# ImpSimple Imperative Programs

In this chapter, we take a more serious look at how to use Coq as a tool 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, in Programming Language Foundations (Software Foundations, volume 2), we develop a theory of program equivalence and introduce Hoare Logic, a popular logic for reasoning about imperative programs.

Set Warnings "-notation-overridden,-parsing,-deprecated-hint-without-locality".
From Coq Require Import Bool.Bool.
From Coq Require Import Init.Nat.
From Coq Require Import Arith.Arith.
From Coq Require Import Arith.EqNat. Import Nat.
From Coq Require Import Lia.
From Coq Require Import Lists.List. Import ListNotations.
From Coq Require Import Strings.String.
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 with variables, and finally a language of commands including assignment, conditionals, sequencing, and loops.

## Syntax

Module AExp.
Abstract syntax trees for 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)
| BNeq (a1 a2 : aexp)
| BLe (a1 a2 : aexp)
| BGt (a1 a2 : aexp)
| BNot (b : bexp)
| BAnd (b1 b2 : bexp).

## Evaluation

Evaluating an arithmetic expression produces a number.
Fixpoint aeval (a : aexp) : nat :=
match a with
| ANum nn
| 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
| BTruetrue
| BFalsefalse
| BEq a1 a2 ⇒ (aeval a1) =? (aeval a2)
| BNeq a1 a2negb ((aeval a1) =? (aeval a2))
| BLe a1 a2 ⇒ (aeval a1) <=? (aeval a2)
| BGt a1 a2negb ((aeval a1) <=? (aeval a2))
| BNot b1negb (beval b1)
| BAnd b1 b2andb (beval b1) (beval b2)
end.

What does the following expression evaluate to?
aeval (APlus (ANum 3) (AMinus (ANum 4) (ANum 1)))
(1) true
(2) false
(3) 0
(4) 3
(5) 6

## Optimization

Fixpoint optimize_0plus (a:aexp) : aexp :=
match a with
| ANum nANum n
| APlus (ANum 0) e2optimize_0plus e2
| APlus e1 e2APlus (optimize_0plus e1) (optimize_0plus e2)
| AMinus e1 e2AMinus (optimize_0plus e1) (optimize_0plus e2)
| AMult e1 e2AMult (optimize_0plus e1) (optimize_0plus e2)
end.

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.

Theorem optimize_0plus_sound: 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

That last proof was getting a little repetitive. Time to learn a few more Coq tricks...

## 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 : ae, aeval ae = aeval ae.
Proof.
try reflexivity. (* This just does reflexivity. *)
Qed.

Theorem silly2 : (P : Prop), PP.
Proof.
intros P HP.
try reflexivity. (* Plain reflexivity would have failed. *)
apply HP. (* We can still finish the proof in some other way. *)
Qed.

### 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:
Lemma foo : 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' : 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': 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.

### The repeat Tactical

The repeat tactical takes another tactic and keeps applying this tactic until it fails or until it succeeds but doesn't make any progress.
Here is an example proving 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.
repeat can loop forever.
Theorem repeat_loop : (m n : nat),
m + n = n + m.
Proof.
intros m n.
(* Uncomment the next line to see the infinite loop occur.  You will
then need to interrupt Coq to make it listen to you again.  (In
Proof General, C-c C-c does this.) *)

## Defining New Tactics

Coq also provides ways of "programming" tactic scripts:
• Ltac: scripting language for tactics (good for more sophisticated proof engineering)
• Tactic Notation for defining tactics with custom concrete syntax
• OCaml tactic scripting API (for wizards)
Ltac is all we need for present purposes.
For example:
Ltac invert H := inversion H; subst; clear H.

## The lia Tactic

The lia tactic implements a decision procedure for a subset of first-order logic called Presburger arithmetic.
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 >), and
• the logical connectives , , ¬, and ,
then invoking lia will either solve the goal or fail, meaning that the goal is actually false. (If the goal is not of this form, lia will fail.)
Example silly_presburger_example : m n o p,
m + nn + oo + 3 = p + 3 →
mp.
Proof.
intros. lia.
Qed.

m + n = n + m.
Proof.
intros. lia.
Qed.

Example add_assoc__lia : m n p,
m + (n + p) = m + n + p.
Proof.
intros. lia.
Qed.

## A Few More Handy Tactics

Finally, here are some miscellaneous tactics that you may find convenient.
• clear H: Delete hypothesis H from the context.
• subst x: Given 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, solve the goal.
• contradiction: Try to find a hypothesis H in the 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 Fixpoints. Another way to think about evaluation -- one that we will see is often more flexible -- is as a relation between expressions and their values. This leads naturally to Inductive definitions like the following...
Module aevalR_first_try.
Inductive aevalR : aexpnatProp :=
| E_ANum (n : nat) :
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).
A standard notation for "evaluates to":
Notation "e '==>' n"
:= (aevalR e n)
(at level 90, left associativity)
: type_scope.

End aevalR_first_try.
With infix notation:
Reserved Notation "e '==>' n" (at level 90, left associativity).

Inductive aevalR : aexpnatProp :=
| 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

For example, the constructor E_APlus...
| E_APlus : (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
There is nothing very deep going on here:
• a group of inference rules corresponds to a single Inductive definition
• each rule's name corresponds to a constructor name
• above the line are premises
• below the line is conclusion
• metavariables like e1 and n1 are implicitly universally quantified
For example, ==> is the smallest relation closed under these rules:
 (E_ANum) ANum n ==> n
 e1 ==> n1 e2 ==> n2 (E_APlus) APlus e1 e2 ==> n1+n2
 e1 ==> n1 e2 ==> n2 (E_AMinus) AMinus e1 e2 ==> n1-n2
 e1 ==> n1 e2 ==> n2 (E_AMult) AMult e1 e2 ==> n1*n2

## Equivalence of the Definitions

It is straightforward to prove that the relational and functional definitions of evaluation agree:
Theorem aeval_iff_aevalR : 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.

Again, we can make the proof quite a bit shorter using some tactics.
Theorem aeval_iff_aevalR' : a n,
(a ==> n) ↔ aeval a = n.
Proof.
(* WORK IN CLASS *) Admitted.

End AExp.

## Computational vs. Relational Definitions

Sometimes relational (rather than functional) definitions are the only reasonable option...
Module aevalR_division.

Inductive aexp : Type :=
| ANum (n : nat)
| APlus (a1 a2 : aexp)
| AMinus (a1 a2 : aexp)
| AMult (a1 a2 : aexp)
| ADiv (a1 a2 : aexp). (* <--- NEW *)
What should aeval return for a division by zero??

Reserved Notation "e '==>' n"
(at level 90, left associativity).

Inductive aevalR : aexpnatProp :=
| 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) : (* <----- NEW *)
(a1 ==> n1) → (a2 ==> n2) → (n2 > 0) →
(mult n2 n3 = n1) → (ADiv a1 a2) ==> n3

where "a '==>' n" := (aevalR a n) : type_scope.
Notice that the evaluation relation has become partial: There are some inputs for which it does not specify an output.
End aevalR_division.

Module aevalR_extended.

A nondeterministic number generator:
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).
What should aeval do with nondeterminism??

Inductive aevalR : aexpnatProp :=
| 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.

Which is better, functional or relational?
• Functional: take advantage of computation.
• Relational: more (easily) expressive.
• Best of both worlds: define both and prove them equivalent.

# Expressions With Variables

Let's return to defining Imp, where 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 use total maps from the Maps chapter.
A machine state (or just state) represents the current values of all variables at some point in the execution of a program.
Definition state := total_map nat.

## Syntax

We can add variables to the arithmetic expressions we had before simply by including 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".

The definition of bexps is unchanged (except that it now refers to the new aexps):
Inductive bexp : Type :=
| BTrue
| BFalse
| BEq (a1 a2 : aexp)
| BNeq (a1 a2 : aexp)
| BLe (a1 a2 : aexp)
| BGt (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 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.
• Declare Custom Entry com tells Coq to create a new "custom grammar" for parsing Imp expressions and programs. The first notation declaration after this tells Coq that anything between <{ and }> should be parsed using the Imp grammar. Again, it is not necessary to understand the details, but it is important to recognize that we are defining new interpretations for some familiar operators like +, -, ×, =, , etc., when they occur between <{ and }>.
Coercion AId : string >-> aexp.
Coercion ANum : nat >-> aexp.

Declare Custom Entry com.
Declare Scope com_scope.

Notation "<{ e }>" := e (at level 0, e custom com at level 99) : com_scope.
Notation "( x )" := x (in custom com, x at level 99) : com_scope.
Notation "x" := x (in custom com at level 0, x constr at level 0) : com_scope.
Notation "f x .. y" := (.. (f x) .. y)
(in custom com at level 0, only parsing,
f constr at level 0, x constr at level 9,
y constr at level 9) : com_scope.
Notation "x + y" := (APlus x y) (in custom com at level 50, left associativity).
Notation "x - y" := (AMinus x y) (in custom com at level 50, left associativity).
Notation "x * y" := (AMult x y) (in custom com at level 40, left associativity).
Notation "'true'" := true (at level 1).
Notation "'true'" := BTrue (in custom com at level 0).
Notation "'false'" := false (at level 1).
Notation "'false'" := BFalse (in custom com at level 0).
Notation "x <= y" := (BLe x y) (in custom com at level 70, no associativity).
Notation "x > y" := (BGt x y) (in custom com at level 70, no associativity).
Notation "x = y" := (BEq x y) (in custom com at level 70, no associativity).
Notation "x <> y" := (BNeq x y) (in custom com at level 70, no associativity).
Notation "x && y" := (BAnd x y) (in custom com at level 80, left associativity).
Notation "'~' b" := (BNot b) (in custom com at level 75, right associativity).

Open Scope com_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 : aexp := <{ 3 + (X × 2) }>.
Definition example_bexp : bexp := <{ true && ~(X ≤ 4) }>.

## Evaluation

Now we need to add an st parameter to both evaluation functions:
Fixpoint aeval (st : state) (* <--- NEW *) (a : aexp) : nat :=
match a with
| ANum nn
| AId xst x (* <--- NEW *)
| <{a1 + a2}> ⇒ (aeval st a1) + (aeval st a2)
| <{a1 - a2}> ⇒ (aeval st a1) - (aeval st a2)
| <{a1 × a2}> ⇒ (aeval st a1) × (aeval st a2)
end.

Fixpoint beval (st : state) (* <--- NEW *) (b : bexp) : bool :=
match b with
| <{true}> ⇒ true
| <{false}> ⇒ false
| <{a1 = a2}> ⇒ (aeval st a1) =? (aeval st a2)
| <{a1a2}> ⇒ negb ((aeval st a1) =? (aeval st a2))
| <{a1a2}> ⇒ (aeval st a1) <=? (aeval st a2)
| <{a1 > a2}> ⇒ negb ((aeval st a1) <=? (aeval st a2))
| <{~ b1}> ⇒ negb (beval st b1)
| <{b1 && b2}> ⇒ andb (beval st b1) (beval st b2)
end.
We can use our notation for total maps in the specific case of states -- i.e., we write the empty state as (_ !-> 0).
Definition empty_st := (_ !-> 0).
Also, we can add a notation for a "singleton state" with just one variable bound to a value.
Notation "x '!->' v" := (x !-> v ; empty_st) (at level 100).

# Commands

Now we are ready define the syntax and behavior of Imp commands (or statements).

## Syntax

Informally, commands c are described by the following BNF grammar.
c := skip | x := a | c ; c | if b then c else c end
| while b do c end
Inductive com : Type :=
| CSkip
| CAsgn (x : string) (a : aexp)
| CSeq (c1 c2 : com)
| CIf (b : bexp) (c1 c2 : com)
| CWhile (b : bexp) (c : com).
As we did for expressions, we can use a few Notation declarations to make reading and writing Imp programs more convenient.
Notation "'skip'" :=
CSkip (in custom com at level 0) : com_scope.
Notation "x := y" :=
(CAsgn x y)
(in custom com at level 0, x constr at level 0,
y at level 85, no associativity) : com_scope.
Notation "x ; y" :=
(CSeq x y)
(in custom com at level 90, right associativity) : com_scope.
Notation "'if' x 'then' y 'else' z 'end'" :=
(CIf x y z)
(in custom com at level 89, x at level 99,
y at level 99, z at level 99) : com_scope.
Notation "'while' x 'do' y 'end'" :=
(CWhile x y)
(in custom com at level 89, x at level 99, y at level 99) : com_scope.
For example, here is the factorial function again, written as a formal Coq definition. When this command terminates, the variable Y will contain the factorial of the initial value of X.
Definition fact_in_coq : com :=
<{ Z := X;
Y := 1;
while Z ≠ 0 do
Y := Y × Z;
Z := Z - 1
end }>.

Print fact_in_coq.

## 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 }>.

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.
Fixpoint ceval_fun_no_while (st : state) (c : com) : state :=
match c with
| <{ skip }> ⇒
st
| <{ x := a }> ⇒
(x !-> (aeval st a) ; st)
| <{ c1 ; c2 }> ⇒
let st' := ceval_fun_no_while st c1 in
ceval_fun_no_while st' c2
| <{ if b then c1 else c2 end}> ⇒
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.

Consider the following "proof object":
```        Fixpoint loop_false (n : nat) : False :=
loop_false n.
```
Accepting such a definition would be catastrophic, so Coq conservatively rejects all nonterminating programs.

## Evaluation as a Relation

Here's a better way: define ceval as a relation rather than a function -- i.e., make its result a Prop rather than a state, similar to what we did for aevalR above.
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 a = n (E_Asgn) st =[ x := a ]=> (x !-> n ; st)
 st =[ c1 ]=> st' st' =[ c2 ]=> st'' (E_Seq) st =[ c1;c2 ]=> st''
 beval st b = true st =[ c1 ]=> st' (E_IfTrue) st =[ if b then c1 else c2 end ]=> st'
 beval st b = false st =[ c2 ]=> st' (E_IfFalse) st =[ if b then c1 else c2 end ]=> 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''
Reserved Notation
"st '=[' c ']=>' st'"
(at level 40, c custom com at level 99,
st constr, st' constr at next level).

Inductive ceval : comstatestateProp :=
| E_Skip : st,
st =[ skip ]=> st
| E_Asgn : st a n x,
aeval st a = n
st =[ x := a ]=> (x !-> n ; st)
| E_Seq : c1 c2 st st' st'',
st =[ c1 ]=> st'
st' =[ c2 ]=> st''
st =[ c1 ; c2 ]=> st''
| E_IfTrue : st st' b c1 c2,
beval st b = true
st =[ c1 ]=> st'
st =[ if b then c1 else c2 end]=> st'
| E_IfFalse : st st' b c1 c2,
beval st b = false
st =[ c2 ]=> st'
st =[ if b then c1 else c2 end]=> st'
| E_WhileFalse : b st c,
beval st b = false
st =[ while b do c end ]=> st
| E_WhileTrue : 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;
if (X ≤ 1)
then Y := 3
else Z := 4
end
]=> (Z !-> 4 ; X !-> 2).
Proof.
(* We must supply the intermediate state *)
apply E_Seq with (X !-> 2).
- (* assignment command *)
apply E_Asgn. reflexivity.
- (* if command *)
apply E_IfFalse.
reflexivity.
apply E_Asgn. reflexivity.
Qed.

Is the following proposition provable?
(c : com) (st st' : state),
st =[ skip ; c ]=> st'
st =[ c ]=> st'
(1) Yes
(2) No
(3) Not sure
Is the following proposition provable?
(c1 c2 : com) (st st' : state),
st =[ c1 ; c2 ]=> st'
st =[ c1 ]=> st
st =[ c2 ]=> st'
(1) Yes
(2) No
(3) Not sure
Is the following proposition provable?
(b : bexp) (c : com) (st st' : state),
st =[ if b then c else c end ]=> st'
st =[ c ]=> st'
(1) Yes
(2) No
(3) Not sure
Is the following proposition provable?
b : bexp,
( st, beval st b = true) →
(c : com) (st : state),
~( st', st =[ while b do c end ]=> st')
(1) Yes
(2) No
(3) Not sure
Is the following proposition provable?
(b : bexp) (c : com) (st : state),
~( st', st =[ while b do c end ]=> st') →
st'', beval st'' b = true
(1) Yes
(2) No
(3) Not sure

## Determinism of Evaluation

Theorem ceval_deterministic: 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_Asgn *) reflexivity.
- (* E_Seq *)
rewrite (IHE1_1 st'0 H1) in ×.
apply IHE1_2. assumption.
- (* E_IfTrue, b evaluates to true *)
apply IHE1. assumption.
- (* E_IfTrue,  b evaluates to false (contradiction) *)
rewrite H in H5. discriminate.
- (* E_IfFalse, b evaluates to true (contradiction) *)
rewrite H in H5. discriminate.
- (* E_IfFalse, b evaluates to false *)
apply IHE1. assumption.
- (* E_WhileFalse, b evaluates to false *)
reflexivity.
- (* E_WhileFalse, b evaluates to true (contradiction) *)
rewrite H in H2. discriminate.
- (* E_WhileTrue, b evaluates to false (contradiction) *)
rewrite H in H4. discriminate.
- (* E_WhileTrue, b evaluates to true *)
rewrite (IHE1_1 st'0 H3) in ×.
apply IHE1_2. assumption. Qed.

#### 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 aexps 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 the first number from the second.
• SMult: Similar, but multiply.
Inductive sinstr : Type :=
| SPush (n : nat)
| 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 it is unspecified what to do when encountering an SPlus, SMinus, or SMult instruction if the stack contains fewer than two elements. In a sense, it is immaterial what we do, since a correct compiler will never emit such a malformed program. But for sake of later exercises, it would be best to skip the offending instruction and continue with the next one.
Fixpoint s_execute (st : state) (stack : list nat)
(prog : list sinstr)
: list nat
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Check s_execute.

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) }>
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, standard (execute_app)

Execution can be decomposed in the following sense: executing stack program p1 ++ p2 is the same as executing p1, taking the resulting stack, and executing p2 from that stack. Prove that fact.
Theorem execute_app : st p1 p2 stack,
s_execute st stack (p1 ++ p2) = s_execute st (s_execute st stack p1) p2.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, standard (stack_compiler_correct)

Now we'll prove the correctness of the compiler implemented in the previous exercise. Begin by proving the following lemma. If it becomes difficult, consider whether your implementation of s_execute or s_compile could be simplified.
Lemma s_compile_correct_aux : st e stack,
s_execute st stack (s_compile e) = aeval st e :: stack.
Proof.
(* FILL IN HERE *) Admitted.
The main theorem should be a very easy corollary of that lemma.
Theorem s_compile_correct : (st : state) (e : aexp),
s_execute st [] (s_compile e) = [ aeval st e ].
Proof.
(* FILL IN HERE *) Admitted.