(** * Equiv: Program Equivalence *)
Require Export 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 [aexp]s and [bexp]s with variables, the definition we want is
clear. We say
that two [aexp]s or [bexp]s are _behaviorally equivalent_ if they
evaluate to the same result _in every state_. *)
Definition aequiv (a1 a2 : aexp) : Prop :=
forall (st:state),
aeval st a1 = aeval st a2.
Definition bequiv (b1 b2 : bexp) : Prop :=
forall (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 :=
forall (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]. 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: {a,b,c,d,e,f,g,h} {i}.
(a)
WHILE X > 0 DO
X ::= X + 1
END
(b)
IFB X = 0 THEN
X ::= X + 1;;
Y ::= 1
ELSE
Y ::= 0
FI;;
X ::= X - Y;;
Y ::= 0
(c)
SKIP
(d)
WHILE X <> 0 DO
X ::= X * Y + 1
END
(e)
Y ::= 0
(f)
Y ::= X + 1;;
WHILE X <> Y DO
Y ::= X + 1
END
(g)
WHILE TRUE DO
SKIP
END
(h)
WHILE X <> X DO
X ::= X + 1
END
(i)
WHILE X <> Y DO
X ::= Y + 1
END
(* FILL IN HERE *)
[] *)
(* ####################################################### *)
(** ** 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: forall c,
cequiv
(SKIP;; c)
c.
Proof. (* WORKED IN CLASS *)
intros c st st'.
split; intros H.
Case "->".
inversion H. subst.
inversion H2. subst.
assumption.
Case "<-".
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: forall c,
cequiv
(c;; SKIP)
c.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** Similarly, here is a simple transformations that simplifies [IFB]
commands: *)
Theorem IFB_true_simple: forall c1 c2,
cequiv
(IFB BTrue THEN c1 ELSE c2 FI)
c1.
Proof.
intros c1 c2.
split; intros H.
Case "->".
inversion H; subst. assumption. inversion H5.
Case "<-".
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: forall b c1 c2,
bequiv b BTrue ->
cequiv
(IFB b THEN c1 ELSE c2 FI)
c1.
Proof.
intros b c1 c2 Hb.
split; intros H.
Case "->".
inversion H; subst.
SCase "b evaluates to true".
assumption.
SCase "b evaluates to false (contradiction)".
unfold bequiv in Hb. simpl in Hb.
rewrite Hb in H5.
inversion H5.
Case "<-".
apply E_IfTrue; try assumption.
unfold bequiv in Hb. simpl in Hb.
rewrite Hb. reflexivity. Qed.
(** **** Exercise: 2 stars (IFB_false) *)
Theorem IFB_false: forall 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: forall 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 : forall b c,
bequiv b BFalse ->
cequiv
(WHILE b DO c END)
SKIP.
Proof.
intros b c Hb. split; intros H.
Case "->".
inversion H; subst.
SCase "E_WhileEnd".
apply E_Skip.
SCase "E_WhileLoop".
rewrite Hb in H2. inversion H2.
Case "<-".
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 : forall 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.
ceval_cases (induction H) Case;
(* 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: *)
Case "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.
Case "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 (WHILE_true) *)
(** Prove the following theorem. _Hint_: You'll want to use
[WHILE_true_nonterm] here. *)
Theorem WHILE_true: forall b c,
bequiv b BTrue ->
cequiv
(WHILE b DO c END)
(WHILE BTrue DO SKIP END).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
Theorem loop_unrolling: forall 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.
Case "->".
inversion Hce; subst.
SCase "loop doesn't run".
apply E_IfFalse. assumption. apply E_Skip.
SCase "loop runs".
apply E_IfTrue. assumption.
apply E_Seq with (st' := st'0). assumption. assumption.
Case "<-".
inversion Hce; subst.
SCase "loop runs".
inversion H5; subst.
apply E_WhileLoop with (st' := st'0).
assumption. assumption. assumption.
SCase "loop doesn't run".
inversion H5; subst. apply E_WhileEnd. assumption. Qed.
(** **** Exercise: 2 stars, optional (seq_assoc) *)
Theorem seq_assoc : forall c1 c2 c3,
cequiv ((c1;;c2);;c3) (c1;;(c2;;c3)).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** ** The Functional Equivalence Axiom *)
(** Finally, let's look at simple equivalences involving assignments.
For example, we might expect to be able to show that [X ::= AId X]
is equivalent to [SKIP]. However, when we try to show it, we get
stuck in an interesting way. *)
Theorem identity_assignment_first_try : forall (X:id),
cequiv (X ::= AId X) SKIP.
Proof.
intros. split; intro H.
Case "->".
inversion H; subst. simpl.
replace (update st X (st X)) with st.
constructor.
(* Stuck... *) Abort.
(** Here we're stuck. The goal looks reasonable, but in fact it is not
provable! If we look back at the set of lemmas we proved about
[update] in the last chapter, we can see that lemma [update_same]
almost does the job, but not quite: it says that the original and
updated states agree at all values, but this is not the same thing
as saying that they are [=] in Coq's sense! *)
(** What is going on here? Recall that our states are just
functions from identifiers to values. For Coq, functions are only
equal when their definitions are syntactically the same, modulo
simplification. (This is the only way we can legally apply the
[refl_equal] constructor of the inductively defined proposition
[eq]!) In practice, for functions built up by repeated uses of the
[update] operation, this means that two functions can be proven
equal only if they were constructed using the _same_ [update]
operations, applied in the same order. In the theorem above, the
sequence of updates on the first parameter [cequiv] is one longer
than for the second parameter, so it is no wonder that the
equality doesn't hold. *)
(** ** *)
(** This problem is actually quite general. If we try to prove other
simple facts, such as
cequiv (X ::= X + 1;;
X ::= X + 1)
(X ::= X + 2)
or
cequiv (X ::= 1;; Y ::= 2)
(y ::= 2;; X ::= 1)
we'll get stuck in the same way: we'll have two functions that
behave the same way on all inputs, but cannot be proven to be [eq]
to each other.
The reasoning principle we would like to use in these situations
is called _functional extensionality_:
forall x, f x = g x
-------------------
f = g
Although this principle is not derivable in Coq's built-in logic,
it is safe to add it as an additional _axiom_. *)
Axiom functional_extensionality : forall {X Y: Type} {f g : X -> Y},
(forall (x: X), f x = g x) -> f = g.
(** It can be shown that adding this axiom doesn't introduce any
inconsistencies into Coq. (In this way, it is similar to adding
one of the classical logic axioms, such as [excluded_middle].) *)
(** With the benefit of this axiom we can prove our theorem. *)
Theorem identity_assignment : forall (X:id),
cequiv
(X ::= AId X)
SKIP.
Proof.
intros. split; intro H.
Case "->".
inversion H; subst. simpl.
replace (update st X (st X)) with st.
constructor.
apply functional_extensionality. intro.
rewrite update_same; reflexivity.
Case "<-".
inversion H; subst.
assert (st' = (update st' X (st' X))).
apply functional_extensionality. intro.
rewrite update_same; reflexivity.
rewrite H0 at 2.
constructor. reflexivity.
Qed.
(** **** Exercise: 2 stars (assign_aequiv) *)
Theorem assign_aequiv : forall 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
[com]s really are _equivalences_ -- i.e., that they are reflexive,
symmetric, and transitive. The proofs are all easy. *)
Lemma refl_aequiv : forall (a : aexp), aequiv a a.
Proof.
intros a st. reflexivity. Qed.
Lemma sym_aequiv : forall (a1 a2 : aexp),
aequiv a1 a2 -> aequiv a2 a1.
Proof.
intros a1 a2 H. intros st. symmetry. apply H. Qed.
Lemma trans_aequiv : forall (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 : forall (b : bexp), bequiv b b.
Proof.
unfold bequiv. intros b st. reflexivity. Qed.
Lemma sym_bequiv : forall (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 : forall (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 : forall (c : com), cequiv c c.
Proof.
unfold cequiv. intros c st st'. apply iff_refl. Qed.
Lemma sym_cequiv : forall (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'.
SCase "Proof of assertion". apply H.
apply iff_sym. assumption.
Qed.
Lemma iff_trans : forall (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 : forall (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 : forall i a1 a1',
aequiv a1 a1' ->
cequiv (CAss i a1) (CAss i a1').
Proof.
intros i a1 a2 Heqv st st'.
split; intros Hceval.
Case "->".
inversion Hceval. subst. apply E_Ass.
rewrite Heqv. reflexivity.
Case "<-".
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 : forall 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.
Case "->".
remember (WHILE b1 DO c1 END) as cwhile eqn:Heqcwhile.
induction Hce; inversion Heqcwhile; subst.
SCase "E_WhileEnd".
apply E_WhileEnd. rewrite <- Hb1e. apply H.
SCase "E_WhileLoop".
apply E_WhileLoop with (st' := st').
SSCase "show loop runs". rewrite <- Hb1e. apply H.
SSCase "body execution".
apply (Hc1e st st'). apply Hce1.
SSCase "subsequent loop execution".
apply IHHce2. reflexivity.
Case "<-".
remember (WHILE b1' DO c1' END) as c'while eqn:Heqc'while.
induction Hce; inversion Heqc'while; subst.
SCase "E_WhileEnd".
apply E_WhileEnd. rewrite -> Hb1e. apply H.
SCase "E_WhileLoop".
apply E_WhileLoop with (st' := st').
SSCase "show loop runs". rewrite -> Hb1e. apply H.
SSCase "body execution".
apply (Hc1e st st'). apply Hce1.
SSCase "subsequent loop execution".
apply IHHce2. reflexivity. Qed.
(** **** Exercise: 3 stars, optional (CSeq_congruence) *)
Theorem CSeq_congruence : forall 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 : forall 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
[aexp]s, [bexp]s, and [com]s. *)
Definition atrans_sound (atrans : aexp -> aexp) : Prop :=
forall (a : aexp),
aequiv a (atrans a).
Definition btrans_sound (btrans : bexp -> bexp) : Prop :=
forall (b : bexp),
bequiv b (btrans b).
Definition ctrans_sound (ctrans : com -> com) : Prop :=
forall (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 n => ANum n
| AId i => AId 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 [bexp]s (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
| BTrue => BTrue
| BFalse => BFalse
| 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 ble_nat n1 n2 then BTrue else BFalse
| (a1', a2') => BLe a1' a2'
end
| BNot b1 =>
match (fold_constants_bexp b1) with
| BTrue => BFalse
| BFalse => BTrue
| 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
| BTrue => fold_constants_com c1
| BFalse => fold_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
| BTrue => WHILE BTrue DO SKIP END
| BFalse => SKIP
| 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.
aexp_cases (induction a) Case; 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.
bexp_cases (induction b) Case;
(* BTrue and BFalse are immediate *)
try reflexivity.
Case "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.
Case "BLe".
(* FILL IN HERE *) admit.
Case "BNot".
simpl. remember (fold_constants_bexp b) as b' eqn:Heqb'.
rewrite IHb.
destruct b'; reflexivity.
Case "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. Qed.
(** [] *)
(** **** 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.
com_cases (induction c) Case; simpl.
Case "SKIP". apply refl_cequiv.
Case "::=". apply CAss_congruence. apply fold_constants_aexp_sound.
Case ";;". apply CSeq_congruence; assumption.
Case "IFB".
assert (bequiv b (fold_constants_bexp b)).
SCase "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).
SCase "b always true".
apply trans_cequiv with c1; try assumption.
apply IFB_true; assumption.
SCase "b always false".
apply trans_cequiv with c2; try assumption.
apply IFB_false; assumption.
Case "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 [aexp]s,
without states.
Write a new version of this function that accounts for variables,
and analogous ones for [bexp]s 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 n => ANum n
| AId i' => if eq_id_dec i i' then u else AId i'
| APlus a1 a2 => APlus (subst_aexp i u a1) (subst_aexp i u a2)
| AMinus a1 a2 => AMinus (subst_aexp i u a1) (subst_aexp i u a2)
| AMult a1 a2 => AMult (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 := forall 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 (update (update empty_state X 1) Y 1) as st1.
remember (update (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' := (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: forall n, var_not_used_in_aexp X (ANum n)
| VNUId: forall Y, X <> Y -> var_not_used_in_aexp X (AId Y)
| VNUPlus: forall 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: forall 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: forall 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 : forall i st a ni,
var_not_used_in_aexp i a ->
aeval (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(++x, x)
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 *)
Tactic Notation "com_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "SKIP" | Case_aux c "::=" | Case_aux c ";;"
| Case_aux c "IFB" | Case_aux c "WHILE" | Case_aux c "HAVOC" ].
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 : forall st : state, SKIP / st || st
| E_Ass : forall (st : state) (a1 : aexp) (n : nat) (X : id),
aeval st a1 = n -> (X ::= a1) / st || update st X n
| E_Seq : forall (c1 c2 : com) (st st' st'' : state),
c1 / st || st' -> c2 / st' || st'' -> (c1 ;; c2) / st || st''
| E_IfTrue : forall (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 : forall (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 : forall (b1 : bexp) (st : state) (c1 : com),
beval st b1 = false -> (WHILE b1 DO c1 END) / st || st
| E_WhileLoop : forall (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').
Tactic Notation "ceval_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "E_Skip" | Case_aux c "E_Ass" | Case_aux c "E_Seq"
| Case_aux c "E_IfTrue" | Case_aux c "E_IfFalse"
| Case_aux c "E_WhileEnd" | Case_aux c "E_WhileLoop"
(* FILL IN HERE *)
].
(** As a sanity check, the following claims should be provable for
your definition: *)
Example havoc_example1 : (HAVOC X) / empty_state || update empty_state X 0.
Proof.
(* FILL IN HERE *) Admitted.
Example havoc_example2 :
(SKIP;; HAVOC Z) / empty_state || 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 := forall 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. *)
(** **** 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 : forall st st', st X <> 0 ->
~ p1 / st || st'.
Proof. (* FILL IN HERE *) Admitted.
Lemma p2_may_diverge : forall 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. In general, it can be quite dangerous to add axioms,
particularly several at once (as they may be mutually
inconsistent). In fact, [functional_extensionality] and
[excluded_middle] can both be assumed without any problems, but
some Coq users prefer to avoid such "heavyweight" general
techniques, and instead craft solutions for specific problems that
stay within Coq's standard 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 :=
forall (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 : forall (st : state),
st ~ st.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, optional (stequiv_sym) *)
Lemma stequiv_sym : forall (st1 st2 : state),
st1 ~ st2 ->
st2 ~ st1.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, optional (stequiv_trans) *)
Lemma stequiv_trans : forall (st1 st2 st3 : state),
st1 ~ st2 ->
st2 ~ st3 ->
st1 ~ st3.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** Another useful fact... *)
(** **** Exercise: 1 star, optional (stequiv_update) *)
Lemma stequiv_update : forall (st1 st2 : state),
st1 ~ st2 ->
forall (X:id) (n:nat),
update st1 X n ~ 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 : forall (st1 st2 : state),
st1 ~ st2 ->
forall (a:aexp), aeval st1 a = aeval st2 a.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars, optional (stequiv_beval) *)
Lemma stequiv_beval : forall (st1 st2 : state),
st1 ~ st2 ->
forall (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: forall (st1 st2 : state),
st1 ~ st2 ->
forall (c: com) (st1': state),
(c / st1 || st1') ->
exists st2' : state,
((c / st2 || st2') /\ st1' ~ st2').
Proof.
intros st1 st2 STEQV c st1' CEV1. generalize dependent st2.
induction CEV1; intros st2 STEQV.
Case "SKIP".
exists st2. split.
constructor.
assumption.
Case ":=".
exists (update st2 x n). split.
constructor. rewrite <- H. symmetry. apply stequiv_aeval.
assumption. apply stequiv_update. assumption.
Case ";".
destruct (IHCEV1_1 st2 STEQV) as [st2' [P1 EQV1]].
destruct (IHCEV1_2 st2' EQV1) as [st2'' [P2 EQV2]].
exists st2''. split.
apply E_Seq with st2'; assumption.
assumption.
Case "IfTrue".
destruct (IHCEV1 st2 STEQV) as [st2' [P EQV]].
exists st2'. split.
apply E_IfTrue. rewrite <- H. symmetry. apply stequiv_beval.
assumption. assumption. assumption.
Case "IfFalse".
destruct (IHCEV1 st2 STEQV) as [st2' [P EQV]].
exists st2'. split.
apply E_IfFalse. rewrite <- H. symmetry. apply stequiv_beval.
assumption. assumption. assumption.
Case "WhileEnd".
exists st2. split.
apply E_WhileEnd. rewrite <- H. symmetry. apply stequiv_beval.
assumption. assumption.
Case "WhileLoop".
destruct (IHCEV1_1 st2 STEQV) as [st2' [P1 EQV1]].
destruct (IHCEV1_2 st2' EQV1) as [st2'' [P2 EQV2]].
exists 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 : forall 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 :=
forall (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' : forall (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... (You can complete the
proof.) *)
Example identity_assignment' :
cequiv' SKIP (X ::= AId X).
Proof.
unfold cequiv'. intros. split; intros.
Case "->".
inversion H; subst; clear H. inversion H0; subst.
apply E_equiv with (update st'0 X (st'0 X)).
constructor. reflexivity. apply stequiv_trans with st'0.
unfold stequiv. intros. apply update_same.
reflexivity. assumption.
Case "<-".
(* 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: forall l1 l2 a1 a2,
l1 <> l2 ->
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.
(** [] *)