(** * 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. *)
(* IMPORTS *)
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.
(* /IMPORTS *)
(* ################################################################# *)
(** * 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 chapter 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
[*], the use of parens to explicitly group subexpressions,
etc.) unspecified. Some additional information (and human
intelligence) would be required to turn this description into
a formal definition, for example when implementing a
compiler.
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, a big 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
important.
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 amount of repetition in this last proof is a little
annoying. And if either the language of arithmetic expressions or
the optimization being proved sound were significantly more
complex, it would start 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.
(** There is no real reason to use [try] in completely manual
proofs like these, but it is very useful for doing automated
proofs in conjunction with the [;] tactical, which we show
next. *)
(* ----------------------------------------------------------------- *)
(** *** The [;] Tactical (Simple Form) *)
(** In its most common form, the [;] tactical takes two tactics as
arguments. The compound tactic [T;T'] first performs [T] and then
performs [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] the current goal *)
destruct n;
(* then [simpl] each resulting subgoal *)
simpl;
(* and do [reflexivity] on each resulting subgoal *)
reflexivity.
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).
(* ... but 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. For
example, here is an informal proof of the optimization 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]. Most 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. [] *)
(** However, 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. 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 shorthand for:
T; [T' | T' | ... | T']
*)
(* ----------------------------------------------------------------- *)
(** *** The [repeat] Tactical *)
(** The [repeat] tactical takes another tactic and keeps applying this
tactic until it fails. Here is an example showing that [10] is in
a long list using repeat. *)
Theorem In10 : In 10 [1;2;3;4;5;6;7;8;9;10].
Proof.
repeat (try (left; reflexivity); right).
Qed.
(** The tactic [repeat T] never fails: if the tactic [T] doesn't apply
to the original goal, then repeat still succeeds without changing
the original goal (i.e., 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 tactic [repeat T] also does not have any upper 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 evaluation
in Coq's term language, Gallina, is guaranteed to terminate,
tactic evaluation is not! This does not affect Coq's logical
consistency, however, since the job of [repeat] and other tactics
is to guide Coq in constructing proofs; if the construction
process diverges, this simply means that we have failed to
construct a proof, not that we have constructed a wrong one. *)
(** **** Exercise: 3 stars (optimize_0plus_b) *)
(** Since the [optimize_0plus] transformation 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
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
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 possible
optimizations on arithmetic and boolean expressions. Write a more
sophisticated optimizer and prove it correct. (You will probably
find it easiest to start small -- add just a single, simple
optimization and prove it correct -- and build up to something
more interesting incrementially.)
(* FILL IN HERE *)
*)
(** [] *)
(* ================================================================= *)
(** ** 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 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 other books on Coq, 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] that takes one
tactic [c] as an argument and is defined to be equivalent to the
tactic [simpl; try c]. Now writing "[simpl_and_try reflexivity.]"
in a proof will be the same as writing "[simpl; try
reflexivity.]" *)
(* ================================================================= *)
(** ** 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 1991 by William Pugh [Pugh 1991].
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.
(* ================================================================= *)
(** ** 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 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 examples 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).
(** It will be convenient to have an 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 reduces confusion by avoiding
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 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 (as well as 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. In informal prose, 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: *)
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.
(** 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: bexp -> bool -> Prop :=
(* FILL IN HERE *)
.
Lemma beval_iff_bevalR : forall b bv,
bevalR b bv <-> beval b = bv.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
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: either way works.
However, there are circumstances where relational definitions of
evaluation work much better than 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.
(** Suppose, instead, that we want to extend the arithmetic operations
by a nondeterministic number generator [any] that, when evaluated,
may yield any number. (Note that this is not the same as making a
_probabilistic_ choice among all possible numbers -- we're not
specifying any particular distribution of results, but just saying
what results are _possible_.) *)
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, since now 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.
(** At this point you maybe wondering: which style should I use by
default? The examples above show that relational definitions are
fundamentally more powerful than functional ones. For situations
like these, where the thing being defined is not easy to express
as a function, or indeed where it is _not_ a function, there is no
choice. But what about when both styles are workable?
One point in favor of relational definitions is that some people
feel they are more elegant and easier to understand. Another is
that Coq automatically generates nice inversion and induction
principles from [Inductive] definitions.
On the other hand, functional definitions can often be more
convenient:
- Functions are by definition deterministic and defined on all
arguments; for a relation we have to show these properties
explicitly if we need them.
- With functions we can also take advantage of Coq's computation
mechanism to simplify expressions during proofs.
Furthermore, functions can be directly "extracted" to executable
code in OCaml or Haskell.
Ultimately, the choice often comes down to either the specifics of
a particular situation or simply a question of taste. Indeed, in
large Coq developments it is common to see a definition given in
_both_ functional and relational styles, plus a lemma stating that
the two coincide, allowing further proofs to switch from one point
of view to the other at will.*)
(* ################################################################# *)
(** * 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_ 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 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 W : id := Id "W".
Definition X : id := Id "X".
Definition Y : id := Id "Y".
Definition Z : id := Id "Z".
(** (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 the chapters
devoped to Imp, this overloading should not cause confusion.) *)
(** The definition of [bexp]s is unchanged (except for 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 are extended to handle
variables in the obvious way, taking a state as an extra
argument: *)
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_ (sometimes called _statements_). *)
(* ================================================================= *)
(** ** Syntax *)
(** Informally, commands [c] are described by the following BNF
grammar. (We choose this slightly awkward concrete syntax for the
sake of being able to define Imp syntax using Coq's Notation
mechanism. In particular, we use [IFB] to avoid conflicting with
the [if] notation from the standard library.)
c ::= SKIP | x ::= a | c ;; c | IFB b THEN c ELSE c FI
| WHILE b DO c END
*)
(**
For example, here's factorial 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 abstract 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. To avoid conflicts with Coq's built-in notations,
we keep this light -- in particular, we don't introduce any
notations for [aexps] and [bexps] to avoid confusion with the
numeric and boolean operators we've already defined. *)
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.
(* ================================================================= *)
(** ** More 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.
(* ################################################################# *)
(** * Evaluating Commands *)
(** 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 OCaml or
Haskell we could add 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!) 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
([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,
of [ceval_fun] cannot be written in Coq -- at least not without
additional tricks and workarounds (see chapter [ImpCEvalFun]
if you're curious about what those might be). *)
(* ================================================================= *)
(** ** Evaluation as a Relation *)
(** Here's a better way: define [ceval] as a _relation_ rather than a
_function_ -- i.e., define it in [Prop] instead of [Type], as we
did for [aevalR] above. *)
(** This is an important change. Besides freeing us from awkward
workarounds, it gives us a lot more flexibility in the definition.
For example, if we add nondeterministic features like [any] to the
language, we want the definition of evaluation to be
nondeterministic -- i.e., not only will it not be total, it will
not even be a function! *)
(** We'll use the notation [c / st \\ st'] for the [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 *)
(** Here is an informal definition of evaluation, presented as inference
rules for readability:
---------------- (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 is trickier than you might expect). *)
Definition pup_to_n : com
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
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 frees us from the artificial
requirement that evaluation should be a total function. But it
also raises a question: Is the second definition of evaluation
really a partial function? Or 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: *)
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 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 following function: *)
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 (where we show the program being evaluated
on the right and the contents of the stack on the left):
[] | 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
should take 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 it should return 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
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
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 that 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
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
(** 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: 4 stars, advanced (stack_compiler_correct) *)
(** Now we'll 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
might find it helpful to go back and change your implementation!)
Prove the following theorem. 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: 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, advanced (break_imp) *)
Module BreakImp.
(** Imperative languages like C and Java often include a [break] or
similar statement for interrupting the execution of loops. In this
exercise we 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 (* <-- new *)
| 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 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. Thus, after executing the
following...
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 result : Type :=
| SContinue : result
| SBreak : result.
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 the innermost 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 [IFB 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 is 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
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 -> result -> state -> Prop :=
| E_Skip : forall st,
CSkip / st \\ SContinue / st
(* FILL IN HERE *)
where "c1 '/' st '\\' s '/' st'" := (ceval c1 st s st').
(** Now prove 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: 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: 2016-11-16 06:03:33 -0500 (Wed, 16 Nov 2016) $ *)