Imp: Simple Imperative Programs
Set Warnings "notationoverridden,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 PLF Require Import Maps.
Arithmetic and Boolean Expressions
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 (BackusNaur 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...
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.
 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.
 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.
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
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.
 reflexivity.
 destruct a1 eqn:Ea1.
+ destruct n eqn:En.
* simpl. apply IHa2.
* simpl. rewrite IHa2. reflexivity.
+
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
+
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
+
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.

simpl. rewrite IHa1. rewrite IHa2. reflexivity.

simpl. rewrite IHa1. rewrite IHa2. reflexivity. Qed.
Coq Automation
Tacticals
The try Tactical
Theorem silly1 : forall ae, aeval ae = aeval ae.
Proof. try reflexivity. Qed.
Theorem silly2 : forall (P : Prop), P > P.
Proof.
intros P HP.
try reflexivity. apply HP. 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)
Lemma foo : forall n, 0 <=? n = true.
Proof.
intros.
destruct n.
 simpl. reflexivity.
 simpl. reflexivity.
Qed.
We can simplify this proof using the ; tactical:
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;
try (simpl; rewrite IHa1; rewrite IHa2; reflexivity).
 reflexivity.

destruct a1 eqn:Ea1;
try (simpl; simpl in IHa1; rewrite IHa1;
rewrite IHa2; reflexivity).
+ 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:
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:
 Suppose a = ANum n for some n. We must show
 Suppose a = APlus a1 a2 for some a1 and a2. We must
show
Theorem optimize_0plus_sound'': forall a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a.
induction a;
try (simpl; rewrite IHa1; rewrite IHa2; reflexivity);
try reflexivity.

destruct a1; try (simpl; simpl in IHa1; rewrite IHa1;
rewrite IHa2; reflexivity).
+ destruct n;
simpl; rewrite IHa2; reflexivity. Qed.
The ; Tactical (General Form)
The repeat Tactical
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.
Since the optimize_0plus transformation doesn't change the value
of aexps, we should be able to apply it to all the aexps that
appear in a bexp without changing the bexp's value. Write a
function that performs this transformation on bexps and prove
it is sound. Use the tacticals we've just seen to make the proof
as elegant as possible.
Exercise: 3 stars, standard (optimize_0plus_b_sound)
Fixpoint optimize_0plus_b (b : bexp) : bexp
. Admitted.
Theorem optimize_0plus_b_sound : forall b,
beval (optimize_0plus_b b) = beval b.
Proof.
Admitted.
☐
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.)
Exercise: 4 stars, standard, optional (optimize)
Defining New Tactic Notations
 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 builtin
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.
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
 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 >,
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
 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.
Evaluation as a Relation
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.
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.
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
(E_APlus) APlus e1 e2 \\ n1+n2
(E_ANum) ANum n \\ n
(E_APlus) APlus e1 e2 \\ n1+n2
(E_AMinus) AMinus e1 e2 \\ n1n2
(E_AMult) AMult e1 e2 \\ n1*n2
Exercise: 1 star, standard, optional (beval_rules)
☐
Equivalence of the Definitions
Theorem aeval_iff_aevalR : forall a n,
(a \\ n) <> aeval a = n.
Proof.
split.

intros H.
induction H; simpl.
+
reflexivity.
+
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
+
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.
+
rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.

generalize dependent n.
induction a;
simpl; intros; subst.
+
apply E_ANum.
+
apply E_APlus.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
+
apply E_AMinus.
apply IHa1. reflexivity.
apply IHa2. reflexivity.
+
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.
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)
Inductive bevalR: bexp > bool > Prop :=
.
Lemma beval_iff_bevalR : forall b bv,
bevalR b bv <> beval b = bv.
Proof.
Admitted.
☐
Computational vs. Relational Definitions
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).
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
 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
 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:
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.
 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.
Expressions With Variables
States
Syntax
Inductive aexp : Type :=
 ANum (n : nat)
 AId (x : string)
 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 bexps is unchanged (except that it now refers
to the new aexps):
Inductive bexp : Type :=
 BTrue
 BFalse
 BEq (a1 a2 : aexp)
 BLe (a1 a2 : aexp)
 BNot (b : bexp)
 BAnd (b1 b2 : bexp).
Notations
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.
We can now write 3 + (X * 2) instead of APlus 3 (AMult X 2),
and true && ~(X <= 4) instead of BAnd true (BNot (BLe X 4)).
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.
Evaluation
Fixpoint aeval (st : state) (a : aexp) : nat :=
match a with
 ANum n => n
 AId x => st x
 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.
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.
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
Syntax
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
 Unset Printing Notations (undo with Set Printing Notations)
 Set Printing Coercions (undo with Unset Printing Coercions)
 Set Printing All (undo with Unset Printing All)
Unset Printing Notations.
Print fact_in_coq.
Set Printing Notations.
Set Printing Coercions.
Print fact_in_coq.
Unset Printing Coercions.
Finding notations
Locate "&&".
Locate ";;".
Locate "WHILE".
Locate ";;".
Locate "WHILE".
Finding identifiers
Locate aexp.
Definition plus2 : com :=
X ::= X + 2.
Definition XtimesYinZ : com :=
Z ::= X * Y.
Definition subtract_slowly_body : com :=
Z ::= Z  1 ;;
X ::= X  1.
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.
Evaluating Commands
Evaluation as a Function (Failed Attempt)
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
end.
Close 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
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
nonterminating function needs to be rejected. Here is
an (invalid!) program showing what would go wrong if Coq
allowed nonterminating 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
Operational Semantics
(E_Skip) st = SKIP => st
(E_Ass) st = x := a1 => (x !> n ; st)
(E_Seq) st = c1;;c2 => st''
(E_IfTrue) st = TEST b1 THEN c1 ELSE c2 FI => st'
(E_IfFalse) st = TEST b1 THEN c1 ELSE c2 FI => st'
(E_WhileFalse) st = WHILE b DO c END => st
(E_WhileTrue) st = WHILE b DO c END => st''
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.
apply E_Seq with (X !> 2).

apply E_Ass. reflexivity.

apply E_IfFalse.
reflexivity.
apply E_Ass. reflexivity.
Qed.
Example ceval_example2:
empty_st =[
X ::= 0;; Y ::= 1;; Z ::= 2
]=> (Z !> 2 ; Y !> 1 ; X !> 0).
Proof.
Admitted.
empty_st =[
X ::= 0;; Y ::= 1;; Z ::= 2
]=> (Z !> 2 ; Y !> 1 ; X !> 0).
Proof.
Admitted.
☐
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).
Exercise: 3 stars, standard, optional (pup_to_n)
Definition pup_to_n : com
. Admitted.
Theorem pup_to_2_ceval :
(X !> 2) =[
pup_to_n
]=> (X !> 0 ; Y !> 3 ; X !> 1 ; Y !> 2 ; Y !> 0 ; X !> 2).
Proof.
Admitted.
☐
Determinism of Evaluation
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.
 reflexivity.
 reflexivity.

assert (st' = st'0) as EQ1.
{ apply IHE1_1; assumption. }
subst st'0.
apply IHE1_2. assumption.

apply IHE1. assumption.

rewrite H in H5. discriminate H5.

rewrite H in H5. discriminate H5.

apply IHE1. assumption.

reflexivity.

rewrite H in H2. discriminate H2.

rewrite H in H4. discriminate H4.

assert (st' = st'0) as EQ1.
{ apply IHE1_1; assumption. }
subst st'0.
apply IHE1_2. assumption. Qed.
Reasoning About Imp Programs
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.
Exercise: 3 stars, standard, recommended (XtimesYinZ_spec)
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.
~(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).
Admitted.
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 :=
.
Theorem no_whiles_eqv:
forall c, no_whiles c = true <> no_whilesR c.
Proof.
Admitted.
☐
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.
Exercise: 4 stars, standard (no_whiles_terminating)
☐
Additional Exercises
Exercise: 3 stars, standard (stack_compiler)
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
. Admitted.
Example s_execute1 :
s_execute empty_st []
[SPush 5; SPush 3; SPush 1; SMinus]
= [2; 5].
Admitted.
Example s_execute2 :
s_execute (X !> 3) [3;4]
[SPush 4; SLoad X; SMult; SPlus]
= [15; 4].
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.
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].
Admitted.
☐
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.
Exercise: 4 stars, advanced (stack_compiler_correct)
Theorem s_compile_correct : forall (st : state) (e : aexp),
s_execute st [] (s_compile e) = [ aeval st e ].
Proof.
Admitted.
☐
Most modern programming languages use a "shortcircuit" 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 shortcircuit
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 shortcircuiting BAnd
would not be equivalent to the original, since it would make more
programs terminate.)
Exercise: 3 stars, standard, optional (short_circuit)
Exercise: 4 stars, advanced (break_imp)
Inductive com : Type :=
 CSkip
 CBreak
 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:
Based on the above description, complete the definition of the
ceval relation.
 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.
Inductive ceval : com > state > result > state > Prop :=
 E_Skip : forall st,
st =[ CSkip ]=> st / SContinue
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.
Admitted.
Theorem while_continue : forall b c st st' s,
st =[ WHILE b DO c END ]=> st' / s >
s = SContinue.
Proof.
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.
Admitted.
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.
Admitted.
st =[ WHILE b DO c END ]=> st' / SContinue >
beval st' b = true >
exists st'', st'' =[ c ]=> st' / SBreak.
Proof.
Admitted.
Theorem ceval_deterministic: forall (c:com) st st1 st2 s1 s2,
st =[ c ]=> st1 / s1 >
st =[ c ]=> st2 / s2 >
st1 = st2 /\ s1 = s2.
Proof.
Admitted.
st =[ c ]=> st1 / s1 >
st =[ c ]=> st2 / s2 >
st1 = st2 /\ s1 = s2.
Proof.
Admitted.
☐