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logical-foundations/Imp.v
Michael Zhang bc27534fa2
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2020-06-03 21:46:06 -05:00

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(** * Imp: Simple Imperative Programs *)
(** In this chapter, we take a more serious look at how to use Coq to
study other things. Our 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 ~(Z = 0) DO
Y ::= Y * Z;;
Z ::= Z - 1
END
*)
(** We concentrate here on defining the _syntax_ and _semantics_ of
Imp; later chapters in _Programming Language Foundations_
(_Software Foundations_, volume 2) develop a theory of
_program equivalence_ and introduce _Hoare Logic_, a widely
used logic for reasoning about imperative programs. *)
Set Warnings "-notation-overridden,-parsing".
From Coq Require Import Bool.Bool.
From Coq Require Import Init.Nat.
From Coq Require Import Arith.Arith.
From Coq Require Import Arith.EqNat.
From Coq Require Import omega.Omega.
From Coq Require Import Lists.List.
From Coq Require Import Strings.String.
Import ListNotations.
From LF Require Import Maps.
(* ################################################################# *)
(** * 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 (n : nat)
| APlus (a1 a2 : aexp)
| AMinus (a1 a2 : aexp)
| AMult (a1 a2 : aexp).
Inductive bexp : Type :=
| BTrue
| BFalse
| BEq (a1 a2 : aexp)
| BLe (a1 a2 : aexp)
| BNot (b : bexp)
| BAnd (b1 b2 : bexp).
(** In this chapter, we'll mostly 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 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 already 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
| ~ b
| b && 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 with an infix
[+]) while leaving other aspects of lexical analysis and
parsing (like the relative precedence of [+], [-], and
[*], the use of parens to group subexpressions, etc.)
unspecified. Some additional information -- and human
intelligence -- would be required to turn this description
into a formal definition, e.g., for implementing a compiler.
The Coq version consistently omits all this information and
concentrates on the abstract syntax only.
- Conversely, 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 kind 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 => (aeval a1) =? (aeval a2)
| BLe a1 a2 => (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 eqn:Ea1.
+ (* a1 = ANum n *) destruct n eqn:En.
* (* 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 (rather than 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, 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, 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 eqn:Ea1;
(* 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 eqn:En;
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 = 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, 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 (i.e., it does not terminate), this simply means
that we have failed to construct a proof, not that we have
constructed a wrong one. *)
(** **** Exercise: 3 stars, standard (optimize_0plus_b_sound)
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 that performs this 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, standard, optional (optimize)
_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 its correctness proof -- 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
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 by William Pugh [Pugh 1991] (in Bib.v).
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 ordering ([<=]), and
- the logical connectives [/\], [\/], [~], and [->],
then invoking [omega] will either solve the goal or fail, meaning
that the goal is actually false. (If the goal is _not_ of this
form, [omega] will also fail.) *)
Example silly_presburger_example : forall m n o p,
m + n <= n + o /\ o + 3 = p + 3 ->
m <= p.
Proof.
intros. omega.
Qed.
(** (Note the [From Coq Require Import omega.Omega.] at the top of
the file.) *)
(* ================================================================= *)
(** ** 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]: For a variable [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] (where [x] is a variable).
- [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 of all of these as we go along. *)
(* ################################################################# *)
(** * 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 n :
aevalR (ANum n) n
| E_APlus (e1 e2: aexp) (n1 n2: nat) :
aevalR e1 n1 ->
aevalR e2 n2 ->
aevalR (APlus e1 e2) (n1 + n2)
| E_AMinus (e1 e2: aexp) (n1 n2: nat) :
aevalR e1 n1 ->
aevalR e2 n2 ->
aevalR (AMinus e1 e2) (n1 - n2)
| E_AMult (e1 e2: aexp) (n1 n2: nat) :
aevalR e1 n1 ->
aevalR e2 n2 ->
aevalR (AMult e1 e2) (n1 * n2).
Module TooHardToRead.
(* A small notational aside. We would previously have written the
definition of [aevalR] like this, with explicit names for the
hypotheses in each case: *)
Inductive aevalR : aexp -> nat -> Prop :=
| E_ANum n :
aevalR (ANum n) n
| E_APlus (e1 e2: aexp) (n1 n2: nat)
(H1 : aevalR e1 n1)
(H2 : aevalR e2 n2) :
aevalR (APlus e1 e2) (n1 + n2)
| E_AMinus (e1 e2: aexp) (n1 n2: nat)
(H1 : aevalR e1 n1)
(H2 : aevalR e2 n2) :
aevalR (AMinus e1 e2) (n1 - n2)
| E_AMult (e1 e2: aexp) (n1 n2: nat)
(H1 : aevalR e1 n1)
(H2 : aevalR e2 n2) :
aevalR (AMult e1 e2) (n1 * n2).
(** Instead, we've chosen to leave the hypotheses anonymous, just
giving their types. This style gives us less control over the
names that Coq chooses during proofs involving [aevalR], but it
makes the definition itself quite a bit lighter. *)
End TooHardToRead.
(** 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]. *)
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 90, left associativity).
Inductive aevalR : aexp -> nat -> Prop :=
| E_ANum (n : nat) :
(ANum n) \\ n
| E_APlus (e1 e2 : aexp) (n1 n2 : nat) :
(e1 \\ n1) -> (e2 \\ n2) -> (APlus e1 e2) \\ (n1 + n2)
| E_AMinus (e1 e2 : aexp) (n1 n2 : nat) :
(e1 \\ n1) -> (e2 \\ n2) -> (AMinus e1 e2) \\ (n1 - n2)
| E_AMult (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 [IndProp] 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
*)
(** **** Exercise: 1 star, standard, optional (beval_rules)
Here, again, is the Coq definition of the [beval] function:
Fixpoint beval (e : bexp) : bool :=
match e with
| BTrue => true
| BFalse => false
| BEq a1 a2 => (aeval a1) =? (aeval a2)
| BLe a1 a2 => (aeval a1) <=? (aeval a2)
| BNot b1 => negb (beval b1)
| BAnd b1 b2 => andb (beval b1) (beval b2)
end.
Write out a corresponding definition of boolean evaluation as a
relation (in inference rule notation). *)
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_beval_rules : option (nat*string) := None.
(** [] *)
(* ================================================================= *)
(** ** 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, standard (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 with division: *)
Inductive aexp : Type :=
| ANum (n : nat)
| APlus (a1 a2 : aexp)
| AMinus (a1 a2 : aexp)
| AMult (a1 a2 : aexp)
| ADiv (a1 a2 : 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 90, left associativity).
Inductive aevalR : aexp -> nat -> Prop :=
| E_ANum (n : nat) :
(ANum n) \\ n
| E_APlus (a1 a2 : aexp) (n1 n2 : nat) :
(a1 \\ n1) -> (a2 \\ n2) -> (APlus a1 a2) \\ (n1 + n2)
| E_AMinus (a1 a2 : aexp) (n1 n2 : nat) :
(a1 \\ n1) -> (a2 \\ n2) -> (AMinus a1 a2) \\ (n1 - n2)
| E_AMult (a1 a2 : aexp) (n1 n2 : nat) :
(a1 \\ n1) -> (a2 \\ n2) -> (AMult a1 a2) \\ (n1 * n2)
| E_ADiv (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.
(** Or suppose 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 probability distribution for the
results, just saying what results are _possible_.) *)
Reserved Notation "e '\\' n" (at level 90, left associativity).
Inductive aexp : Type :=
| AAny (* <--- NEW *)
| ANum (n : nat)
| APlus (a1 a2 : aexp)
| AMinus (a1 a2 : aexp)
| AMult (a1 a2 : 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 (n : nat) :
AAny \\ n (* <--- NEW *)
| E_ANum (n : nat) :
(ANum n) \\ n
| E_APlus (a1 a2 : aexp) (n1 n2 : nat) :
(a1 \\ n1) -> (a2 \\ n2) -> (APlus a1 a2) \\ (n1 + n2)
| E_AMinus (a1 a2 : aexp) (n1 n2 : nat) :
(a1 \\ n1) -> (a2 \\ n2) -> (AMinus a1 a2) \\ (n1 - n2)
| E_AMult (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? In the examples we've just seen, relational definitions
turned out to be more useful 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
real choice. But what about when both styles are workable?
One point in favor of relational definitions is that they can be
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" from Gallina 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 *)
(** 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 maps from the [Maps] chapter, and
[string]s will be used to represent 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 strings to [nat], and will use [0] as default value
in the store. For more complex programming languages, the state
might have more structure. *)
Definition state := total_map nat.
(* ================================================================= *)
(** ** Syntax *)
(** We can add variables to the arithmetic expressions we had before by
simply adding one more constructor: *)
Inductive aexp : Type :=
| ANum (n : nat)
| AId (x : string) (* <--- NEW *)
| APlus (a1 a2 : aexp)
| AMinus (a1 a2 : aexp)
| AMult (a1 a2 : aexp).
(** Defining a few variable names as notational shorthands will make
examples easier to read: *)
Definition W : string := "W".
Definition X : string := "X".
Definition Y : string := "Y".
Definition Z : string := "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
developed to Imp, this overloading should not cause confusion.) *)
(** The definition of [bexp]s is unchanged (except that it now refers
to the new [aexp]s): *)
Inductive bexp : Type :=
| BTrue
| BFalse
| BEq (a1 a2 : aexp)
| BLe (a1 a2 : aexp)
| BNot (b : bexp)
| BAnd (b1 b2 : bexp).
(* ================================================================= *)
(** ** Notations *)
(** To make Imp programs easier to read and write, we introduce some
notations and implicit coercions.
You do not need to understand exactly what these declarations do.
Briefly, though, the [Coercion] declaration in Coq stipulates that
a function (or constructor) can be implicitly used by the type
system to coerce a value of the input type to a value of the
output type. For instance, the coercion declaration for [AId]
allows us to use plain strings when an [aexp] is expected; the
string will implicitly be wrapped with [AId]. *)
(** The notations below are declared in specific _notation scopes_, in
order to avoid conflicts with other interpretations of the same
symbols. Again, it is not necessary to understand the details,
but it is important to recognize that we are defining _new_
intepretations for some familiar operators like [+], [-], [*],
[=., [<=], etc. *)
Coercion AId : string >-> aexp.
Coercion ANum : nat >-> aexp.
Definition bool_to_bexp (b : bool) : bexp :=
if b then BTrue else BFalse.
Coercion bool_to_bexp : bool >-> bexp.
Bind Scope imp_scope with aexp.
Bind Scope imp_scope with bexp.
Delimit Scope imp_scope with imp.
Notation "x + y" := (APlus x y) (at level 50, left associativity) : imp_scope.
Notation "x - y" := (AMinus x y) (at level 50, left associativity) : imp_scope.
Notation "x * y" := (AMult x y) (at level 40, left associativity) : imp_scope.
Notation "x <= y" := (BLe x y) (at level 70, no associativity) : imp_scope.
Notation "x = y" := (BEq x y) (at level 70, no associativity) : imp_scope.
Notation "x && y" := (BAnd x y) (at level 40, left associativity) : imp_scope.
Notation "'~' b" := (BNot b) (at level 75, right associativity) : imp_scope.
Definition example_aexp := (3 + (X * 2))%imp : aexp.
Definition example_bexp := (true && ~(X <= 4))%imp : bexp.
(** One downside of these coercions is that they can make it a little
harder for humans to calculate the types of expressions. If you
get confused, try doing [Set Printing Coercions] to see exactly
what is going on. *)
Set Printing Coercions.
Print example_bexp.
(* ===> example_bexp = bool_to_bexp true && ~ (AId X <= ANum 4) *)
Unset Printing Coercions.
(* ================================================================= *)
(** ** 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 => (aeval st a1) =? (aeval st a2)
| BLe a1 a2 => (aeval st a1) <=? (aeval st a2)
| BNot b1 => negb (beval st b1)
| BAnd b1 b2 => andb (beval st b1) (beval st b2)
end.
(** We specialize our notation for total maps to the specific case of
states, i.e. using [(_ !-> 0)] as empty state. *)
Definition empty_st := (_ !-> 0).
(** Now we can add a notation for a "singleton state" with just one
variable bound to a value. *)
Notation "a '!->' x" := (t_update empty_st a x) (at level 100).
Example aexp1 :
aeval (X !-> 5) (3 + (X * 2))%imp
= 13.
Proof. reflexivity. Qed.
Example bexp1 :
beval (X !-> 5) (true && ~(X <= 4))%imp
= 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.
c ::= SKIP | x ::= a | c ;; c | TEST b THEN c ELSE c FI
| WHILE b DO c END
(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 [TEST] to avoid conflicting with
the [if] and [IF] notations from the standard library.)
For example, here's factorial in Imp:
Z ::= X;;
Y ::= 1;;
WHILE ~(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
| CAss (x : string) (a : aexp)
| CSeq (c1 c2 : com)
| CIf (b : bexp) (c1 c2 : com)
| CWhile (b : bexp) (c : com).
(** As for expressions, we can use a few [Notation] declarations to
make reading and writing Imp programs more convenient. *)
Bind Scope imp_scope with com.
Notation "'SKIP'" :=
CSkip : imp_scope.
Notation "x '::=' a" :=
(CAss x a) (at level 60) : imp_scope.
Notation "c1 ;; c2" :=
(CSeq c1 c2) (at level 80, right associativity) : imp_scope.
Notation "'WHILE' b 'DO' c 'END'" :=
(CWhile b c) (at level 80, right associativity) : imp_scope.
Notation "'TEST' c1 'THEN' c2 'ELSE' c3 'FI'" :=
(CIf c1 c2 c3) (at level 80, right associativity) : imp_scope.
(** For example, here is the factorial function again, written as a
formal definition to Coq: *)
Definition fact_in_coq : com :=
(Z ::= X;;
Y ::= 1;;
WHILE ~(Z = 0) DO
Y ::= Y * Z;;
Z ::= Z - 1
END)%imp.
(* ================================================================= *)
(** ** Desugaring notations *)
(** Coq offers a rich set of features to manage the increasing
complexity of the objects we work with, such as coercions
and notations. However, their heavy usage can make for quite
overwhelming syntax. It is often instructive to "turn off"
those features to get a more elementary picture of things,
using the following commands:
- [Unset Printing Notations] (undo with [Set Printing Notations])
- [Set Printing Coercions] (undo with [Unset Printing Coercions])
- [Set Printing All] (undo with [Unset Printing All])
These commands can also be used in the middle of a proof,
to elaborate the current goal and context.
*)
Unset Printing Notations.
Print fact_in_coq.
(* ===>
fact_in_coq =
CSeq (CAss Z X)
(CSeq (CAss Y (S O))
(CWhile (BNot (BEq Z O))
(CSeq (CAss Y (AMult Y Z))
(CAss Z (AMinus Z (S O))))))
: com *)
Set Printing Notations.
Set Printing Coercions.
Print fact_in_coq.
(* ===>
fact_in_coq =
(Z ::= AId X;;
Y ::= ANum 1;;
WHILE ~ (AId Z = ANum 0) DO
Y ::= AId Y * AId Z;;
Z ::= AId Z - ANum 1
END)%imp
: com *)
Unset Printing Coercions.
(* ================================================================= *)
(** ** The [Locate] command *)
(* ----------------------------------------------------------------- *)
(** *** Finding notations *)
(** When faced with unknown notation, use [Locate] with a _string_
containing one of its symbols to see its possible
interpretations. *)
Locate "&&".
(* ===>
Notation "x && y" := andb x y : bool_scope (default interpretation) *)
Locate ";;".
(* ===>
Notation "c1 ;; c2" := CSeq c1 c2 : imp_scope (default interpretation) *)
Locate "WHILE".
(* ===>
Notation "'WHILE' b 'DO' c 'END'" := CWhile b c : imp_scope
(default interpretation) *)
(* ----------------------------------------------------------------- *)
(** *** Finding identifiers *)
(** When used with an identifier, the command [Locate] prints
the full path to every value in scope with the same name.
This is useful to troubleshoot problems due to variable
shadowing. *)
Locate aexp.
(* ===>
Inductive Top.aexp
Inductive Top.AExp.aexp
(shorter name to refer to it in current context is AExp.aexp)
Inductive Top.aevalR_division.aexp
(shorter name to refer to it in current context is aevalR_division.aexp)
Inductive Top.aevalR_extended.aexp
(shorter name to refer to it in current context is aevalR_extended.aexp)
*)
(* ================================================================= *)
(** ** More Examples *)
(** Assignment: *)
Definition plus2 : com :=
X ::= X + 2.
Definition XtimesYinZ : com :=
Z ::= X * Y.
Definition subtract_slowly_body : com :=
Z ::= Z - 1 ;;
X ::= X - 1.
(* ----------------------------------------------------------------- *)
(** *** Loops *)
Definition subtract_slowly : com :=
(WHILE ~(X = 0) DO
subtract_slowly_body
END)%imp.
Definition subtract_3_from_5_slowly : com :=
X ::= 3 ;;
Z ::= 5 ;;
subtract_slowly.
(* ----------------------------------------------------------------- *)
(** *** An infinite loop: *)
Definition loop : com :=
WHILE true 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. *)
(** The following declaration is needed to be able to use the
notations in match patterns. *)
Open Scope imp_scope.
Fixpoint ceval_fun_no_while (st : state) (c : com)
: state :=
match c with
| SKIP =>
st
| x ::= a1 =>
(x !-> (aeval st a1) ; st)
| c1 ;; c2 =>
let st' := ceval_fun_no_while st c1 in
ceval_fun_no_while st' c2
| TEST 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.
Close Scope imp_scope.
(** 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 [st =[ c ]=> st'] for the [ceval] relation:
[st =[ c ]=> 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)
st =[ SKIP ]=> st
aeval st a1 = n
-------------------------------- (E_Ass)
st =[ x := a1 ]=> (x !-> n ; st)
st =[ c1 ]=> st'
st' =[ c2 ]=> st''
--------------------- (E_Seq)
st =[ c1;;c2 ]=> st''
beval st b1 = true
st =[ c1 ]=> st'
--------------------------------------- (E_IfTrue)
st =[ TEST b1 THEN c1 ELSE c2 FI ]=> st'
beval st b1 = false
st =[ c2 ]=> st'
--------------------------------------- (E_IfFalse)
st =[ TEST b1 THEN c1 ELSE c2 FI ]=> st'
beval st b = false
----------------------------- (E_WhileFalse)
st =[ WHILE b DO c END ]=> st
beval st b = true
st =[ c ]=> st'
st' =[ WHILE b DO c END ]=> st''
-------------------------------- (E_WhileTrue)
st =[ WHILE b DO c END ]=> st''
*)
(** Here is the formal definition. Make sure you understand
how it corresponds to the inference rules. *)
Reserved Notation "st '=[' c ']=>' st'"
(at level 40).
Inductive ceval : com -> state -> state -> Prop :=
| E_Skip : forall st,
st =[ SKIP ]=> st
| E_Ass : forall st a1 n x,
aeval st a1 = n ->
st =[ x ::= a1 ]=> (x !-> n ; st)
| E_Seq : forall c1 c2 st st' st'',
st =[ c1 ]=> st' ->
st' =[ c2 ]=> st'' ->
st =[ c1 ;; c2 ]=> st''
| E_IfTrue : forall st st' b c1 c2,
beval st b = true ->
st =[ c1 ]=> st' ->
st =[ TEST b THEN c1 ELSE c2 FI ]=> st'
| E_IfFalse : forall st st' b c1 c2,
beval st b = false ->
st =[ c2 ]=> st' ->
st =[ TEST b THEN c1 ELSE c2 FI ]=> st'
| E_WhileFalse : forall b st c,
beval st b = false ->
st =[ WHILE b DO c END ]=> st
| E_WhileTrue : forall st st' st'' b c,
beval st b = true ->
st =[ c ]=> st' ->
st' =[ WHILE b DO c END ]=> st'' ->
st =[ WHILE b DO c END ]=> st''
where "st =[ c ]=> st'" := (ceval c 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:
empty_st =[
X ::= 2;;
TEST X <= 1
THEN Y ::= 3
ELSE Z ::= 4
FI
]=> (Z !-> 4 ; X !-> 2).
Proof.
(* We must supply the intermediate state *)
apply E_Seq with (X !-> 2).
- (* assignment command *)
apply E_Ass. reflexivity.
- (* if command *)
apply E_IfFalse.
reflexivity.
apply E_Ass. reflexivity.
Qed.
(** **** Exercise: 2 stars, standard (ceval_example2) *)
Example ceval_example2:
empty_st =[
X ::= 0;; Y ::= 1;; Z ::= 2
]=> (Z !-> 2 ; Y !-> 1 ; X !-> 0).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, standard, optional (pup_to_n)
Write an Imp program that sums the numbers from [1] to
[X] (inclusive: [1 + 2 + ... + X]) in the variable [Y].
Prove that this program executes as intended for [X] = [2]
(this is trickier than you might expect). *)
Definition pup_to_n : com
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Theorem pup_to_2_ceval :
(X !-> 2) =[
pup_to_n
]=> (X !-> 0 ; Y !-> 3 ; X !-> 1 ; Y !-> 2 ; Y !-> 0 ; X !-> 2).
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,
st =[ c ]=> st1 ->
st =[ c ]=> 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. discriminate H5.
- (* E_IfFalse, b1 evaluates to true (contradiction) *)
rewrite H in H5. discriminate H5.
- (* E_IfFalse, b1 evaluates to false *)
apply IHE1. assumption.
- (* E_WhileFalse, b1 evaluates to false *)
reflexivity.
- (* E_WhileFalse, b1 evaluates to true (contradiction) *)
rewrite H in H2. discriminate H2.
- (* E_WhileTrue, b1 evaluates to false (contradiction) *)
rewrite H in H4. discriminate H4.
- (* E_WhileTrue, 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 more systematic and powerful techniques for
reasoning about Imp programs in _Programming Language
Foundations_, but we can get some distance just working with the
bare definitions. This section explores some examples. *)
Theorem plus2_spec : forall st n st',
st X = n ->
st =[ plus2 ]=> 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, standard, recommended (XtimesYinZ_spec)
State and prove a specification of [XtimesYinZ]. *)
(* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_XtimesYinZ_spec : option (nat*string) := None.
(** [] *)
(** **** Exercise: 3 stars, standard, recommended (loop_never_stops) *)
Theorem loop_never_stops : forall st st',
~(st =[ loop ]=> st').
Proof.
intros st st' contra. unfold loop in contra.
remember (WHILE true DO SKIP END)%imp 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
[discriminate]). *)
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, standard (no_whiles_eqv)
Consider the following function: *)
Open Scope imp_scope.
Fixpoint no_whiles (c : com) : bool :=
match c with
| SKIP =>
true
| _ ::= _ =>
true
| c1 ;; c2 =>
andb (no_whiles c1) (no_whiles c2)
| TEST _ THEN ct ELSE cf FI =>
andb (no_whiles ct) (no_whiles cf)
| WHILE _ DO _ END =>
false
end.
Close Scope imp_scope.
(** 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, standard (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 *)
(* Do not modify the following line: *)
Definition manual_grade_for_no_whiles_terminating : option (nat*string) := None.
(** [] *)
(* ################################################################# *)
(** * Additional Exercises *)
(** **** Exercise: 3 stars, standard (stack_compiler)
Old 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 written 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 goal 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 (n : nat)
| SLoad (x : string)
| SPlus
| SMinus
| SMult.
(** 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_st []
[SPush 5; SPush 3; SPush 1; SMinus]
= [2; 5].
(* FILL IN HERE *) Admitted.
Example s_execute2 :
s_execute (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 (X - (2 * Y))%imp
= [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, standard, 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]. (N.b. This is only true because expression
evaluation in Imp is rather simple. In a bigger language where
evaluating an expression might diverge, the short-circuiting [BAnd]
would _not_ be equivalent to the original, since it would make more
programs terminate.) *)
(* FILL IN HERE
[] *)
Module BreakImp.
(** **** Exercise: 4 stars, advanced (break_imp)
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
| CBreak (* <--- NEW *)
| CAss (x : string) (a : aexp)
| CSeq (c1 c2 : com)
| CIf (b : bexp) (c1 c2 : com)
| CWhile (b : bexp) (c : 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 "'TEST' 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
| SBreak.
Reserved Notation "st '=[' c ']=>' st' '/' s"
(at level 40, st' at next level).
(** Intuitively, [st =[ c ]=> st' / s] 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 "[st =[ c ]=> st' / s]" relation is very
similar to the one we gave above for the regular evaluation
relation ([st =[ c ]=> 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 [TEST 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,
st =[ CSkip ]=> st / SContinue
(* FILL IN HERE *)
where "st '=[' c ']=>' st' '/' s" := (ceval c st s st').
(** Now prove the following properties of your definition of [ceval]: *)
Theorem break_ignore : forall c st st' s,
st =[ BREAK;; c ]=> st' / s ->
st = st'.
Proof.
(* FILL IN HERE *) Admitted.
Theorem while_continue : forall b c st st' s,
st =[ WHILE b DO c END ]=> st' / s ->
s = SContinue.
Proof.
(* FILL IN HERE *) Admitted.
Theorem while_stops_on_break : forall b c st st',
beval st b = true ->
st =[ c ]=> st' / SBreak ->
st =[ WHILE b DO c END ]=> st' / SContinue.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, advanced, optional (while_break_true) *)
Theorem while_break_true : forall b c st st',
st =[ WHILE b DO c END ]=> st' / SContinue ->
beval st' b = true ->
exists st'', st'' =[ c ]=> st' / SBreak.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 4 stars, advanced, optional (ceval_deterministic) *)
Theorem ceval_deterministic: forall (c:com) st st1 st2 s1 s2,
st =[ c ]=> st1 / s1 ->
st =[ c ]=> st2 / s2 ->
st1 = st2 /\ s1 = s2.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
End BreakImp.
(** **** Exercise: 4 stars, standard, 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
[] *)
(* Wed Jan 9 12:02:46 EST 2019 *)
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