(** * Imp: Simple Imperative Programs *)
(** In this chapter, we begin a new direction that will continue for
the rest of the course. Up to now most of our attention has been
focused on various aspects of Coq itself, while from now on we'll
mostly be using Coq to formalize other things. (We'll continue to
pause from time to time to introduce a few additional aspects of
Coq.)
Our first case study is a _simple imperative programming language_
called Imp, embodying a tiny core fragment of conventional
mainstream languages such as C and Java. 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_, a widely used logic for
reasoning about imperative programs. *)
Require Import Coq.Bool.Bool.
Require Import Coq.Arith.Arith.
Require Import Coq.Arith.EqNat.
Require Import Coq.omega.Omega.
Require Import Coq.Lists.List.
Import ListNotations.
Require Import Maps.
Require Import SfLib. (* for [admit] *)
(* ####################################################### *)
(** * 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:
a ::= nat
| a + a
| a - a
| a * a
b ::= true
| false
| a = a
| a <= a
| not b
| b and b
*)
(** 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 (a : aexp) : nat :=
match a 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 (b : bexp) : bool :=
match b with
| BTrue => true
| BFalse => false
| BEq a1 a2 => beq_nat (aeval a1) (aeval a2)
| BLe a1 a2 => leb (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 (a:aexp) : aexp :=
match a 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 a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a. induction a.
- (* ANum *) reflexivity.
- (* APlus *) destruct a1.
+ (* a1 = ANum n *) destruct n.
* (* n = 0 *) simpl. apply IHa2.
* (* n <> 0 *) simpl. rewrite IHa2. reflexivity.
+ (* a1 = APlus a1_1 a1_2 *)
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
+ (* a1 = AMinus a1_1 a1_2 *)
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
+ (* a1 = AMult a1_1 a1_2 *)
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
- (* AMinus *)
simpl. rewrite IHa1. rewrite IHa2. reflexivity.
- (* AMult *)
simpl. rewrite IHa1. rewrite IHa2. 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 [try] Tactical *)
(** 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 most commonly used form, the [;] tactical takes two 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, leb 0 n = true.
Proof.
intros.
destruct n.
(* Leaves two subgoals, which are discharged identically... *)
- (* n=0 *) simpl. reflexivity.
- (* n=Sn' *) simpl. reflexivity.
Qed.
(** We can simplify this proof using the [;] tactical: *)
Lemma foo' : forall n, leb 0 n = true.
Proof.
intros.
destruct n; (* [destruct] the current goal *)
simpl; (* then [simpl] each resulting subgoal *)
reflexivity. (* and 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 a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a.
induction a;
(* Most cases follow directly by the IH *)
try (simpl; rewrite IHa1; rewrite IHa2; reflexivity).
(* The remaining cases -- ANum and APlus -- are different *)
- (* ANum *) reflexivity.
- (* APlus *)
destruct a1;
(* Again, most cases follow directly by the IH *)
try (simpl; simpl in IHa1; rewrite IHa1;
rewrite IHa2; 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. *)
+ (* a1 = ANum n *) destruct n;
simpl; rewrite IHa2; 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 [a],
aeval (optimize_0plus a) = aeval a.
_Proof_: By induction on [a]. The [AMinus] and [AMult] cases
follow directly from the IH. The remaining cases are as follows:
- Suppose [a = 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 [a = APlus a1 a2] for some [a1] and [a2]. We
must show
aeval (optimize_0plus (APlus a1 a2))
= aeval (APlus a1 a2).
Consider the possible forms of [a1]. For most of them,
[optimize_0plus] simply calls itself recursively for the
subexpressions and rebuilds a new expression of the same form
as [a1]; in these cases, the result follows directly from the
IH.
The interesting case is when [a1 = ANum n] for some [n].
If [n = ANum 0], then
optimize_0plus (APlus a1 a2) = optimize_0plus a2
and the IH for [a2] 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 [a = 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 a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a.
induction a;
(* Most cases follow directly by the IH *)
try (simpl; rewrite IHa1; rewrite IHa2; reflexivity);
(* ... or are immediate by definition *)
try reflexivity.
(* The interesting case is when a = APlus a1 a2. *)
- (* APlus *)
destruct a1; try (simpl; simpl in IHa1; rewrite IHa1;
rewrite IHa2; reflexivity).
+ (* a1 = ANum n *) destruct n;
simpl; rewrite IHa2; reflexivity. Qed.
(* ####################################################### *)
(** *** The [;] Tactical (General Form) *)
(** The [;] tactical also has a more general form than the simple
[T;T'] we've seen above, 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 [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 In10 : In 10 [1;2;3;4;5;6;7;8;9;10].
Proof.
repeat (right; try (left; reflexivity)).
Qed.
(* Print In10. *)
(** 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 In10' : In 10 [1;2;3;4;5;6;7;8;9;10].
Proof.
repeat (left; reflexivity).
repeat (right; try (left; reflexivity)).
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! *)
(* ####################################################### *)
(** ** 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 [a] is a variable of type [aexp], then doing
aexp_cases (induction a) Case
will perform an induction on [a] (the same as if we had just typed
[induction a]) 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 a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a.
induction a;
try (simpl; rewrite IHa1; rewrite IHa2; 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. *)
- (* APlus *)
destruct a1;
try (simpl; simpl in IHa1;
rewrite IHa1; rewrite IHa2; reflexivity).
+ (* ANum *) destruct n;
simpl; rewrite IHa2; reflexivity. Qed.
(** **** Exercise: 3 stars (optimize_0plus_b) *)
(** 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. *)
Fixpoint optimize_0plus_b (b : bexp) : bexp :=
(* FILL IN HERE *) admit.
Theorem optimize_0plus_b_sound : forall b,
beval (optimize_0plus_b b) = beval b.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 4 stars, optional (optimizer) *)
(** _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.
(* FILL IN HERE *)
*)
(** [] *)
(* ####################################################### *)
(** ** 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. *)
Require Import Coq.omega.Omega.
Example silly_presburger_example : forall m n o p,
m + n <= n + o /\ o + 3 = p + 3 ->
m <= p.
Proof.
intros. omega.
Qed.
(** Leibniz wrote, "It is unworthy of excellent men to lose
hours like slaves in the labor of calculation which could be
relegated to anyone else if machines were used." We recommend
using the omega tactic whenever possible. *)
(* ####################################################### *)
(** ** 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
[Fixpoint]s. 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 slash
as the closest approximation in [.v] files.) *)
Notation "e '\\' n"
:= (aevalR e n) (at level 50, left associativity)
: 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.
(* ####################################################### *)
(** ** Inference Rule Notation *)
(** In informal discussions, it is convenient to 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 (we have already seen them in the
Prop chapter). *)
(** 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. (Such variables are often called _metavariables_ to
distinguish them from the variables of the language we are
defining. At the moment, our arithmetic expressions don't include
variables, but we'll soon be adding them.) 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
*)
(* ####################################################### *)
(** ** Equivalence of the Definitions *)
(** 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.
- (* -> *)
intros H.
induction H; simpl.
+ (* E_ANum *)
reflexivity.
+ (* E_APlus *)
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
+ (* E_AMinus *)
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
+ (* E_AMult *)
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
- (* <- *)
generalize dependent n.
induction a;
simpl; intros; subst.
+ (* ANum *)
apply E_ANum.
+ (* APlus *)
apply E_APlus.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
+ (* AMinus *)
apply E_AMinus.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
+ (* AMult *)
apply E_AMult.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
Qed.
(** Note: if you're reading the HTML file, you'll see an empty square box instead
of a proof for this theorem.
You can click on this box to "unfold" the text to see the proof.
Click on the unfolded to text to "fold" it back up to a box. We'll be using
this style frequently from now on to help keep the HTML easier to read.
The full proofs always appear in the .v files. *)
(** We can make the proof quite a bit shorter by making more
use of tacticals... *)
Theorem aeval_iff_aevalR' : forall a n,
(a \\ n) <-> aeval a = n.
Proof.
(* WORKED IN CLASS *)
split.
- (* -> *)
intros H; induction H; subst; reflexivity.
- (* <- *)
generalize dependent n.
induction a; simpl; intros; subst; constructor;
try apply IHa1; try apply IHa2; reflexivity.
Qed.
(** **** Exercise: 3 stars (bevalR) *)
(** Write a relation [bevalR] in the same style as
[aevalR], and prove that it is equivalent to [beval].*)
(*
Inductive bevalR:
(* FILL IN HERE *)
*)
(** [] *)
End AExp.
(* ####################################################### *)
(** ** Computational vs. Relational Definitions *)
(** 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
computations mechanism.
However, there are circumstances where relational definitions of
evaluation are preferable to functional ones. *)
Module aevalR_division.
(** For example, suppose that we wanted to extend the arithmetic
operations by considering also a division operation:*)
Inductive aexp : Type :=
| ANum : nat -> aexp
| APlus : aexp -> aexp -> aexp
| AMinus : aexp -> aexp -> aexp
| AMult : aexp -> aexp -> aexp
| ADiv : aexp -> aexp -> aexp. (* <--- new *)
(** Extending the definition of [aeval] to handle this new operation
would not be straightforward (what should we return as the result
of [ADiv (ANum 5) (ANum 0)]?). But extending [aevalR] is
straightforward. *)
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 (a1 a2: aexp) (n1 n2 : nat),
(a1 \\ n1) -> (a2 \\ n2) -> (APlus a1 a2) \\ (n1 + n2)
| E_AMinus : forall (a1 a2: aexp) (n1 n2 : nat),
(a1 \\ n1) -> (a2 \\ n2) -> (AMinus a1 a2) \\ (n1 - n2)
| E_AMult : forall (a1 a2: aexp) (n1 n2 : nat),
(a1 \\ n1) -> (a2 \\ n2) -> (AMult a1 a2) \\ (n1 * n2)
| E_ADiv : forall (a1 a2: aexp) (n1 n2 n3: nat),
(a1 \\ n1) -> (a2 \\ n2) -> (n2 > 0) ->
(mult n2 n3 = n1) -> (ADiv a1 a2) \\ n3
where "a '\\' n" := (aevalR a n) : type_scope.
End aevalR_division.
Module aevalR_extended.
(** *** Adding nondeterminism *)
(** Suppose, instead, that we want to extend the arithmetic operations
by a nondeterministic number generator [any]:*)
Reserved Notation "e '\\' n" (at level 50, left associativity).
Inductive aexp : Type :=
| AAny : aexp (* <--- NEW *)
| ANum : nat -> aexp
| APlus : aexp -> aexp -> aexp
| AMinus : aexp -> aexp -> aexp
| AMult : aexp -> aexp -> aexp.
(** Again, extending [aeval] would be tricky (because evaluation is
_not_ a deterministic function from expressions to numbers), but
extending [aevalR] is no problem: *)
Inductive aevalR : aexp -> nat -> Prop :=
| E_Any : forall (n:nat),
AAny \\ n (* <--- new *)
| E_ANum : forall (n:nat),
(ANum n) \\ n
| E_APlus : forall (a1 a2: aexp) (n1 n2 : nat),
(a1 \\ n1) -> (a2 \\ n2) -> (APlus a1 a2) \\ (n1 + n2)
| E_AMinus : forall (a1 a2: aexp) (n1 n2 : nat),
(a1 \\ n1) -> (a2 \\ n2) -> (AMinus a1 a2) \\ (n1 - n2)
| E_AMult : forall (a1 a2: aexp) (n1 n2 : nat),
(a1 \\ n1) -> (a2 \\ n2) -> (AMult a1 a2) \\ (n1 * n2)
where "a '\\' n" := (aevalR a n) : type_scope.
End aevalR_extended.
(* ####################################################### *)
(** * 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. *)
(* ####################################################### *)
(** ** States *)
(** Since we'll want to look variables up to find out their current
values, we'll reuse the type [id] from the [Maps] chapter for the
type of variables in Imp.
A _machine state_ (or just _state_) represents the current values
of _all_ the variables at some point in the execution of a
program. *)
(** For simplicity, we assume that the state is defined for
_all_ variables, even though any given program is only going to
mention a finite number of them. The state captures all of the
information stored in memory. For Imp programs, because each
variable stores only a natural number, we can represent the state
as a mapping from identifiers to [nat]. For more complex
programming languages, the state might have more structure.
*)
Definition state := total_map nat.
Definition empty_state : state :=
t_empty 0.
(* ################################################### *)
(** ** 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.
(** 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.
(* ################################################### *)
(** ** Evaluation *)
(** The arith and boolean evaluators can be extended to handle
variables in the obvious way: *)
Fixpoint aeval (st : state) (a : aexp) : nat :=
match a 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) (b : bexp) : bool :=
match b with
| BTrue => true
| BFalse => false
| BEq a1 a2 => beq_nat (aeval st a1) (aeval st a2)
| BLe a1 a2 => leb (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 (t_update empty_state X 5)
(APlus (ANum 3) (AMult (AId X) (ANum 2)))
= 13.
Proof. reflexivity. Qed.
Example bexp1 :
beval (t_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_ (often called _statements_). *)
(* ################################################### *)
(** ** Syntax *)
(** Informally, commands [c] are described by the following BNF
grammar:
c ::= SKIP
| x ::= a
| c ;; c
| IFB b THEN c ELSE c FI
| WHILE b DO c END
*)
(**
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.
(** 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' c1 'THEN' c2 'ELSE' c3 'FI'" :=
(CIf c1 c2 c3) (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.
(* ################################################################ *)
(** * 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 as a 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
| x ::= a1 =>
t_update st x (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 b DO c 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 b DO c END =>
if (beval st b)
then ceval_fun st (c; WHILE b DO c 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, it doesn't always
terminate: for example, 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 (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] above. *)
(** This is an important change. Besides freeing us from the awkward
workarounds that would be needed to define evaluation as a
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:
[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']".
*)
(** *** Operational Semantics
---------------- (E_Skip)
SKIP / st \\ st
aeval st a1 = n
-------------------------------- (E_Ass)
x := a1 / st \\ (t_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 b = false
------------------------------ (E_WhileEnd)
WHILE b DO c END / st \\ st
beval st b = true
c / st \\ st'
WHILE b DO c END / st' \\ st''
--------------------------------- (E_WhileLoop)
WHILE b DO c 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 \\ (t_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' b c1 c2,
beval st b = true ->
c1 / st \\ st' ->
(IFB b THEN c1 ELSE c2 FI) / st \\ st'
| E_IfFalse : forall st st' b c1 c2,
beval st b = false ->
c2 / st \\ st' ->
(IFB b THEN c1 ELSE c2 FI) / st \\ st'
| E_WhileEnd : forall b st c,
beval st b = false ->
(WHILE b DO c END) / st \\ st
| E_WhileLoop : forall st st' st'' b c,
beval st b = true ->
c / st \\ st' ->
(WHILE b DO c END) / st' \\ st'' ->
(WHILE b DO c END) / st \\ st''
where "c1 '/' st '\\' st'" := (ceval c1 st st').
(** 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
\\ (t_update (t_update empty_state X 2) Z 4).
Proof.
(* We must supply the intermediate state *)
apply E_Seq with (t_update empty_state X 2).
- (* assignment command *)
apply E_Ass. reflexivity.
- (* if command *)
apply E_IfFalse.
reflexivity.
apply E_Ass. reflexivity. Qed.
(** **** Exercise: 2 stars (ceval_example2) *)
Example ceval_example2:
(X ::= ANum 0;; Y ::= ANum 1;; Z ::= ANum 2) / empty_state \\
(t_update (t_update (t_update empty_state X 0) Y 1) Z 2).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, advanced (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 / (t_update empty_state X 2) \\
t_update (t_update (t_update (t_update (t_update (t_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.
induction E1;
intros st2 E2; inversion E2; subst.
- (* E_Skip *) reflexivity.
- (* E_Ass *) reflexivity.
- (* E_Seq *)
assert (st' = st'0) as EQ1.
{ (* Proof of assertion *) apply IHE1_1; assumption. }
subst st'0.
apply IHE1_2. assumption.
- (* E_IfTrue, b1 evaluates to true *)
apply IHE1. assumption.
- (* E_IfTrue, b1 evaluates to false (contradiction) *)
rewrite H in H5. inversion H5.
- (* E_IfFalse, b1 evaluates to true (contradiction) *)
rewrite H in H5. inversion H5.
- (* E_IfFalse, b1 evaluates to false *)
apply IHE1. assumption.
- (* E_WhileEnd, b1 evaluates to false *)
reflexivity.
- (* E_WhileEnd, b1 evaluates to true (contradiction) *)
rewrite H in H2. inversion H2.
- (* E_WhileLoop, b1 evaluates to false (contradiction) *)
rewrite H in H4. inversion H4.
- (* E_WhileLoop, b1 evaluates to true *)
assert (st' = st'0) as EQ1.
{ (* Proof of assertion *) apply IHE1_1; assumption. }
subst st'0.
apply IHE1_2. assumption. Qed.
(* ####################################################### *)
(** * Reasoning About Imp 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. *)
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 t_update_eq. Qed.
(** **** Exercise: 3 stars, recommended (XtimesYinZ_spec) *)
(** State and prove a specification of [XtimesYinZ]. *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 3 stars, recommended (loop_never_stops) *)
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 eqn:Heqloopdef.
(* 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.
(** [] *)
(** **** Exercise: 3 stars (no_whilesR) *)
(** Consider the definition of the [no_whiles] boolean predicate below: *)
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.
(** This predicate yields [true] just on 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 (no_whiles_terminating) *)
(** Imp programs that don't involve while loops always terminate.
State and prove a theorem [no_whiles_terminating] that says this. *)
(** (Use either [no_whiles] or [no_whilesR], as you prefer.) *)
(* FILL IN HERE *)
(** [] *)
(* ####################################################### *)
(** * Additional Exercises *)
(** **** Exercise: 3 stars (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.
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 what we do, since our compiler will never emit such a
malformed program. *)
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 (t_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.
(** After you've defined [s_compile], prove the following to test
that it works. *)
Example s_compile1 :
s_compile (AMinus (AId X) (AMult (ANum 2) (AId Y)))
= [SLoad X; SPush 2; SLoad Y; SMult; SMinus].
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, advanced (stack_compiler_correct) *)
(** The task of this exercise is to prove the correctness of the
compiler implemented in the previous exercise. Remember that
the specification left unspecified what to do when encountering an
[SPlus], [SMinus], or [SMult] instruction if the stack contains
less than two elements. (In order to make your correctness proof
easier you may find it useful to go back and change your
implementation!)
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; the main
theorem will then be a simple corollary of this lemma. *)
Theorem s_compile_correct : forall (st : state) (e : aexp),
s_execute st [] (s_compile e) = [ aeval st e ].
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 5 stars, advanced (break_imp) *)
Module BreakImp.
(** Imperative languages such as C or Java often have a [break] or
similar statement for interrupting the execution of loops. In this
exercise we will consider how to add [break] to Imp.
First, we need to enrich the language of commands with an
additional case. *)
Inductive com : Type :=
| CSkip : com
| CBreak : com
| CAss : id -> aexp -> com
| CSeq : com -> com -> com
| CIf : bexp -> com -> com -> com
| CWhile : bexp -> com -> com.
Notation "'SKIP'" :=
CSkip.
Notation "'BREAK'" :=
CBreak.
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' c1 'THEN' c2 'ELSE' c3 'FI'" :=
(CIf c1 c2 c3) (at level 80, right associativity).
(** Next, we need to define the behavior of [BREAK]. Informally,
whenever [BREAK] is executed in a sequence of commands, it stops
the execution of that sequence and signals that the innermost
enclosing loop (if any) should terminate. If there aren't any
enclosing loops, then the whole program simply terminates. The
final state should be the same as the one in which the [BREAK]
statement was executed.
One important point is what to do when there are multiple loops
enclosing a given [BREAK]. In those cases, [BREAK] should only
terminate the _innermost_ loop where it occurs. Thus, after
executing the following piece of code...
X ::= 0;;
Y ::= 1;;
WHILE 0 <> Y DO
WHILE TRUE DO
BREAK
END;;
X ::= 1;;
Y ::= Y - 1
END
... the value of [X] should be [1], and not [0].
One way of expressing this behavior is to add another parameter to
the evaluation relation that specifies whether evaluation of a
command executes a [BREAK] statement: *)
Inductive status : Type :=
| SContinue : status
| SBreak : status.
Reserved Notation "c1 '/' st '\\' s '/' st'"
(at level 40, st, s at level 39).
(** Intuitively, [c / st \\ s / st'] means that, if [c] is started in
state [st], then it terminates in state [st'] and either signals
that any surrounding loop (or the whole program) should exit
immediately ([s = SBreak]) or that execution should continue
normally ([s = SContinue]).
The definition of the "[c / st \\ s / st']" relation is very
similar to the one we gave above for the regular evaluation
relation ([c / st \\ st']) -- we just need to handle the
termination signals appropriately:
- If the command is [SKIP], then the state doesn't change, and
execution of any enclosing loop can continue normally.
- If the command is [BREAK], the state stays unchanged, but we
signal a [SBreak].
- If the command is an assignment, then we update the binding for
that variable in the state accordingly and signal that execution
can continue normally.
- If the command is of the form [IF b THEN c1 ELSE c2 FI], then
the state is updated as in the original semantics of Imp, except
that we also propagate the signal from the execution of
whichever branch was taken.
- If the command is a sequence [c1 ; c2], we first execute
[c1]. If this yields a [SBreak], we skip the execution of [c2]
and propagate the [SBreak] signal to the surrounding context;
the resulting state should be the same as the one obtained by
executing [c1] alone. Otherwise, we execute [c2] on the state
obtained after executing [c1], and propagate the signal that was
generated there.
- Finally, for a loop of the form [WHILE b DO c END], the
semantics is almost the same as before. The only difference is
that, when [b] evaluates to true, we execute [c] and check the
signal that it raises. If that signal is [SContinue], then the
execution proceeds as in the original semantics. Otherwise, we
stop the execution of the loop, and the resulting state is the
same as the one resulting from the execution of the current
iteration. In either case, since [BREAK] only terminates the
innermost loop, [WHILE] signals [SContinue]. *)
(** Based on the above description, complete the definition of the
[ceval] relation. *)
Inductive ceval : com -> state -> status -> state -> Prop :=
| E_Skip : forall st,
CSkip / st \\ SContinue / st
(* FILL IN HERE *)
where "c1 '/' st '\\' s '/' st'" := (ceval c1 st s st').
(** Now the following properties of your definition of [ceval]: *)
Theorem break_ignore : forall c st st' s,
(BREAK;; c) / st \\ s / st' ->
st = st'.
Proof.
(* FILL IN HERE *) Admitted.
Theorem while_continue : forall b c st st' s,
(WHILE b DO c END) / st \\ s / st' ->
s = SContinue.
Proof.
(* FILL IN HERE *) Admitted.
Theorem while_stops_on_break : forall b c st st',
beval st b = true ->
c / st \\ SBreak / st' ->
(WHILE b DO c END) / st \\ SContinue / st'.
Proof.
(* FILL IN HERE *) Admitted.
(** **** Exercise: 3 stars, advanced, optional (while_break_true) *)
Theorem while_break_true : forall b c st st',
(WHILE b DO c END) / st \\ SContinue / st' ->
beval st' b = true ->
exists st'', c / st'' \\ SBreak / st'.
Proof.
(* FILL IN HERE *) Admitted.
(** **** Exercise: 4 stars, advanced, optional (ceval_deterministic) *)
Theorem ceval_deterministic: forall (c:com) st st1 st2 s1 s2,
c / st \\ s1 / st1 ->
c / st \\ s2 / st2 ->
st1 = st2 /\ s1 = s2.
Proof.
(* FILL IN HERE *) Admitted.
End BreakImp.
(** [] *)
(** **** 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, 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 *)
(** [] *)
(* <$Date: 2016-02-05 09:55:10 -0500 (Fri, 05 Feb 2016) $ *)