 2)) (AId X))
  = AMult (ANum 3) (AId X).
Proof. reflexivity. Qed.

Note that this version of constant folding doesn't eliminate trivial additions, etc. — we are focusing attention on a single optimization for the sake of simplicity. It is not hard to incorporate other ways of simplifying expressions; the definitions and proofs just get longer.

Example fold_aexp_ex2 :
    fold_constants_aexp
      (AMinus (AId X) (APlus (AMult (ANum 0) (ANum 6)) (AId Y)))
  = AMinus (AId X) (APlus (ANum 0) (AId Y)).
Proof. reflexivity. Qed.

Not only can we lift fold_constants_aexp to bexps (in the BEq and BLe cases), we can also find constant boolean expressions and reduce them in-place.

Fixpoint fold_constants_bexp (b : bexp) : bexp :=
  match b with
  | BTrueBTrue
  | BFalseBFalse
  | BEq a1 a2
      match (fold_constants_aexp a1, fold_constants_aexp a2) with
      | (ANum n1, ANum n2) ⇒ if beq_nat n1 n2 then BTrue else BFalse
      | (a1', a2') ⇒ BEq a1' a2'
      end
  | BLe a1 a2
      match (fold_constants_aexp a1, fold_constants_aexp a2) with
      | (ANum n1, ANum n2) ⇒ if leb n1 n2 then BTrue else BFalse
      | (a1', a2') ⇒ BLe a1' a2'
      end
  | BNot b1
      match (fold_constants_bexp b1) with
      | BTrueBFalse
      | BFalseBTrue
      | b1'BNot b1'
      end
  | BAnd b1 b2
      match (fold_constants_bexp b1, fold_constants_bexp b2) with
      | (BTrue, BTrue) ⇒ BTrue
      | (BTrue, BFalse) ⇒ BFalse
      | (BFalse, BTrue) ⇒ BFalse
      | (BFalse, BFalse) ⇒ BFalse
      | (b1', b2') ⇒ BAnd b1' b2'
      end
  end.

Example fold_bexp_ex1 :
    fold_constants_bexp (BAnd BTrue (BNot (BAnd BFalse BTrue)))
  = BTrue.
Proof. reflexivity. Qed.

Example fold_bexp_ex2 :
    fold_constants_bexp
      (BAnd (BEq (AId X) (AId Y))
            (BEq (ANum 0)
                 (AMinus (ANum 2) (APlus (ANum 1) (ANum 1)))))
  = BAnd (BEq (AId X) (AId Y)) BTrue.
Proof. reflexivity. Qed.

To fold constants in a command, we apply the appropriate folding functions on all embedded expressions.

Fixpoint fold_constants_com (c : com) : com :=
  match c with
  | SKIP
      SKIP
  | i ::= a
      CAss i (fold_constants_aexp a)
  | c1 ;; c2
      (fold_constants_com c1) ;; (fold_constants_com c2)
  | IFB b THEN c1 ELSE c2 FI
      match fold_constants_bexp b with
      | BTruefold_constants_com c1
      | BFalsefold_constants_com c2
      | b'IFB b' THEN fold_constants_com c1
                     ELSE fold_constants_com c2 FI
      end
  | WHILE b DO c END
      match fold_constants_bexp b with
      | BTrueWHILE BTrue DO SKIP END
      | BFalseSKIP
      | b'WHILE b' DO (fold_constants_com c) END
      end
  end.

Example fold_com_ex1 :
  fold_constants_com
    (* Original program: *)
    (X ::= APlus (ANum 4) (ANum 5);;
     Y ::= AMinus (AId X) (ANum 3);;
     IFB BEq (AMinus (AId X) (AId Y)) (APlus (ANum 2) (ANum 4)) THEN
       SKIP
     ELSE
       Y ::= ANum 0
     FI;;
     IFB BLe (ANum 0) (AMinus (ANum 4) (APlus (ANum 2) (ANum 1))) THEN
       Y ::= ANum 0
     ELSE
       SKIP
     FI;;
     WHILE BEq (AId Y) (ANum 0) DO
       X ::= APlus (AId X) (ANum 1)
     END)
  = (* After constant folding: *)
    (X ::= ANum 9;;
     Y ::= AMinus (AId X) (ANum 3);;
     IFB BEq (AMinus (AId X) (AId Y)) (ANum 6) THEN
       SKIP
     ELSE
       (Y ::= ANum 0)
     FI;;
     Y ::= ANum 0;;
     WHILE BEq (AId Y) (ANum 0) DO
       X ::= APlus (AId X) (ANum 1)
     END).
Proof. reflexivity. Qed.

Soundness of Constant Folding

Now we need to show that what we've done is correct.
Here's the proof for arithmetic expressions:

Theorem fold_constants_aexp_sound :
  atrans_sound fold_constants_aexp.
Proof.
  unfold atrans_sound. intros a. unfold aequiv. intros st.
  induction a; simpl;
    (* ANum and AId follow immediately *)
    try reflexivity;
    (* APlus, AMinus, and AMult follow from the IH
       and the observation that
              aeval st (APlus a1 a2)
            = ANum ((aeval st a1) + (aeval st a2))
            = aeval st (ANum ((aeval st a1) + (aeval st a2)))
       (and similarly for AMinus/minus and AMult/mult) *)

    try (destruct (fold_constants_aexp a1);
         destruct (fold_constants_aexp a2);
         rewrite IHa1; rewrite IHa2; reflexivity). Qed.

Exercise: 3 stars, optional (fold_bexp_Eq_informal)

Here is an informal proof of the BEq case of the soundness argument for boolean expression constant folding. Read it carefully and compare it to the formal proof that follows. Then fill in the BLe case of the formal proof (without looking at the BEq case, if possible).
Theorem: The constant folding function for booleans, fold_constants_bexp, is sound.
Proof: We must show that b is equivalent to fold_constants_bexp, for all boolean expressions b. Proceed by induction on b. We show just the case where b has the form BEq a1 a2.
In this case, we must show
       beval st (BEq a1 a2)
     = beval st (fold_constants_bexp (BEq a1 a2)).
There are two cases to consider:
  • First, suppose fold_constants_aexp a1 = ANum n1 and fold_constants_aexp a2 = ANum n2 for some n1 and n2.
    In this case, we have
        fold_constants_bexp (BEq a1 a2)
      = if beq_nat n1 n2 then BTrue else BFalse
    and
        beval st (BEq a1 a2)
      = beq_nat (aeval st a1) (aeval st a2).
    By the soundness of constant folding for arithmetic expressions (Lemma fold_constants_aexp_sound), we know
        aeval st a1
      = aeval st (fold_constants_aexp a1)
      = aeval st (ANum n1)
      = n1
    and
        aeval st a2
      = aeval st (fold_constants_aexp a2)
      = aeval st (ANum n2)
      = n2,
    so
        beval st (BEq a1 a2)
      = beq_nat (aeval a1) (aeval a2)
      = beq_nat n1 n2.
    Also, it is easy to see (by considering the cases n1 = n2 and n1 n2 separately) that
        beval st (if beq_nat n1 n2 then BTrue else BFalse)
      = if beq_nat n1 n2 then beval st BTrue else beval st BFalse
      = if beq_nat n1 n2 then true else false
      = beq_nat n1 n2.
    So
        beval st (BEq a1 a2)
      = beq_nat n1 n2.
      = beval st (if beq_nat n1 n2 then BTrue else BFalse),
    as required.
  • Otherwise, one of fold_constants_aexp a1 and fold_constants_aexp a2 is not a constant. In this case, we must show
        beval st (BEq a1 a2)
      = beval st (BEq (fold_constants_aexp a1)
                      (fold_constants_aexp a2)),
    which, by the definition of beval, is the same as showing
        beq_nat (aeval st a1) (aeval st a2)
      = beq_nat (aeval st (fold_constants_aexp a1))
                (aeval st (fold_constants_aexp a2)).
    But the soundness of constant folding for arithmetic expressions (fold_constants_aexp_sound) gives us
      aeval st a1 = aeval st (fold_constants_aexp a1)
      aeval st a2 = aeval st (fold_constants_aexp a2),
    completing the case.

Theorem fold_constants_bexp_sound:
  btrans_sound fold_constants_bexp.
Proof.
  unfold btrans_sound. intros b. unfold bequiv. intros st.
  induction b;
    (* BTrue and BFalse are immediate *)
    try reflexivity.
  - (* BEq *)
    (* Doing induction when there are a lot of constructors makes
       specifying variable names a chore, but Coq doesn't always
       choose nice variable names.  We can rename entries in the
       context with the rename tactic: rename a into a1 will
       change a to a1 in the current goal and context. *)

    rename a into a1. rename a0 into a2. simpl.
    remember (fold_constants_aexp a1) as a1' eqn:Heqa1'.
    remember (fold_constants_aexp a2) as a2' eqn:Heqa2'.
    replace (aeval st a1) with (aeval st a1') by
       (subst a1'; rewrite fold_constants_aexp_sound; reflexivity).
    replace (aeval st a2) with (aeval st a2') by
       (subst a2'; rewrite fold_constants_aexp_sound; reflexivity).
    destruct a1'; destruct a2'; try reflexivity.
      (* The only interesting case is when both a1 and a2
         become constants after folding *)

      simpl. destruct (beq_nat n n0); reflexivity.
  - (* BLe *)
    (* FILL IN HERE *) admit.
  - (* BNot *)
    simpl. remember (fold_constants_bexp b) as b' eqn:Heqb'.
    rewrite IHb.
    destruct b'; reflexivity.
  - (* BAnd *)
    simpl.
    remember (fold_constants_bexp b1) as b1' eqn:Heqb1'.
    remember (fold_constants_bexp b2) as b2' eqn:Heqb2'.
    rewrite IHb1. rewrite IHb2.
    destruct b1'; destruct b2'; reflexivity.
(* FILL IN HERE *) Admitted.

Exercise: 3 stars (fold_constants_com_sound)

Complete the WHILE case of the following proof.

Theorem fold_constants_com_sound :
  ctrans_sound fold_constants_com.
Proof.
  unfold ctrans_sound. intros c.
  induction c; simpl.
  - (* SKIP *) apply refl_cequiv.
  - (* ::= *) apply CAss_congruence. apply fold_constants_aexp_sound.
  - (* ;; *) apply CSeq_congruence; assumption.
  - (* IFB *)
    assert (bequiv b (fold_constants_bexp b)).
    { (* Pf of assertion *) apply fold_constants_bexp_sound. }
    destruct (fold_constants_bexp b) eqn:Heqb;
      (* If the optimization doesn't eliminate the if, then the result
         is easy to prove from the IH and fold_constants_bexp_sound *)

      try (apply CIf_congruence; assumption).
    + (* b always true *)
      apply trans_cequiv with c1; try assumption.
      apply IFB_true; assumption.
    + (* b always false *)
      apply trans_cequiv with c2; try assumption.
      apply IFB_false; assumption.
  - (* WHILE *)
    (* FILL IN HERE *) Admitted.

Soundness of (0 + n) Elimination, Redux

Exercise: 4 stars, advanced, optional (optimize_0plus)

Recall the definition optimize_0plus from Imp.v:
    Fixpoint optimize_0plus (e:aexp) : aexp :=
      match e with
      | ANum n ⇒
          ANum n
      | APlus (ANum 0) e2 ⇒
          optimize_0plus e2
      | APlus e1 e2 ⇒
          APlus (optimize_0plus e1) (optimize_0plus e2)
      | AMinus e1 e2 ⇒
          AMinus (optimize_0plus e1) (optimize_0plus e2)
      | AMult e1 e2 ⇒
          AMult (optimize_0plus e1) (optimize_0plus e2)
      end.
Note that this function is defined over the old aexps, without states.
Write a new version of this function that accounts for variables, and analogous ones for bexps and commands:
     optimize_0plus_aexp
     optimize_0plus_bexp
     optimize_0plus_com
Prove that these three functions are sound, as we did for fold_constants_*. (Make sure you use the congruence lemmas in the proof of optimize_0plus_com — otherwise it will be long!)
Then define an optimizer on commands that first folds constants (using fold_constants_com) and then eliminates 0 + n terms (using optimize_0plus_com).
  • Give a meaningful example of this optimizer's output.
  • Prove that the optimizer is sound. (This part should be very easy.)

(* FILL IN HERE *)

Proving That Programs Are Not Equivalent

Suppose that c1 is a command of the form X ::= a1;; Y ::= a2 and c2 is the command X ::= a1;; Y ::= a2', where a2' is formed by substituting a1 for all occurrences of X in a2. For example, c1 and c2 might be:
       c1  =  (X ::= 42 + 53;;
               Y ::= Y + X)
       c2  =  (X ::= 42 + 53;;
               Y ::= Y + (42 + 53))
Clearly, this particular c1 and c2 are equivalent. Is this true in general?
We will see in a moment that it is not, but it is worthwhile to pause, now, and see if you can find a counter-example on your own.
Here, formally, is the function that substitutes an arithmetic expression for each occurrence of a given variable in another expression:

Fixpoint subst_aexp (i : id) (u : aexp) (a : aexp) : aexp :=
  match a with
  | ANum nANum n
  | AId i'if beq_id i i' then u else AId i'
  | APlus a1 a2APlus (subst_aexp i u a1) (subst_aexp i u a2)
  | AMinus a1 a2AMinus (subst_aexp i u a1) (subst_aexp i u a2)
  | AMult a1 a2AMult (subst_aexp i u a1) (subst_aexp i u a2)
  end.

Example subst_aexp_ex :
  subst_aexp X (APlus (ANum 42) (ANum 53)) (APlus (AId Y) (AId X)) =
  (APlus (AId Y) (APlus (ANum 42) (ANum 53))).
Proof. reflexivity. Qed.

And here is the property we are interested in, expressing the claim that commands c1 and c2 as described above are always equivalent.

Definition subst_equiv_property := i1 i2 a1 a2,
  cequiv (i1 ::= a1;; i2 ::= a2)
         (i1 ::= a1;; i2 ::= subst_aexp i1 a1 a2).

Sadly, the property does not always hold.
Theorem: It is not the case that, for all i1, i2, a1, and a2,
         cequiv (i1 ::= a1;; i2 ::= a2)
                (i1 ::= a1;; i2 ::= subst_aexp i1 a1 a2).
Proof: Suppose, for a contradiction, that for all i1, i2, a1, and a2, we have
      cequiv (i1 ::= a1;; i2 ::= a2)
             (i1 ::= a1;; i2 ::= subst_aexp i1 a1 a2).
Consider the following program:
         X ::= APlus (AId X) (ANum 1);; Y ::= AId X
Note that
         (X ::= APlus (AId X) (ANum 1);; Y ::= AId X)
         / empty_state  st1,
where st1 = { X 1, Y 1 }.
By our assumption, we know that
        cequiv (X ::= APlus (AId X) (ANum 1);; Y ::= AId X)
               (X ::= APlus (AId X) (ANum 1);; Y ::= APlus (AId X) (ANum 1))
so, by the definition of cequiv, we have
        (X ::= APlus (AId X) (ANum 1);; Y ::= APlus (AId X) (ANum 1))
        / empty_state  st1.
But we can also derive
        (X ::= APlus (AId X) (ANum 1);; Y ::= APlus (AId X) (ANum 1))
        / empty_state  st2,
where st2 = { X 1, Y 2 }. Note that st1 st2; this is a contradiction, since ceval is deterministic!

Theorem subst_inequiv :
  ¬ subst_equiv_property.
Proof.
  unfold subst_equiv_property.
  intros Contra.

  (* Here is the counterexample: assuming that subst_equiv_property
     holds allows us to prove that these two programs are
     equivalent... *)

  remember (X ::= APlus (AId X) (ANum 1);;
            Y ::= AId X)
      as c1.
  remember (X ::= APlus (AId X) (ANum 1);;
            Y ::= APlus (AId X) (ANum 1))
      as c2.
  assert (cequiv c1 c2) by (subst; apply Contra).

  (* ... allows us to show that the command c2 can terminate
     in two different final states:
        st1 = {X |-> 1, Y |-> 1}
        st2 = {X |-> 1, Y |-> 2}. *)

  remember (t_update (t_update empty_state X 1) Y 1) as st1.
  remember (t_update (t_update empty_state X 1) Y 2) as st2.
  assert (H1: c1 / empty_state st1);
  assert (H2: c2 / empty_state st2);
  try (subst;
       apply E_Seq with (st' := (t_update empty_state X 1));
       apply E_Ass; reflexivity).
  apply H in H1.

  (* Finally, we use the fact that evaluation is deterministic
     to obtain a contradiction. *)

  assert (Hcontra: st1 = st2)
    by (apply (ceval_deterministic c2 empty_state); assumption).
  assert (Hcontra': st1 Y = st2 Y)
    by (rewrite Hcontra; reflexivity).
  subst. inversion Hcontra'. Qed.

Exercise: 4 stars, optional (better_subst_equiv)

The equivalence we had in mind above was not complete nonsense — it was actually almost right. To make it correct, we just need to exclude the case where the variable X occurs in the right-hand-side of the first assignment statement.

Inductive var_not_used_in_aexp (X:id) : aexp Prop :=
  | VNUNum: n, var_not_used_in_aexp X (ANum n)
  | VNUId: Y, XY var_not_used_in_aexp X (AId Y)
  | VNUPlus: a1 a2,
      var_not_used_in_aexp X a1
      var_not_used_in_aexp X a2
      var_not_used_in_aexp X (APlus a1 a2)
  | VNUMinus: a1 a2,
      var_not_used_in_aexp X a1
      var_not_used_in_aexp X a2
      var_not_used_in_aexp X (AMinus a1 a2)
  | VNUMult: a1 a2,
      var_not_used_in_aexp X a1
      var_not_used_in_aexp X a2
      var_not_used_in_aexp X (AMult a1 a2).

Lemma aeval_weakening : i st a ni,
  var_not_used_in_aexp i a
  aeval (t_update st i ni) a = aeval st a.
Proof.
  (* FILL IN HERE *) Admitted.

Using var_not_used_in_aexp, formalize and prove a correct verson of subst_equiv_property.

(* FILL IN HERE *)

Exercise: 3 stars, optional (inequiv_exercise)

Prove that an infinite loop is not equivalent to SKIP

Theorem inequiv_exercise:
  ¬ cequiv (WHILE BTrue DO SKIP END) SKIP.
Proof.
  (* FILL IN HERE *) Admitted.

Extended exercise: Non-deterministic Imp

As we have seen (in theorem ceval_deterministic in the Imp chapter), Imp's evaluation relation is deterministic. However, non-determinism is an important part of the definition of many real programming languages. For example, in many imperative languages (such as C and its relatives), the order in which function arguments are evaluated is unspecified. The program fragment
      x = 0;;
      f(++xx)
might call f with arguments (1, 0) or (1, 1), depending how the compiler chooses to order things. This can be a little confusing for programmers, but it gives the compiler writer useful freedom.
In this exercise, we will extend Imp with a simple non-deterministic command and study how this change affects program equivalence. The new command has the syntax HAVOC X, where X is an identifier. The effect of executing HAVOC X is to assign an arbitrary number to the variable X, non-deterministically. For example, after executing the program:
      HAVOC Y;;
      Z ::= Y * 2
the value of Y can be any number, while the value of Z is twice that of Y (so Z is always even). Note that we are not saying anything about the probabilities of the outcomes — just that there are (infinitely) many different outcomes that can possibly happen after executing this non-deterministic code.
In a sense a variable on which we do HAVOC roughly corresponds to an unitialized variable in the C programming language. After the HAVOC the variable holds a fixed but arbitrary number. Most sources of nondeterminism in language definitions are there precisely because programmers don't care which choice is made (and so it is good to leave it open to the compiler to choose whichever will run faster).
We call this new language Himp (``Imp extended with HAVOC'').

Module Himp.

To formalize the language, we first add a clause to the definition of commands.

Inductive com : Type :=
  | CSkip : com
  | CAss : id aexp com
  | CSeq : com com com
  | CIf : bexp com com com
  | CWhile : bexp com com
  | CHavoc : id com. (* <---- new *)

Notation "'SKIP'" :=
  CSkip.
Notation "X '::=' a" :=
  (CAss X a) (at level 60).
Notation "c1 ;; c2" :=
  (CSeq c1 c2) (at level 80, right associativity).
Notation "'WHILE' b 'DO' c 'END'" :=
  (CWhile b c) (at level 80, right associativity).
Notation "'IFB' e1 'THEN' e2 'ELSE' e3 'FI'" :=
  (CIf e1 e2 e3) (at level 80, right associativity).
Notation "'HAVOC' l" := (CHavoc l) (at level 60).

Exercise: 2 stars (himp_ceval)

Now, we must extend the operational semantics. We have provided a template for the ceval relation below, specifying the big-step semantics. What rule(s) must be added to the definition of ceval to formalize the behavior of the HAVOC command?

Reserved Notation "c1 '/' st '' st'" (at level 40, st at level 39).

Inductive ceval : com state state Prop :=
  | E_Skip : st : state, SKIP / st st
  | E_Ass : (st : state) (a1 : aexp) (n : nat) (X : id),
            aeval st a1 = n (X ::= a1) / st t_update st X n
  | E_Seq : (c1 c2 : com) (st st' st'' : state),
            c1 / st st' c2 / st' st'' (c1 ;; c2) / st st''
  | E_IfTrue : (st st' : state) (b1 : bexp) (c1 c2 : com),
               beval st b1 = true
               c1 / st st' (IFB b1 THEN c1 ELSE c2 FI) / st st'
  | E_IfFalse : (st st' : state) (b1 : bexp) (c1 c2 : com),
                beval st b1 = false
                c2 / st st' (IFB b1 THEN c1 ELSE c2 FI) / st st'
  | E_WhileEnd : (b1 : bexp) (st : state) (c1 : com),
                 beval st b1 = false (WHILE b1 DO c1 END) / st st
  | E_WhileLoop : (st st' st'' : state) (b1 : bexp) (c1 : com),
                  beval st b1 = true
                  c1 / st st'
                  (WHILE b1 DO c1 END) / st' st''
                  (WHILE b1 DO c1 END) / st st''
(* FILL IN HERE *)

  where "c1 '/' st '' st'" := (ceval c1 st st').

As a sanity check, the following claims should be provable for your definition:

Example havoc_example1 : (HAVOC X) / empty_state t_update empty_state X 0.
Proof.
(* FILL IN HERE *) Admitted.

Example havoc_example2 :
  (SKIP;; HAVOC Z) / empty_state t_update empty_state Z 42.
Proof.
(* FILL IN HERE *) Admitted.
Finally, we repeat the definition of command equivalence from above:

Definition cequiv (c1 c2 : com) : Prop := st st' : state,
  c1 / st st' c2 / st st'.

This definition still makes perfect sense in the case of always-terminating programs, so let's apply it to prove some non-deterministic programs equivalent or non-equivalent.
Why only "in the case of always-terminating programs"? Is this a good place to add Phil's note that the big-step semantics cannot distinguish always-terminating from sometimes-terminating programs and the requested pointer to the discussion in Smallstep.v?

Exercise: 3 stars (havoc_swap)

Are the following two programs equivalent?

Definition pXY :=
  HAVOC X;; HAVOC Y.

Definition pYX :=
  HAVOC Y;; HAVOC X.

If you think they are equivalent, prove it. If you think they are not, prove that.

Theorem pXY_cequiv_pYX :
  cequiv pXY pYX ¬cequiv pXY pYX.
Proof. (* FILL IN HERE *) Admitted.

Exercise: 4 stars, optional (havoc_copy)

Are the following two programs equivalent?

Definition ptwice :=
  HAVOC X;; HAVOC Y.

Definition pcopy :=
  HAVOC X;; Y ::= AId X.

If you think they are equivalent, then prove it. If you think they are not, then prove that. (Hint: You may find the assert tactic useful.)

Theorem ptwice_cequiv_pcopy :
  cequiv ptwice pcopy ¬cequiv ptwice pcopy.
Proof. (* FILL IN HERE *) Admitted.
The definition of program equivalence we are using here has some subtle consequences on programs that may loop forever. What cequiv says is that the set of possible terminating outcomes of two equivalent programs is the same. However, in a language with non-determinism, like Himp, some programs always terminate, some programs always diverge, and some programs can non-deterministically terminate in some runs and diverge in others. The final part of the following exercise illustrates this phenomenon.

Exercise: 5 stars, advanced (p1_p2_equiv)

Prove that p1 and p2 are equivalent. In this and the following exercises, try to understand why the cequiv definition has the behavior it has on these examples.

Definition p1 : com :=
  WHILE (BNot (BEq (AId X) (ANum 0))) DO
    HAVOC Y;;
    X ::= APlus (AId X) (ANum 1)
  END.

Definition p2 : com :=
  WHILE (BNot (BEq (AId X) (ANum 0))) DO
    SKIP
  END.

Intuitively, the programs have the same termination behavior: either they loop forever, or they terminate in the same state they started in. We can capture the termination behavior of p1 and p2 individually with these lemmas:

Lemma p1_may_diverge : st st', st X ≠ 0
  ¬ p1 / st st'.
Proof. (* FILL IN HERE *) Admitted.

Lemma p2_may_diverge : st st', st X ≠ 0
  ¬ p2 / st st'.
Proof.
(* FILL IN HERE *) Admitted.

You should use these lemmas to prove that p1 and p2 are actually equivalent.

Theorem p1_p2_equiv : cequiv p1 p2.
Proof. (* FILL IN HERE *) Admitted.

Exercise: 4 stars, advanced (p3_p4_inquiv)

Prove that the following programs are not equivalent.

Definition p3 : com :=
  Z ::= ANum 1;;
  WHILE (BNot (BEq (AId X) (ANum 0))) DO
    HAVOC X;;
    HAVOC Z
  END.

Definition p4 : com :=
  X ::= (ANum 0);;
  Z ::= (ANum 1).

Theorem p3_p4_inequiv : ¬ cequiv p3 p4.
Proof. (* FILL IN HERE *) Admitted.

Exercise: 5 stars, advanced, optional (p5_p6_equiv)


Definition p5 : com :=
  WHILE (BNot (BEq (AId X) (ANum 1))) DO
    HAVOC X
  END.

Definition p6 : com :=
  X ::= ANum 1.

Theorem p5_p6_equiv : cequiv p5 p6.
Proof. (* FILL IN HERE *) Admitted.

End Himp.

Doing Without Extensionality (Optional)

Purists might object to using the functional_extensionality axiom as we have here (e.g. in the proof of the t_update_same lemma). In general, it can be dangerous to add axioms willy nilly, particularly several at once (as they may be mutually inconsistent). In fact, it is known that functional_extensionality and excluded_middle can both be assumed without any problems; nevertheless, some Coq users prefer to avoid such "heavyweight" general techniques and instead try to craft solutions for specific problems that stay within Coq's built-in logic.
For our particular problem here, rather than extending the definition of equality to do what we want on functions representing states, we could instead give an explicit notion of equivalence on states. For example:

Definition stequiv (st1 st2 : state) : Prop :=
  (X:id), st1 X = st2 X.

Notation "st1 '~~' st2" := (stequiv st1 st2) (at level 30).

It is easy to prove that stequiv is an equivalence (i.e., it is reflexive, symmetric, and transitive), so it partitions the set of all states into equivalence classes.

Exercise: 1 star, optional (stequiv_refl)

Lemma stequiv_refl : (st : state),
  st ~~ st.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, optional (stequiv_sym)

Lemma stequiv_sym : (st1 st2 : state),
  st1 ~~ st2
  st2 ~~ st1.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, optional (stequiv_trans)

Lemma stequiv_trans : (st1 st2 st3 : state),
  st1 ~~ st2
  st2 ~~ st3
  st1 ~~ st3.
Proof.
  (* FILL IN HERE *) Admitted.
Another useful fact...

Exercise: 1 star, optional (stequiv_t_update)

Lemma stequiv_t_update : (st1 st2 : state),
  st1 ~~ st2
  (X:id) (n:nat),
  t_update st1 X n ~~ t_update st2 X n.
Proof.
  (* FILL IN HERE *) Admitted.
It is then straightforward to show that aeval and beval behave uniformly on all members of an equivalence class:

Exercise: 2 stars, optional (stequiv_aeval)

Lemma stequiv_aeval : (st1 st2 : state),
  st1 ~~ st2
  (a:aexp), aeval st1 a = aeval st2 a.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, optional (stequiv_beval)

Lemma stequiv_beval : (st1 st2 : state),
  st1 ~~ st2
  (b:bexp), beval st1 b = beval st2 b.
Proof.
  (* FILL IN HERE *) Admitted.
We can also characterize the behavior of ceval on equivalent states (this result is a bit more complicated to write down because ceval is a relation).

Lemma stequiv_ceval: (st1 st2 : state),
  st1 ~~ st2
  (c: com) (st1': state),
    (c / st1 st1')
    st2' : state,
    ((c / st2 st2') st1' ~~ st2').
Proof.
  intros st1 st2 STEQV c st1' CEV1. generalize dependent st2.
  induction CEV1; intros st2 STEQV.
  - (* SKIP *)
    st2. split.
      constructor.
      assumption.
  - (* := *)
    (t_update st2 x n). split.
       constructor. rewrite H. symmetry. apply stequiv_aeval.
       assumption. apply stequiv_t_update. assumption.
  - (* ; *)
    destruct (IHCEV1_1 st2 STEQV) as [st2' [P1 EQV1]].
    destruct (IHCEV1_2 st2' EQV1) as [st2'' [P2 EQV2]].
    st2''. split.
      apply E_Seq with st2'; assumption.
      assumption.
  - (* IfTrue *)
    destruct (IHCEV1 st2 STEQV) as [st2' [P EQV]].
    st2'. split.
      apply E_IfTrue. rewrite H. symmetry. apply stequiv_beval.
      assumption. assumption. assumption.
  - (* IfFalse *)
    destruct (IHCEV1 st2 STEQV) as [st2' [P EQV]].
    st2'. split.
     apply E_IfFalse. rewrite H. symmetry. apply stequiv_beval.
     assumption. assumption. assumption.
  - (* WhileEnd *)
    st2. split.
      apply E_WhileEnd. rewrite H. symmetry. apply stequiv_beval.
      assumption. assumption.
  - (* WhileLoop *)
    destruct (IHCEV1_1 st2 STEQV) as [st2' [P1 EQV1]].
    destruct (IHCEV1_2 st2' EQV1) as [st2'' [P2 EQV2]].
    st2''. split.
      apply E_WhileLoop with st2'. rewrite H. symmetry.
      apply stequiv_beval. assumption. assumption. assumption.
      assumption.
Qed.

Now we need to redefine cequiv to use ¬ instead of =. It is not completely trivial to do this in a way that keeps the definition simple and symmetric, but here is one approach (thanks to Andrew McCreight). We first define a looser variant of that "folds in" the notion of equivalence.

Reserved Notation "c1 '/' st ''' st'" (at level 40, st at level 39).

Inductive ceval' : com state state Prop :=
  | E_equiv : c st st' st'',
    c / st st'
    st' ~~ st''
    c / st ' st''
  where "c1 '/' st ''' st'" := (ceval' c1 st st').

Now the revised definition of cequiv' looks familiar:

Definition cequiv' (c1 c2 : com) : Prop :=
  (st st' : state),
    (c1 / st ' st') (c2 / st ' st').

A sanity check shows that the original notion of command equivalence is at least as strong as this new one. (The converse is not true, naturally.)

Lemma cequiv__cequiv' : (c1 c2: com),
  cequiv c1 c2 cequiv' c1 c2.
Proof.
  unfold cequiv, cequiv'; split; intros.
    inversion H0 ; subst. apply E_equiv with st'0.
    apply (H st st'0); assumption. assumption.
    inversion H0 ; subst. apply E_equiv with st'0.
    apply (H st st'0). assumption. assumption.
Qed.

Exercise: 2 stars, optional (identity_assignment')

Finally, here is our example once more... Notice that we use the t_update_same_no_ext lemma in order to avoid invoking functional extensionality. (You can complete the proofs.)

Theorem t_update_same_no_ext : X x1 x2 (m : total_map X),
  t_update m x1 (m x1) x2 = m x2.
Proof.
  (* FILL IN HERE *) Admitted.

Example identity_assignment' :
  cequiv' SKIP (X ::= AId X).
Proof.
    unfold cequiv'. intros. split; intros.
    - (* -> *)
      inversion H; subst; clear H. inversion H0; subst.
      apply E_equiv with (t_update st'0 X (st'0 X)).
      constructor. { reflexivity. }
      assert (t_update st'0 X (st'0 X) ~~ st'0) as H.
      { unfold stequiv. intros Y.
        apply t_update_same_no_ext. }
      apply (stequiv_trans _ _ _ H H1).
    - (* <- *)
      (* FILL IN HERE *) Admitted.
On the whole, this explicit equivalence approach is considerably harder to work with than relying on functional extensionality. (Coq does have an advanced mechanism called "setoids" that makes working with equivalences somewhat easier, by allowing them to be registered with the system so that standard rewriting tactics work for them almost as well as for equalities.) But it is worth knowing about, because it applies even in situations where the equivalence in question is not over functions. For example, if we chose to represent state mappings as binary search trees, we would need to use an explicit equivalence of this kind.

Additional Exercises

Exercise: 4 stars, optional (for_while_equiv)

This exercise extends the optional add_for_loop exercise from Imp.v, where you were asked to extend the language of commands with C-style for loops. Prove that the command:
      for (c1 ; b ; c2) {
          c3
      }
is equivalent to:
       c1 ;
       WHILE b DO
         c3 ;
         c2
       END
(* FILL IN HERE *)

Exercise: 3 stars, optional (swap_noninterfering_assignments)

Theorem swap_noninterfering_assignments: l1 l2 a1 a2,
  l1l2
  var_not_used_in_aexp l1 a2
  var_not_used_in_aexp l2 a1
  cequiv
    (l1 ::= a1;; l2 ::= a2)
    (l2 ::= a2;; l1 ::= a1).
Proof.
(* Hint: You'll need functional_extensionality *)
(* FILL IN HERE *) Admitted.

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