(** * Imp: Simple Imperative Programs *)
(* $Date: 2012-09-08 20:51:57 -0400 (Sat, 08 Sep 2012) $ *)
(** In this chapter, we begin a new direction that will
continue for the rest of the course. Up to now we've been mostly
studying Coq itself, but from now on we'll mostly be using Coq to
formalize other things.
Our first case study is a _simple imperative programming language_
called Imp. Here is a familiar mathematical function written in
Imp.
Z ::= X;
Y ::= 1;
WHILE not (Z = 0) DO
Y ::= Y * Z;
Z ::= Z - 1
END
*)
(** This chapter looks at how to define the _syntax_ and _semantics_
of Imp; the chapters that follow develop a theory of _program
equivalence_ and introduce _Hoare Logic_, the best-known logic for
reasoning about imperative programs. *)
(* ####################################################### *)
(** *** Sflib *)
(** A minor technical point: Instead of asking Coq to import our
earlier definitions from chapter [Logic], we import a small library
called [Sflib.v], containing just a few definitions and theorems
from earlier chapters that we'll actually use in the rest of the
course. This change should be nearly invisible, since most of what's
missing from Sflib has identical definitions in the Coq standard
library. The main reason for doing it is to tidy the global Coq
environment so that, for example, it is easier to search for
relevant theorems. *)
Require Export SfLib.
(* ####################################################### *)
(** * Arithmetic and Boolean Expressions *)
(** We'll present Imp in three parts: first a core language of
_arithmetic and boolean expressions_, then an extension of these
expressions with _variables_, and finally a language of _commands_
including assignment, conditions, sequencing, and loops. *)
(* ####################################################### *)
(** ** Syntax *)
Module AExp.
(** These two definitions specify the _abstract syntax_ of
arithmetic and boolean expressions. *)
Inductive aexp : Type :=
| ANum : nat -> aexp
| APlus : aexp -> aexp -> aexp
| AMinus : aexp -> aexp -> aexp
| AMult : aexp -> aexp -> aexp.
Inductive bexp : Type :=
| BTrue : bexp
| BFalse : bexp
| BEq : aexp -> aexp -> bexp
| BLe : aexp -> aexp -> bexp
| BNot : bexp -> bexp
| BAnd : bexp -> bexp -> bexp.
(** In this chapter, we'll elide the translation from the
concrete syntax that a programmer would actually write to these
abstract syntax trees -- the process that, for example, would
translate the string ["1+2*3"] to the AST [APlus (ANum
1) (AMult (ANum 2) (ANum 3))]. The optional chapter [ImpParser]
develops a simple implementation of a lexical analyzer and parser
that can perform this translation. You do _not_ need to
understand that file to understand this one, but if you haven't
taken a course where these techniques are covered (e.g., a
compilers course) you may want to skim it. *)
(** For comparison, here's a conventional BNF (Backus-Naur Form)
grammar defining the same abstract syntax:
aexp ::= nat
| aexp '+' aexp
| aexp '-' aexp
| aexp '*' aexp
bexp ::= true
| false
| aexp '=' aexp
| aexp '<=' aexp
| bexp 'and' bexp
| 'not' bexp
*)
(** Compared to the Coq version above...
- The BNF is more informal -- for example, it gives some
suggestions about the surface syntax of expressions (like the
fact that the addition operation is written [+] and is an
infix symbol) while leaving other aspects of lexical analysis
and parsing (like the relative precedence of [+], [-], and
[*]) unspecified. Some additional information -- and human
intelligence -- would be required to turn this description
into a formal definition (when implementing a compiler, for
example).
The Coq version consistently omits all this information and
concentrates on the abstract syntax only.
- On the other hand, the BNF version is lighter and
easier to read. Its informality makes it flexible, which is
a huge advantage in situations like discussions at the
blackboard, where conveying general ideas is more important
than getting every detail nailed down precisely.
Indeed, there are dozens of BNF-like notations and people
switch freely among them, usually without bothering to say which
form of BNF they're using because there is no need to: a
rough-and-ready informal understanding is all that's
needed. *)
(** It's good to be comfortable with both sorts of notations: informal
ones for communicating between humans and formal ones for carrying
out implementations and proofs. *)
(* ####################################################### *)
(** ** Evaluation *)
(** _Evaluating_ an arithmetic expression produces a number. *)
Fixpoint aeval (e : aexp) : nat :=
match e with
| ANum n => n
| APlus a1 a2 => (aeval a1) + (aeval a2)
| AMinus a1 a2 => (aeval a1) - (aeval a2)
| AMult a1 a2 => (aeval a1) * (aeval a2)
end.
Example test_aeval1:
aeval (APlus (ANum 2) (ANum 2)) = 4.
Proof. reflexivity. Qed.
(** Similarly, evaluating a boolean expression yields a boolean. *)
Fixpoint beval (e : bexp) : bool :=
match e with
| BTrue => true
| BFalse => false
| BEq a1 a2 => beq_nat (aeval a1) (aeval a2)
| BLe a1 a2 => ble_nat (aeval a1) (aeval a2)
| BNot b1 => negb (beval b1)
| BAnd b1 b2 => andb (beval b1) (beval b2)
end.
(* ####################################################### *)
(** ** Optimization *)
(** We haven't defined very much yet, but we can already get
some mileage out of the definitions. Suppose we define a function
that takes an arithmetic expression and slightly simplifies it,
changing every occurrence of [0+e] (i.e., [(APlus (ANum 0) e])
into just [e]. *)
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.
(** To make sure our optimization is doing the right thing we
can test it on some examples and see if the output looks OK. *)
Example test_optimize_0plus:
optimize_0plus (APlus (ANum 2)
(APlus (ANum 0)
(APlus (ANum 0) (ANum 1))))
= APlus (ANum 2) (ANum 1).
Proof. reflexivity. Qed.
(** But if we want to be sure the optimization is correct --
i.e., that evaluating an optimized expression gives the same
result as the original -- we should prove it. *)
Theorem optimize_0plus_sound: forall e,
aeval (optimize_0plus e) = aeval e.
Proof.
intros e. induction e.
Case "ANum". reflexivity.
Case "APlus". destruct e1.
SCase "e1 = ANum n". destruct n.
SSCase "n = 0". simpl. apply IHe2.
SSCase "n <> 0". simpl. rewrite IHe2. reflexivity.
SCase "e1 = APlus e1_1 e1_2".
simpl. simpl in IHe1. rewrite IHe1.
rewrite IHe2. reflexivity.
SCase "e1 = AMinus e1_1 e1_2".
simpl. simpl in IHe1. rewrite IHe1.
rewrite IHe2. reflexivity.
SCase "e1 = AMult e1_1 e1_2".
simpl. simpl in IHe1. rewrite IHe1.
rewrite IHe2. reflexivity.
Case "AMinus".
simpl. rewrite IHe1. rewrite IHe2. reflexivity.
Case "AMult".
simpl. rewrite IHe1. rewrite IHe2. reflexivity. Qed.
(* ####################################################### *)
(** * Coq Automation *)
(** The repetition in this last proof is starting to be a little
annoying. If either the language of arithmetic expressions or the
optimization being proved sound were significantly more complex,
it would begin to be a real problem.
So far, we've been doing all our proofs using just a small handful
of Coq's tactics and completely ignoring its powerful facilities
for constructing parts of proofs automatically. This section
introduces some of these facilities, and we will see more over the
next several chapters. Getting used to them will take some
energy -- Coq's automation is a power tool -- but it will allow us
to scale up our efforts to more complex definitions and more
interesting properties without becoming overwhelmed by boring,
repetitive, low-level details. *)
(* ####################################################### *)
(** ** Tacticals *)
(** _Tacticals_ is Coq's term for tactics that take other tactics as
arguments -- "higher-order tactics," if you will. *)
(* ####################################################### *)
(** *** The [repeat] Tactical *)
(** The [repeat] tactical takes another tactic and keeps applying
this tactic until the tactic fails. Here is an example showing
that [100] is even using repeat. *)
Theorem ev100 : ev 100.
Proof.
repeat (apply ev_SS). (* applies ev_SS 50 times,
then [apply ev_SS] fails *)
apply ev_0.
Qed.
(* Print ev100. *)
(** The [repeat T] tactic never fails; if the tactic [T] doesn't apply
to the original goal, then repeat still succeeds without changing
the original goal (it repeats zero times). *)
Theorem ev100' : ev 100.
Proof.
repeat (apply ev_0). (* doesn't fail, applies ev_0 zero times *)
repeat (apply ev_SS). apply ev_0. (* we can continue the proof *)
Qed.
(** The [repeat T] tactic does not have any bound on the number of
times it applies [T]. If [T] is a tactic that always succeeds then
repeat [T] will loop forever (e.g. [repeat simpl] loops forever
since [simpl] always succeeds). While Coq's term language is
guaranteed to terminate, Coq's tactic language is not. *)
(* ####################################################### *)
(** *** The [try] Tactical *)
(** A similar tactical is [try]: If [T] is a tactic, then [try T]
is a tactic that is just like [T] except that, if [T] fails,
[try T] _successfully_ does nothing at all (instead of failing). *)
Theorem silly1 : forall ae, aeval ae = aeval ae.
Proof. try reflexivity. (* this just does [reflexivity] *) Qed.
Theorem silly2 : forall (P : Prop), P -> P.
Proof.
intros P HP.
try reflexivity. (* just [reflexivity] would have failed *)
apply HP. (* we can still finish the proof in some other way *)
Qed.
(** Using [try] in a completely manual proof is a bit silly, but we'll
see below that [try] is very useful for doing automated proofs
in conjunction with the [;] tactical. *)
(* ####################################################### *)
(** *** The [;] Tactical (Simple Form) *)
(** In its simplest and most commonly used form, the [;] tactical
takes 2 tactics as argument: [T;T'] first performs the tactic [T]
and then performs the tactic [T'] on _each subgoal_ generated by
[T]. *)
(** For example, consider the following trivial lemma: *)
Lemma foo : forall n, ble_nat 0 n = true.
Proof.
intros.
destruct n.
(* Leaves two subgoals... *)
Case "n=0". simpl. reflexivity.
Case "n=Sn'". simpl. reflexivity.
(* ... which are discharged similarly *)
Qed.
(** We can simplify this proof using the [;] tactical: *)
Lemma foo' : forall n, ble_nat 0 n = true.
Proof.
intros.
destruct n; (* [destruct] the current goal *)
simpl; (* then [simpl] each resulting subgoal *)
reflexivity. (* then do [reflexivity] on each resulting subgoal *)
Qed.
(** Using [try] and [;] together, we can get rid of the repetition in
the proof that was bothering us a little while ago. *)
Theorem optimize_0plus_sound': forall e,
aeval (optimize_0plus e) = aeval e.
Proof.
intros e.
induction e;
(* Most cases follow directly by the IH *)
try (simpl; rewrite IHe1; rewrite IHe2; reflexivity).
(* The remaining cases -- ANum and APlus -- are more
interesting... *)
Case "ANum". reflexivity.
Case "APlus".
destruct e1;
(* Again, most cases follow directly by the IH *)
try (simpl; simpl in IHe1; rewrite IHe1;
rewrite IHe2; reflexivity).
(* The interesting case, on which the [try...] does nothing,
is when [e1 = ANum n]. In this case, we have to destruct
[n] (to see whether the optimization applies) and rewrite
with the induction hypothesis. *)
SCase "e1 = ANum n". destruct n;
simpl; rewrite IHe2; reflexivity. Qed.
(** Coq experts often use this "[...; try... ]" idiom after a
tactic like [induction] to take care of many similar cases all at
once. Naturally, this practice has an analog in informal
proofs. *)
(** Here is an informal proof of this theorem that
matches the structure of the formal one:
_Theorem_: For all arithmetic expressions [e],
aeval (optimize_0plus e) = aeval e.
_Proof_: By induction on [e]. The [AMinus] and [AMult] cases
follow directly from the IH. The remaining cases are as follows:
- Suppose [e = ANum n] for some [n]. We must show
aeval (optimize_0plus (ANum n)) = aeval (ANum n).
This is immediate from the definition of [optimize_0plus].
- Suppose [e = APlus e1 e2] for some [e1] and [e2]. We
must show
aeval (optimize_0plus (APlus e1 e2))
= aeval (APlus e1 e2).
Consider the possible forms of [e1]. For most of them,
[optimize_0plus] simply calls itself recursively for the
subexpressions and rebuilds a new expression of the same form
as [e1]; in these cases, the result follows directly from the
IH.
The interesting case is when [e1 = ANum n] for some [n].
If [n = ANum 0], then
optimize_0plus (APlus e1 e2) = optimize_0plus e2
and the IH for [e2] is exactly what we need. On the other
hand, if [n = S n'] for some [n'], then again [optimize_0plus]
simply calls itself recursively, and the result follows from
the IH. [] *)
(** This proof can still be improved: the first case (for [e = ANum
n]) is very trivial -- even more trivial than the cases that we
said simply followed from the IH -- yet we have chosen to write it
out in full. It would be better and clearer to drop it and just
say, at the top, "Most cases are either immediate or direct from
the IH. The only interesting case is the one for [APlus]..." We
can make the same improvement in our formal proof too. Here's how
it looks: *)
Theorem optimize_0plus_sound'': forall e,
aeval (optimize_0plus e) = aeval e.
Proof.
intros e.
induction e;
(* Most cases follow directly by the IH *)
try (simpl; rewrite IHe1; rewrite IHe2; reflexivity);
(* ... or are immediate by definition *)
try reflexivity.
(* The interesting case is when e = APlus e1 e2. *)
Case "APlus".
destruct e1; try (simpl; simpl in IHe1; rewrite IHe1;
rewrite IHe2; reflexivity).
SCase "e1 = ANum n". destruct n;
simpl; rewrite IHe2; reflexivity. Qed.
(* ####################################################### *)
(** *** The [;] Tactical (General Form) *)
(** The [;] tactical has a more general than the simple [T;T'] we've
seen above, and which is sometimes also useful.
If [T], [T1], ..., [Tn] are tactics, then
T; [T1 | T2 | ... | Tn]
is a tactic that first performs [T] and then performs [T1] on the
first subgoal generated by [T], performs [T2] on the second
subgoal, etc.
So [T;T'] is just special notation for the case when all of the
[Ti]'s are the same tactic; i.e. [T;T'] is just a shorthand for:
T; [T' | T' | ... | T']
The form [T;T'] is used most often in practice. *)
(* ####################################################### *)
(** ** Defining New Tactic Notations *)
(** Coq also provides several ways of "programming" tactic scripts.
- The [Tactic Notation] idiom illustrated below gives a handy
way to define "shorthand tactics" that bundle several tactics
into a single command.
- For more sophisticated programming, Coq offers a small
built-in programming language called [Ltac] with primitives
that can examine and modify the proof state. The details are
a bit too complicated to get into here (and it is generally
agreed that [Ltac] is not the most beautiful part of Coq's
design!), but they can be found in the reference manual, and
there are many examples of [Ltac] definitions in the Coq
standard library that you can use as examples.
- There is also an OCaml API, which can be used to build tactics
that access Coq's internal structures at a lower level, but
this is seldom worth the trouble for ordinary Coq users.
The [Tactic Notation] mechanism is the easiest to come to grips with,
and it offers plenty of power for many purposes. Here's an example.
*)
Tactic Notation "simpl_and_try" tactic(c) :=
simpl;
try c.
(** This defines a new tactical called [simpl_and_try] which
takes one tactic [c] as an argument, and is defined to be
equivalent to the tactic [simpl; try c]. For example, writing
"[simpl_and_try reflexivity.]" in a proof would be the same as
writing "[simpl; try reflexivity.]" *)
(** The next subsection gives a more sophisticated use of this
feature... *)
(* ####################################################### *)
(** *** Bulletproofing Case Analyses *)
(** Being able to deal with most of the cases of an [induction]
or [destruct] all at the same time is very convenient, but it can
also be a little confusing. One problem that often comes up is
that _maintaining_ proofs written in this style can be difficult.
For example, suppose that, later, we extended the definition of
[aexp] with another constructor that also required a special
argument. The above proof might break because Coq generated the
subgoals for this constructor before the one for [APlus], so that,
at the point when we start working on the [APlus] case, Coq is
actually expecting the argument for a completely different
constructor. What we'd like is to get a sensible error message
saying "I was expecting the [AFoo] case at this point, but the
proof script is talking about [APlus]." Here's a nice trick (due
to Aaron Bohannon) that smoothly achieves this. *)
Tactic Notation "aexp_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "ANum" | Case_aux c "APlus"
| Case_aux c "AMinus" | Case_aux c "AMult" ].
(** ([Case_aux] implements the common functionality of [Case],
[SCase], [SSCase], etc. For example, [Case "foo"] is defined as
[Case_aux Case "foo".) *)
(** For example, if [e] is a variable of type [aexp], then doing
aexp_cases (induction e) Case
will perform an induction on [e] (the same as if we had just typed
[induction e]) and _also_ add a [Case] tag to each subgoal
generated by the [induction], labeling which constructor it comes
from. For example, here is yet another proof of
[optimize_0plus_sound], using [aexp_cases]: *)
Theorem optimize_0plus_sound''': forall e,
aeval (optimize_0plus e) = aeval e.
Proof.
intros e.
aexp_cases (induction e) Case;
try (simpl; rewrite IHe1; rewrite IHe2; reflexivity);
try reflexivity.
(* At this point, there is already an ["APlus"] case name
in the context. The [Case "APlus"] here in the proof
text has the effect of a sanity check: if the "Case"
string in the context is anything _other_ than ["APlus"]
(for example, because we added a clause to the definition
of [aexp] and forgot to change the proof) we'll get a
helpful error at this point telling us that this is now
the wrong case. *)
Case "APlus".
aexp_cases (destruct e1) SCase;
try (simpl; simpl in IHe1; rewrite IHe1; rewrite IHe2; reflexivity).
SCase "ANum". destruct n;
simpl; rewrite IHe2; reflexivity. Qed.
(** **** Exercise: 3 stars (optimize_0plus_b) *)
(* EXPECTED *)
(** Since the [optimize_0plus] tranformation doesn't change the value
of [aexp]s, we should be able to apply it to all the [aexp]s that
appear in a [bexp] without changing the [bexp]'s value. Write a
function which performs that transformation on [bexp]s, and prove
it is sound. Use the tacticals we've just seen to make the proof
as elegant as possible. *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 4 stars, optional (optimizer) *)
(* NO SOLUTION *)
(** _Design exercise_: The optimization implemented by our
[optimize_0plus] function is only one of many imaginable
optimizations on arithmetic and boolean expressions. Write a more
sophisticated optimizer and prove it correct.
*)
(** [] *)
(* ####################################################### *)
(** ** The [omega] Tactic *)
(** The [omega] tactic implements a decision procedure for a subset of
first-order logic called _Presburger arithmetic_. It is based on
the Omega algorithm invented in 1992 by William Pugh.
If the goal is a universally quantified formula made out of
- numeric constants, addition ([+] and [S]), subtraction ([-]
and [pred]), and multiplication by constants (this is what
makes it Presburger arithmetic),
- equality ([=] and [<>]) and inequality ([<=]), and
- the logical connectives [/\], [\/], [~], and [->],
then invoking [omega] will either solve the goal or tell you that
it is actually false. *)
Example silly_presburger_example : forall m n o p,
m + n <= n + o /\ o + 3 = p + 3 ->
m <= p.
Proof.
intros. omega.
Qed.
(** Andrew Appel calls this the "Santa Claus tactic." We'll see
examples of its use below. *)
(* ####################################################### *)
(** ** A Few More Handy Tactics *)
(** Finally, here are some miscellaneous tactics that you may find
convenient.
- [clear H]: Delete hypothesis [H] from the context.
- [subst x]: Find an assumption [x = e] or [e = x] in the
context, replace [x] with [e] throughout the context and
current goal, and clear the assumption.
- [subst]: Substitute away _all_ assumptions of the form [x = e]
or [e = x].
- [rename... into...]: Change the name of a hypothesis in the
proof context. For example, if the context includes a variable
named [x], then [rename x into y] will change all occurrences
of [x] to [y].
- [assumption]: Try to find a hypothesis [H] in the context that
exactly matches the goal; if one is found, behave just like
[apply H].
- [contradiction]: Try to find a hypothesis [H] in the current
context that is logically equivalent to [False]. If one is
found, solve the goal.
- [constructor]: Try to find a constructor [c] (from some
[Inductive] definition in the current environment) that can be
applied to solve the current goal. If one is found, behave
like [apply c].
We'll see many examples of these in the proofs below. *)
(* ####################################################### *)
(** * Evaluation as a Relation *)
(** We have presented [aeval] and [beval] as functions defined by
[Fixpoints]. Another way to think about evaluation, one that we
will see is often more flexible, is as a _relation_ between
expressions and their values. This leads naturally to [Inductive]
definitions like the following one for arithmetic
expressions... *)
Module aevalR_first_try.
Inductive aevalR : aexp -> nat -> Prop :=
| E_ANum : forall (n: nat),
aevalR (ANum n) n
| E_APlus : forall (e1 e2: aexp) (n1 n2: nat),
aevalR e1 n1 ->
aevalR e2 n2 ->
aevalR (APlus e1 e2) (n1 + n2)
| E_AMinus: forall (e1 e2: aexp) (n1 n2: nat),
aevalR e1 n1 ->
aevalR e2 n2 ->
aevalR (AMinus e1 e2) (n1 - n2)
| E_AMult : forall (e1 e2: aexp) (n1 n2: nat),
aevalR e1 n1 ->
aevalR e2 n2 ->
aevalR (AMult e1 e2) (n1 * n2).
(** As is often the case with relations, we'll find it
convenient to define infix notation for [aevalR]. We'll write [e
|| n] to mean that arithmetic expression [e] evaluates to value
[n]. (This notation is one place where the limitation to ASCII
symbols becomes a little bothersome. The standard notation for
the evaluation relation is a double down-arrow. We'll typeset it
like this in the HTML version of the notes and use a double
vertical bar as the closest approximation in [.v] files.) *)
Notation "e '||' n" := (aevalR e n) : type_scope.
End aevalR_first_try.
(** In fact, Coq provides a way to use this notation in the definition
of [aevalR] itself. This avoids situations where we're working on
a proof involving statements in the form [e || n] but we have to
refer back to a definition written using the form [aevalR e n].
We do this by first "reserving" the notation, then giving the
definition together with a declaration of what the notation
means. *)
Reserved Notation "e '||' n" (at level 50, left associativity).
Inductive aevalR : aexp -> nat -> Prop :=
| E_ANum : forall (n:nat),
(ANum n) || n
| E_APlus : forall (e1 e2: aexp) (n1 n2 : nat),
(e1 || n1) -> (e2 || n2) -> (APlus e1 e2) || (n1 + n2)
| E_AMinus : forall (e1 e2: aexp) (n1 n2 : nat),
(e1 || n1) -> (e2 || n2) -> (AMinus e1 e2) || (n1 - n2)
| E_AMult : forall (e1 e2: aexp) (n1 n2 : nat),
(e1 || n1) -> (e2 || n2) -> (AMult e1 e2) || (n1 * n2)
where "e '||' n" := (aevalR e n) : type_scope.
Tactic Notation "aevalR_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "E_ANum" | Case_aux c "E_APlus"
| Case_aux c "E_AMinus" | Case_aux c "E_AMult" ].
(** It is straightforward to prove that the relational and functional
definitions of evaluation agree on all possible arithmetic
expressions... *)
Theorem aeval_iff_aevalR : forall a n,
(a || n) <-> aeval a = n.
Proof.
split.
Case "->".
intros H.
aevalR_cases (induction H) SCase; simpl.
SCase "E_ANum".
reflexivity.
SCase "E_APlus".
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
SCase "E_AMinus".
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
SCase "E_AMult".
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
Case "<-".
generalize dependent n.
aexp_cases (induction a) SCase;
simpl; intros; subst.
SCase "ANum".
apply E_ANum.
SCase "APlus".
apply E_APlus.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
SCase "AMinus".
apply E_AMinus.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
SCase "AMult".
apply E_AMult.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
Qed.
(** We can make the proof quite a bit shorter by making more
aggressive use of tacticals... *)
Theorem aeval_iff_aevalR' : forall a n,
(a || n) <-> aeval a = n.
Proof.
(* WORKED IN CLASS *)
split.
Case "->".
intros H; induction H; subst; reflexivity.
Case "<-".
generalize dependent n.
induction a; simpl; intros; subst; constructor;
try apply IHa1; try apply IHa2; reflexivity.
Qed.
(** **** Exercise: 3 stars (bevalR) *)
(* EXPECTED *)
(** Write a relation [bevalR] in the same style as
[aevalR], and prove that it is equivalent to [beval].*)
(*
Inductive bevalR:
*)
(* FILL IN HERE *)
(** [] *)
(** For the definitions of evaluation for arithmetic and boolean
expressions, the choice of whether to use functional or relational
definitions is mainly a matter of taste. In general, Coq has
somewhat better support for working with relations. On the other
hand, in some sense function definitions carry more information,
because functions are necessarily deterministic and defined on all
arguments; for a relation we have to show these properties
explicitly if we need them. Functions also take advantage of Coq's
computational mechanism.
However, there are circumstances where relational definitions of
evaluation are greatly preferable to functional ones, as we'll see
shortly. *)
(* ####################################################### *)
(** ** Inference Rule Notation *)
(** In informal discussions, it is convenient write the rules for
[aevalR] and similar relations in the more readable "graphical"
form of _inference rules_, where the premises above the line
justify the conclusion below the line. For example, the
constructor [E_APlus]...
| E_APlus : forall (e1 e2: aexp) (n1 n2: nat),
aevalR e1 n1 ->
aevalR e2 n2 ->
aevalR (APlus e1 e2) (n1 + n2)
...would be written like this as an inference rule:
e1 || n1
e2 || n2
-------------------- (E_APlus)
APlus e1 e2 || n1+n2
Formally, there is nothing very deep about inference rules: they
are just implications. You can read the rule name on the right as
the name of the constructor and read each of the linebreaks
between the premises above the line and the line itself as [->].
All the variables mentioned in the rule ([e1], [n1], etc.) are
implicitly bound by universal quantifiers at the beginning. The
whole collection of rules is understood as being wrapped in an
[Inductive] declaration (informally, this is either elided or else
indicated by saying something like "Let [aevalR] be the smallest
relation closed under the following rules...").
For example, [||] is the smallest relation closed under these
rules:
----------- (E_ANum)
ANum n || n
e1 || n1
e2 || n2
-------------------- (E_APlus)
APlus e1 e2 || n1+n2
e1 || n1
e2 || n2
--------------------- (E_AMinus)
AMinus e1 e2 || n1-n2
e1 || n1
e2 || n2
-------------------- (E_AMult)
AMult e1 e2 || n1*n2
*)
End AExp.
(* ####################################################### *)
(** * Expressions With Variables *)
(** Let's turn our attention back to defining Imp. The next thing we
need to do is to enrich our arithmetic and boolean expressions
with variables. To keep things simple, we'll assume that all
variables are global and that they only hold numbers. *)
(* ##################################################### *)
(** ** Identifiers *)
(** To begin, we'll need to formalize _identifiers_ such as program
variables. We could use strings for this -- or, in a real
compiler, fancier structures like pointers into a symbol table.
But for simplicity let's just use natural numbers as identifiers.
(We hide this section in a module because these definitions are
actually in [SfLib], but we want to repeat them here so that we
can explain them.) *)
Module Id.
(** We define a new inductive datatype [Id] so that we won't confuse
identifiers and numbers. *)
Inductive id : Type :=
Id : nat -> id.
Definition beq_id X1 X2 :=
match (X1, X2) with
(Id n1, Id n2) => beq_nat n1 n2
end.
(** After we "wrap" numbers as identifiers in this way, it is
convenient to recapitulate a few properties of numbers as
analogous properties of identifiers, so that we can work with
identifiers in definitions and proofs abstractly, without
unwrapping them to expose the underlying numbers. Since all we
need to know about identifiers is whether they are the same or
different, just a few basic facts are all we need. *)
Theorem beq_id_refl : forall X,
true = beq_id X X.
Proof.
intros. destruct X.
apply beq_nat_refl. Qed.
(** **** Exercise: 1 star, optional (beq_id_eq) *)
(* EXPECTED *)
(** For this and the following exercises, do not use induction, but
rather apply similar results already proved for natural numbers.
Some of the tactics mentioned above may prove useful. *)
Theorem beq_id_eq : forall i1 i2,
true = beq_id i1 i2 -> i1 = i2.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, optional (beq_id_false_not_eq) *)
(* EXPECTED *)
Theorem beq_id_false_not_eq : forall i1 i2,
beq_id i1 i2 = false -> i1 <> i2.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, optional (not_eq_beq_id_false) *)
Theorem not_eq_beq_id_false : forall i1 i2,
i1 <> i2 -> beq_id i1 i2 = false.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, optional (beq_id_sym) *)
Theorem beq_id_sym: forall i1 i2,
beq_id i1 i2 = beq_id i2 i1.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
End Id.
(* ####################################################### *)
(** ** States *)
(** A _state_ represents the current values of all the variables at
some point in the execution of a program. *)
(** For simplicity (to avoid dealing with partial functions), we
let the state be defined for _all_ variables, even though any
given program is only going to mention a finite number of them. *)
Definition state := id -> nat.
Definition empty_state : state :=
fun _ => 0.
Definition update (st : state) (X:id) (n : nat) : state :=
fun X' => if beq_id X X' then n else st X'.
(** For proofs involving states, we'll need several simple properties
of [update]. *)
(** **** Exercise: 1 star (update_eq) *)
Theorem update_eq : forall n X st,
(update st X n) X = n.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star (update_neq) *)
Theorem update_neq : forall V2 V1 n st,
beq_id V2 V1 = false ->
(update st V2 n) V1 = (st V1).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, recommended (update_shadow) *)
Theorem update_shadow : forall x1 x2 k1 k2 (f : state),
(update (update f k2 x1) k2 x2) k1 = (update f k2 x2) k1.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars (update_same) *)
Theorem update_same : forall x1 k1 k2 (f : state),
f k1 = x1 ->
(update f k1 x1) k2 = f k2.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars (update_permute) *)
Theorem update_permute : forall x1 x2 k1 k2 k3 f,
beq_id k2 k1 = false ->
(update (update f k2 x1) k1 x2) k3 = (update (update f k1 x2) k2 x1) k3.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ################################################### *)
(** ** Syntax *)
(** We can add variables to the arithmetic expressions we had before by
simply adding one more constructor: *)
Inductive aexp : Type :=
| ANum : nat -> aexp
| AId : id -> aexp (* <----- NEW *)
| APlus : aexp -> aexp -> aexp
| AMinus : aexp -> aexp -> aexp
| AMult : aexp -> aexp -> aexp.
Tactic Notation "aexp_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "ANum" | Case_aux c "AId" | Case_aux c "APlus"
| Case_aux c "AMinus" | Case_aux c "AMult" ].
(** Defining a few variable names as notational shorthands will make
examples easier to read: *)
Definition X : id := Id 0.
Definition Y : id := Id 1.
Definition Z : id := Id 2.
(** (This convention for naming program variables ([X], [Y],
[Z]) clashes a bit with our earlier use of uppercase letters for
types. Since we're not using polymorphism heavily in this part of
the course, this overloading should not cause confusion.) *)
(** The definition of [bexp]s is the same as before (using the new
[aexp]s): *)
Inductive bexp : Type :=
| BTrue : bexp
| BFalse : bexp
| BEq : aexp -> aexp -> bexp
| BLe : aexp -> aexp -> bexp
| BNot : bexp -> bexp
| BAnd : bexp -> bexp -> bexp.
Tactic Notation "bexp_cases" tactic(first) ident(c) :=
first;
[ Case_aux c "BTrue" | Case_aux c "BFalse" | Case_aux c "BEq"
| Case_aux c "BLe" | Case_aux c "BNot" | Case_aux c "BAnd" ].
(* ################################################### *)
(** ** Evaluation *)
(** The arith and boolean evaluators can be extended to handle
variables in the obvious way: *)
Fixpoint aeval (st : state) (e : aexp) : nat :=
match e with
| ANum n => n
| AId X => st X (* <----- NEW *)
| APlus a1 a2 => (aeval st a1) + (aeval st a2)
| AMinus a1 a2 => (aeval st a1) - (aeval st a2)
| AMult a1 a2 => (aeval st a1) * (aeval st a2)
end.
Fixpoint beval (st : state) (e : bexp) : bool :=
match e with
| BTrue => true
| BFalse => false
| BEq a1 a2 => beq_nat (aeval st a1) (aeval st a2)
| BLe a1 a2 => ble_nat (aeval st a1) (aeval st a2)
| BNot b1 => negb (beval st b1)
| BAnd b1 b2 => andb (beval st b1) (beval st b2)
end.
Example aexp1 :
aeval (update empty_state X 5)
(APlus (ANum 3) (AMult (AId X) (ANum 2)))
= 13.
Proof. reflexivity. Qed.
Example bexp1 :
beval (update empty_state X 5)
(BAnd BTrue (BNot (BLe (AId X) (ANum 4))))
= true.
Proof. reflexivity. Qed.
(* ####################################################### *)
(** * Commands *)
(** Now we are ready define the syntax and behavior of Imp
_commands_ (or _statements_). *)
(* ################################################### *)
(** ** Syntax *)
(** Informally, commands are described by the following BNF
grammar:
com ::= 'SKIP'
| X '::=' aexp
| com ';' com
| 'WHILE' bexp 'DO' com 'END'
| 'IFB' bexp 'THEN' com 'ELSE' com 'FI'
For example, here's the factorial function in Imp.
Z ::= X;
Y ::= 1;
WHILE not (Z = 0) DO
Y ::= Y * Z;
Z ::= Z - 1
END
When this command terminates, the variable [Y] will contain the
factorial of the initial value of [X].
*)
(** Here is the formal definition of the syntax of commands: *)
Inductive com : Type :=
| CSkip : com
| CAss : id -> aexp -> com
| CSeq : com -> com -> com
| CIf : bexp -> com -> com -> com
| CWhile : bexp -> com -> com.
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" ].
(** As usual, we can use a few [Notation] declarations to make things
more readable. We need to be a bit careful to avoid conflicts
with Coq's built-in notations, so we'll keep this light -- in
particular, we won't introduce any notations for [aexps] and
[bexps] to avoid confusion with the numerical and boolean
operators we've already defined. We use the keyword [IFB] for
conditionals instead of [IF], for similar reasons. *)
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).
(** For example, here is the factorial function again, written as a
formal definition to Coq: *)
Definition fact_in_coq : com :=
Z ::= AId X;
Y ::= ANum 1;
WHILE BNot (BEq (AId Z) (ANum 0)) DO
Y ::= AMult (AId Y) (AId Z);
Z ::= AMinus (AId Z) (ANum 1)
END.
(* ####################################################### *)
(** ** Examples *)
(** Assignment: *)
Definition plus2 : com :=
X ::= (APlus (AId X) (ANum 2)).
Definition XtimesYinZ : com :=
Z ::= (AMult (AId X) (AId Y)).
Definition subtract_slowly_body : com :=
Z ::= AMinus (AId Z) (ANum 1) ;
X ::= AMinus (AId X) (ANum 1).
(** Loops: *)
Definition subtract_slowly : com :=
WHILE BNot (BEq (AId X) (ANum 0)) DO
subtract_slowly_body
END.
Definition subtract_3_from_5_slowly : com :=
X ::= ANum 3 ;
Z ::= ANum 5 ;
subtract_slowly.
(** An infinite loop: *)
Definition loop : com :=
WHILE BTrue DO
SKIP
END.
(** Factorial again (broken up into smaller pieces this time, for
convenience when we come back to proving things about it
later). *)
Definition fact_body : com :=
Y ::= AMult (AId Y) (AId Z) ;
Z ::= AMinus (AId Z) (ANum 1).
Definition fact_loop : com :=
WHILE BNot (BEq (AId Z) (ANum 0)) DO
fact_body
END.
Definition fact_com : com :=
Z ::= AId X ;
Y ::= ANum 1 ;
fact_loop.
(* ################################################################ *)
(** * Evaluation *)
(** Next we need to define what it means to evaluate an Imp command.
The fact that [WHILE] loops don't necessarily terminate makes defining
an evaluation function tricky ... *)
(* #################################### *)
(** ** Evaluation Function (Failed Attempt) *)
(** Here's an attempt at defining an evaluation function for commands,
omitting the [WHILE] case. *)
Fixpoint ceval_fun_no_while (st : state) (c : com) : state :=
match c with
| SKIP =>
st
| l ::= a1 =>
update st l (aeval st a1)
| c1 ; c2 =>
let st' := ceval_fun_no_while st c1 in
ceval_fun_no_while st' c2
| IFB b THEN c1 ELSE c2 FI =>
if (beval st b)
then ceval_fun_no_while st c1
else ceval_fun_no_while st c2
| WHILE b1 DO c1 END =>
st (* bogus *)
end.
(** In a traditional functional programming language like ML or
Haskell we could write the WHILE case as follows:
<<
Fixpoint ceval_fun (st : state) (c : com) : state :=
match c with
...
| WHILE b1 DO c1 END =>
if (beval st b1)
then ceval_fun st (c1; WHILE b1 DO c1 END)
else st
end.
>>
Coq doesn't accept such a definition ([Error: Cannot guess
decreasing argument of fix]) because the function we want to
define is not guaranteed to terminate. Indeed, the full version of
the [ceval_fun] function applied to the [loop] program above would
never terminate. Since Coq is not just a functional programming
language, but also a consistent logic, any potentially
non-terminating function needs to be rejected. Here is an
invalid(!) Coq program showing what would go wrong if Coq allowed
non-terminating recursive functions:
<<
Fixpoint loop_false (n : nat) : False := loop_false n.
>>
That is, propositions like [False] would become
provable (e.g. [loop_false 0] would be a proof of [False]), which
would be a disaster for Coq's logical consistency.
Thus, because it doesn't terminate on all inputs, the full version
of [ceval_fun] cannot be written in Coq -- at least not without
additional tricks, which make everything much more complicated
(see chapter [ImpCEvalFun] if curious). *)
(* #################################### *)
(** ** Evaluation as a Relation *)
(** Here's a better way: we define [ceval] as a _relation_ rather than
a _function_ -- i.e., we define it in [Prop] instead of [Type], as
we did for [aevalR] and [bevalR] above. *)
(** This is an important change. Besides freeing us from the
hacks that would be needed to define an evaluation function, it
gives us a lot more flexibility in the definition. For example,
if we added concurrency features to the language, we'd want the
definition of evaluation to be non-deterministic -- i.e., not only
would it not be total, it would not even be a partial function! *)
(** We'll use the notation [c / st || st'] for our [ceval] relation,
that is [c / st || st'] means that executing program [c] in a
starting state [st] results in an ending state [st']. This can be
pronounced "[c] takes state [st] to [st']".
---------------- (E_Skip)
SKIP / st || st
aeval st a1 = n
-------------------------------- (E_Ass)
X := a1 / st || (update st X n)
c1 / st || st'
c2 / st' || st''
------------------- (E_Seq)
c1;c2 / st || st''
beval st b1 = true
c1 / st || st'
------------------------------------- (E_IfTrue)
IF b1 THEN c1 ELSE c2 FI / st || st'
beval st b1 = false
c2 / st || st'
------------------------------------- (E_IfFalse)
IF b1 THEN c1 ELSE c2 FI / st || st'
beval st b1 = false
------------------------------ (E_WhileEnd)
WHILE b1 DO c1 END / st || st
beval st b1 = true
c1 / st || st'
WHILE b1 DO c1 END / st' || st''
--------------------------------- (E_WhileLoop)
WHILE b1 DO c1 END / st || st''
*)
(** Here is the formal definition. (Make sure you understand
how it corresponds to the inference rules.) *)
Reserved Notation "c1 '/' st '||' st'" (at level 40, st at level 39).
Inductive ceval : com -> state -> state -> Prop :=
| E_Skip : forall st,
SKIP / st || st
| E_Ass : forall st a1 n X,
aeval st a1 = n ->
(X ::= a1) / st || (update st X n)
| E_Seq : forall c1 c2 st st' st'',
c1 / st || st' ->
c2 / st' || st'' ->
(c1 ; c2) / st || st''
| E_IfTrue : forall st st' b1 c1 c2,
beval st b1 = true ->
c1 / st || st' ->
(IFB b1 THEN c1 ELSE c2 FI) / st || st'
| E_IfFalse : forall st st' b1 c1 c2,
beval st b1 = false ->
c2 / st || st' ->
(IFB b1 THEN c1 ELSE c2 FI) / st || st'
| E_WhileEnd : forall b1 st c1,
beval st b1 = false ->
(WHILE b1 DO c1 END) / st || st
| E_WhileLoop : forall st st' st'' b1 c1,
beval st b1 = true ->
c1 / st || st' ->
(WHILE b1 DO c1 END) / st' || st'' ->
(WHILE b1 DO c1 END) / st || st''
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" ].
(** The cost of defining evaluation as a relation instead of a
function is that we now need to construct _proofs_ that some
program evaluates to some result state, rather than just letting
Coq's computation mechanism do it for us. *)
Example ceval_example1:
(X ::= ANum 2;
IFB BLe (AId X) (ANum 1)
THEN Y ::= ANum 3
ELSE Z ::= ANum 4
FI)
/ empty_state
|| (update (update empty_state X 2) Z 4).
Proof.
(* We must supply the intermediate state *)
apply E_Seq with (update empty_state X 2).
Case "assignment command".
apply E_Ass. reflexivity.
Case "if command".
apply E_IfFalse.
reflexivity.
apply E_Ass. reflexivity. Qed.
(** **** Exercise: 2 stars (ceval_example2) *)
(* EXPECTED *)
Example ceval_example2:
(X ::= ANum 0; Y ::= ANum 1; Z ::= ANum 2) / empty_state ||
(update (update (update empty_state X 0) Y 1) Z 2).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, optional (pup_to_n) *)
(** Write an Imp program that sums the numbers from [1] to
[X] (inclusive: [1 + 2 + ... + X]) in the variable [Y].
Prove that this program executes as intended for X = 2
(this latter part is trickier than you might expect). *)
Definition pup_to_n : com :=
(* FILL IN HERE *) admit.
Theorem pup_to_2_ceval :
pup_to_n / (update empty_state X 2) ||
update (update (update (update (update (update empty_state
X 2) Y 0) Y 2) X 1) Y 3) X 0.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ####################################################### *)
(** ** Determinism of Evaluation *)
(** Changing from a computational to a relational definition of
evaluation is a good move because it allows us to escape from the
artificial requirement (imposed by Coq's restrictions on
[Fixpoint] definitions) that evaluation should be a total
function. But it also raises a question: Is the second definition
of evaluation actually a partial function? That is, is it
possible that, beginning from the same state [st], we could
evaluate some command [c] in different ways to reach two different
output states [st'] and [st'']?
In fact, this cannot happen: [ceval] is a partial function.
Here's the proof: *)
Theorem ceval_deterministic: forall c st st1 st2,
c / st || st1 ->
c / st || st2 ->
st1 = st2.
Proof.
intros c st st1 st2 E1 E2.
generalize dependent st2.
ceval_cases (induction E1) Case;
intros st2 E2; inversion E2; subst.
Case "E_Skip". reflexivity.
Case "E_Ass". reflexivity.
Case "E_Seq".
assert (st' = st'0) as EQ1.
SCase "Proof of assertion". apply IHE1_1; assumption.
subst st'0.
apply IHE1_2. assumption.
Case "E_IfTrue".
SCase "b1 evaluates to true".
apply IHE1. assumption.
SCase "b1 evaluates to false (contradiction)".
rewrite H in H5. inversion H5.
Case "E_IfFalse".
SCase "b1 evaluates to true (contradiction)".
rewrite H in H5. inversion H5.
SCase "b1 evaluates to false".
apply IHE1. assumption.
Case "E_WhileEnd".
SCase "b1 evaluates to true".
reflexivity.
SCase "b1 evaluates to false (contradiction)".
rewrite H in H2. inversion H2.
Case "E_WhileLoop".
SCase "b1 evaluates to true (contradiction)".
rewrite H in H4. inversion H4.
SCase "b1 evaluates to false".
assert (st' = st'0) as EQ1.
SSCase "Proof of assertion". apply IHE1_1; assumption.
subst st'0.
apply IHE1_2. assumption. Qed.
(* ####################################################### *)
(** * Reasoning About Programs *)
(** We'll get much deeper into systematic techniques for reasoning
about Imp programs in the following chapters, but we can do quite
a bit just working with the bare definitions. This section
explores some examples. *)
(** ** Basic Examples *)
Theorem plus2_spec : forall st n st',
st X = n ->
plus2 / st || st' ->
st' X = n + 2.
Proof.
intros st n st' HX Heval.
(* Inverting Heval essentially forces Coq to expand one
step of the ceval computation - in this case revealing
that st' must be st extended with the new value of X,
since plus2 is an assignment *)
inversion Heval. subst. clear Heval. simpl.
apply update_eq. Qed.
(** **** Exercise: 3 stars, recommended (XtimesYinZ_spec) *)
(* EXPECTED *)
(** State and prove a specification of [XtimesYinZ]. *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 3 stars, recommended (loop_never_stops) *)
(* EXPECTED *)
Theorem loop_never_stops : forall st st',
~(loop / st || st').
Proof.
intros st st' contra. unfold loop in contra.
remember (WHILE BTrue DO SKIP END) as loopdef.
(* Proceed by induction on the assumed derivation showing that
[loopdef] terminates. Most of the cases are immediately
contradictory (and so can be solved in one step with
[inversion]). *)
(* FILL IN HERE *) Admitted.
(** [] *)
Fixpoint no_whiles (c : com) : bool :=
match c with
| SKIP => true
| _ ::= _ => true
| c1 ; c2 => andb (no_whiles c1) (no_whiles c2)
| IFB _ THEN ct ELSE cf FI => andb (no_whiles ct) (no_whiles cf)
| WHILE _ DO _ END => false
end.
(** **** Exercise: 3 stars, optional (no_whilesR) *)
(** The [no_whiles] property yields [true] on just those programs that
have no while loops. Using [Inductive], write a property
[no_whilesR] such that [no_whilesR c] is provable exactly when [c]
is a program with no while loops. Then prove its equivalence
with [no_whiles]. *)
Inductive no_whilesR: com -> Prop :=
(* FILL IN HERE *)
.
Theorem no_whiles_eqv:
forall c, no_whiles c = true <-> no_whilesR c.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 4 stars, optional (no_whiles_terminating) *)
(** Imp programs that don't involve while loops always terminate.
State and prove a theorem that says this. *)
(** (Use either [no_whiles] or [no_whilesR], as you prefer.) *)
(* FILL IN HERE *)
(** [] *)
(** ** Proving a Program Correct (Optional) *)
(** Recall the factorial program: *)
Print fact_body. Print fact_loop. Print fact_com.
(** Here is an alternative "mathematical" definition of the factorial
function: *)
Fixpoint real_fact (n:nat) : nat :=
match n with
| O => 1
| S n' => n * (real_fact n')
end.
(** We would like to show that they agree -- if we start [fact_com] in
a state where variable [X] contains some number [x], then it will
terminate in a state where variable [Y] contains the factorial of
[x].
To show this, we rely on the critical idea of a _loop
invariant_. *)
Definition fact_invariant (x:nat) (st:state) :=
(st Y) * (real_fact (st Z)) = real_fact x.
Theorem fact_body_preserves_invariant: forall st st' x,
fact_invariant x st ->
st Z <> 0 ->
fact_body / st || st' ->
fact_invariant x st'.
Proof.
unfold fact_invariant, fact_body.
intros st st' x Hm HZnz He.
inversion He; subst; clear He.
inversion H1; subst; clear H1.
inversion H4; subst; clear H4.
unfold update. simpl.
(* Show that st Z = S z' for some z' *)
destruct (st Z) as [| z'].
apply ex_falso_quodlibet. apply HZnz. reflexivity.
rewrite <- Hm. rewrite <- mult_assoc.
replace (S z' - 1) with z' by omega.
reflexivity. Qed.
Theorem fact_loop_preserves_invariant : forall st st' x,
fact_invariant x st ->
fact_loop / st || st' ->
fact_invariant x st'.
Proof.
intros st st' x H Hce.
remember fact_loop as c.
ceval_cases (induction Hce) Case;
inversion Heqc; subst; clear Heqc.
Case "E_WhileEnd".
(* trivial when the loop doesn't run... *)
assumption.
Case "E_WhileLoop".
(* if the loop does run, we know that fact_body preserves
fact_invariant -- we just need to assemble the pieces *)
apply IHHce2.
apply fact_body_preserves_invariant with st;
try assumption.
intros Contra. simpl in H0; subst.
rewrite Contra in H0. inversion H0.
reflexivity. Qed.
Theorem guard_false_after_loop: forall b c st st',
(WHILE b DO c END) / st || st' ->
beval st' b = false.
Proof.
intros b c st st' Hce.
remember (WHILE b DO c END) as cloop.
ceval_cases (induction Hce) Case;
inversion Heqcloop; subst; clear Heqcloop.
Case "E_WhileEnd".
assumption.
Case "E_WhileLoop".
apply IHHce2. reflexivity. Qed.
(** Patching it all together... *)
Theorem fact_com_correct : forall st st' x,
st X = x ->
fact_com / st || st' ->
st' Y = real_fact x.
Proof.
intros st st' x HX Hce.
inversion Hce; subst; clear Hce.
inversion H1; subst; clear H1.
inversion H4; subst; clear H4.
inversion H1; subst; clear H1.
rename st' into st''. simpl in H5.
(* The invariant is true before the loop runs... *)
remember (update (update st Z (st X)) Y 1) as st'.
assert (fact_invariant (st X) st').
subst. unfold fact_invariant, update. simpl. omega.
(* ...so when the loop is done running, the invariant
is maintained *)
assert (fact_invariant (st X) st'').
apply fact_loop_preserves_invariant with st'; assumption.
unfold fact_invariant in H0.
(* Finally, if the loop terminated, then Z is 0; so Y must be
factorial of X *)
apply guard_false_after_loop in H5. simpl in H5.
destruct (st'' Z).
Case "st'' Z = 0". simpl in H0. omega.
Case "st'' Z > 0 (impossible)". inversion H5.
Qed.
(** One might wonder whether all this work with poking at states and
unfolding definitions could be ameliorated with some more powerful
lemmas and/or more uniform reasoning principles... Indeed, this is
exactly the topic of the upcoming [Hoare] chapter! *)
(** **** Exercise: 4 stars, optional (subtract_slowly_spec) *)
(** Prove a specification for subtract_slowly, using the above
specification of [fact_com] and the invariant below as
guides. *)
Definition ss_invariant (x:nat) (z:nat) (st:state) :=
minus (st Z) (st X) = minus z x.
(* FILL IN HERE *)
(** [] *)
(* ####################################################### *)
(** * Additional Exercises *)
(** **** Exercise: 4 stars, optional (add_for_loop) *)
(** Add C-style [for] loops to the language of commands, update the
[ceval] definition to define the semantics of [for] loops, and add
cases for [for] loops as needed so that all the proofs in this file
are accepted by Coq.
A [for] loop should be parameterized by (a) a statement executed
initially, (b) a test that is run on each iteration of the loop to
determine whether the loop should continue, (c) a statement
executed at the end of each loop iteration, and (d) a statement
that makes up the body of the loop. (You don't need to worry
about making up a concrete Notation for [for] loops, but feel free
to play with this too if you like.) *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 3 stars, optional (short_circuit) *)
(** Most modern programming languages use a "short-circuit" evaluation
rule for boolean [and]: to evaluate [BAnd b1 b2], first evaluate
[b1]. If it evaluates to [false], then the entire [BAnd]
expression evaluates to [false] immediately, without evaluating
[b2]. Otherwise, [b2] is evaluated to determine the result of the
[BAnd] expression.
Write an alternate version of [beval] that performs short-circuit
evaluation of [BAnd] in this manner, and prove that it is
equivalent to [beval]. *)
(* FILL IN HERE *)
(** **** Exercise: 4 stars, recommended (stack_compiler) *)
(** HP Calculators, programming languages like Forth and Postscript,
and abstract machines like the Java Virtual Machine all evaluate
arithmetic expressions using a stack. For instance, the expression
<<
(2*3)+(3*(4-2))
>>
would be entered as
<<
2 3 * 3 4 2 - * +
>>
and evaluated like this:
<<
[] | 2 3 * 3 4 2 - * +
[2] | 3 * 3 4 2 - * +
[3, 2] | * 3 4 2 - * +
[6] | 3 4 2 - * +
[3, 6] | 4 2 - * +
[4, 3, 6] | 2 - * +
[2, 4, 3, 6] | - * +
[2, 3, 6] | * +
[6, 6] | +
[12] |
>>
The task of this exercise is to write a small compiler that
translates [aexp]s into stack machine instructions, and to prove its
correctness.
The instruction set for our stack language will consist of the
following instructions:
- [SPush n]: Push the number [n] on the stack.
- [SLoad X]: Load the identifier [X] from the store and push it
on the stack
- [SPlus]: Pop the two top numbers from the stack, add them, and
push the result onto the stack.
- [SMinus]: Similar, but subtract.
- [SMult]: Similar, but multiply.
*)
Inductive sinstr : Type :=
| SPush : nat -> sinstr
| SLoad : id -> sinstr
| SPlus : sinstr
| SMinus : sinstr
| SMult : sinstr.
(** Write a function to evaluate programs in the stack language. It
takes as input a state, a stack represented as a list of
numbers (top stack item is the head of the list), and a program
represented as a list of instructions, and returns the stack after
executing the program. Test your function on the examples below.
Note that the specification leaves unspecified what to do when
encountering an [SPlus], [SMinus], or [SMult] instruction if the
stack contains less than two elements. In a sense it is
immaterial, since our compiler will never emit such a malformed
program. However, when you do the correctness proof you may find
some choices makes the proof easier than others. *)
Fixpoint s_execute (st : state) (stack : list nat)
(prog : list sinstr)
: list nat :=
(* FILL IN HERE *) admit.
Example s_execute1 :
s_execute empty_state []
[SPush 5, SPush 3, SPush 1, SMinus]
= [2, 5].
(* FILL IN HERE *) Admitted.
Example s_execute2 :
s_execute (update empty_state X 3) [3,4]
[SPush 4, SLoad X, SMult, SPlus]
= [15, 4].
(* FILL IN HERE *) Admitted.
(** Next, write a function which compiles an [aexp] into a stack
machine program. The effect of running the program should be the
same as pushing the value of the expression on the stack. *)
Fixpoint s_compile (e : aexp) : list sinstr :=
(* FILL IN HERE *) admit.
(*
Example s_compile1 :
s_compile (AMinus (AId X) (AMult (ANum 2) (AId Y)))
= [SLoad X, SPush 2, SLoad Y, SMult, SMinus].
Proof. reflexivity. Qed.
*)
(** Finally, prove the following theorem, stating that the [compile]
function behaves correctly. You will need to start by stating a
more general lemma to get a usable induction hypothesis. *)
(* FILL IN HERE *)
Theorem s_compile_correct : forall (st : state) (e : aexp),
s_execute st [] (s_compile e) = [ aeval st e ].
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)