# Software Technology: Simple Imperative Programs

Acknowledgements

This material is taken verbatim from Software Foundations V4.0

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Assumed Knowledge: * Can reason with inductively designed propositions in Coq.
Learning Outcomes: * TODO
Grab the Coq source file Imp.v
In this chapter, we begin a new direction that will continue for the rest of the course. Up to now most of our attention has been focused on various aspects of Coq itself, while from now on we'll mostly be using Coq to formalize other things. (We'll continue to pause from time to time to introduce a few additional aspects of Coq.)
Our first case study is a simple imperative programming language called Imp, embodying a tiny core fragment of conventional mainstream languages such as C and Java. Here is a familiar mathematical function written in Imp.
Z ::= X;;
Y ::= 1;;
WHILE not (Z = 0) DO
Y ::= Y * Z;;
Z ::= Z - 1
END
This chapter looks at how to define the syntax and semantics of Imp; the chapters that follow develop a theory of program equivalence and introduce Hoare Logic, a widely used logic for reasoning about imperative programs.

Require Import Coq.Bool.Bool.
Require Import Coq.Arith.Arith.
Require Import Coq.Arith.EqNat.
Require Import Coq.omega.Omega.
Require Import Coq.Lists.List.
Import ListNotations.
Require Import Maps.

# Arithmetic and Boolean Expressions

We'll present Imp in three parts: first a core language of arithmetic and boolean expressions, then an extension of these expressions with variables, and finally a language of commands including assignment, conditions, sequencing, and loops.

## Syntax

Module AExp.

These two definitions specify the abstract syntax of arithmetic and boolean expressions.

Inductive aexp : Type :=
| ANum : nat aexp
| APlus : aexp aexp aexp
| AMinus : aexp aexp aexp
| AMult : aexp aexp aexp.

Inductive bexp : Type :=
| BTrue : bexp
| BFalse : bexp
| BEq : aexp aexp bexp
| BLe : aexp aexp bexp
| BNot : bexp bexp
| BAnd : bexp bexp bexp.

In this chapter, we'll elide the translation from the concrete syntax that a programmer would actually write to these abstract syntax trees — the process that, for example, would translate the string "1+2*3" to the AST APlus (ANum 1) (AMult (ANum 2) (ANum 3)). The optional chapter ImpParser develops a simple implementation of a lexical analyzer and parser that can perform this translation. You do not need to understand that chapter to understand this one, but if you haven't taken a course where these techniques are covered (e.g., a compilers course) you may want to skim it.
For comparison, here's a conventional BNF (Backus-Naur Form) grammar defining the same abstract syntax:
a ::= nat
| a + a
| a - a
| a * a

b ::= true
| false
| a = a
| a ≤ a
| not b
| b and b
Compared to the Coq version above...
• The BNF is more informal — for example, it gives some suggestions about the surface syntax of expressions (like the fact that the addition operation is written + and is an infix symbol) while leaving other aspects of lexical analysis and parsing (like the relative precedence of +, -, and *, the use of parens to explicitly group subexpressions, etc.) unspecified. Some additional information (and human intelligence) would be required to turn this description into a formal definition, for example when implementing a compiler.
The Coq version consistently omits all this information and concentrates on the abstract syntax only.
• On the other hand, the BNF version is lighter and easier to read. Its informality makes it flexible, which is a huge advantage in situations like discussions at the blackboard, where conveying general ideas is more important than getting every detail nailed down precisely.
Indeed, there are dozens of BNF-like notations and people switch freely among them, usually without bothering to say which form of BNF they're using because there is no need to: a rough-and-ready informal understanding is all that's 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 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 a2beq_nat (aeval a1) (aeval a2)
| BLe a1 a2leb (aeval a1) (aeval a2)
| BNot b1negb (beval b1)
| BAnd b1 b2andb (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: a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a. induction a.
- (* ANum *) reflexivity.
- (* APlus *) destruct a1.
+ (* a1 = ANum n *) destruct n.
* (* n = 0 *) simpl. apply IHa2.
* (* n <> 0 *) simpl. rewrite IHa2. reflexivity.
+ (* a1 = APlus a1_1 a1_2 *)
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
+ (* a1 = AMinus a1_1 a1_2 *)
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
+ (* a1 = AMult a1_1 a1_2 *)
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
- (* AMinus *)
simpl. rewrite IHa1. rewrite IHa2. reflexivity.
- (* AMult *)
simpl. rewrite IHa1. rewrite IHa2. reflexivity. Qed.

# Coq Automation

The repetition in this last proof is starting to be a little annoying. If either the language of arithmetic expressions or the optimization being proved sound were significantly more complex, it would begin to be a real problem.
So far, we've been doing all our proofs using just a small handful of Coq's tactics and completely ignoring its powerful facilities for constructing parts of proofs automatically. This section introduces some of these facilities, and we will see more over the next several chapters. Getting used to them will take some energy — Coq's automation is a power tool — but it will allow us to scale up our efforts to more complex definitions and more interesting properties without becoming overwhelmed by boring, repetitive, low-level details.

## Tacticals

Tacticals is Coq's term for tactics that take other tactics as arguments — "higher-order tactics," if you will.

### The try Tactical

If T is a tactic, then try T is a tactic that is just like T except that, if T fails, try T successfully does nothing at all (instead of failing).

Theorem silly1 : ae, aeval ae = aeval ae.
Proof. try reflexivity. (* this just does reflexivity *) Qed.

Theorem silly2 : (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 we'll see below that it is very useful for doing automated proofs in conjunction with the ; tactical.

### 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 : n, leb 0 n = true.
Proof.
intros.
destruct n.
(* Leaves two subgoals, which are discharged
identically...  *)

- (* n=0 *) simpl. reflexivity.
- (* n=Sn' *) simpl. reflexivity.
Qed.

We can simplify this proof using the ; tactical:

Lemma foo' : n, leb 0 n = true.
Proof.
intros.
(* destruct the current goal *)
destruct n;
(* then simpl each resulting subgoal *)
simpl;
(* and do reflexivity on each resulting subgoal *)
reflexivity.
Qed.

Using try and ; together, we can get rid of the repetition in the proof that was bothering us a little while ago.

Theorem optimize_0plus_sound': a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a.
induction a;
(* Most cases follow directly by the IH... *)
try (simpl; rewrite IHa1; rewrite IHa2; reflexivity).
(* ... but the remaining cases -- ANum and APlus --
are different: *)

- (* ANum *) reflexivity.
- (* APlus *)
destruct a1;
(* Again, most cases follow directly by the IH: *)
try (simpl; simpl in IHa1; rewrite IHa1;
rewrite IHa2; reflexivity).
(* The interesting case, on which the try...
does nothing, is when e1 = ANum n. In this
case, we have to destruct n (to see whether
the optimization applies) and rewrite with the
induction hypothesis. *)

+ (* a1 = ANum n *) destruct n;
simpl; rewrite IHa2; reflexivity. Qed.

Coq experts often use this "...; try... " idiom after a tactic like induction to take care of many similar cases all at once. Naturally, this practice has an analog in informal proofs.
Here is an informal proof of this theorem that matches the structure of the formal one:
Theorem: For all arithmetic expressions a,
aeval (optimize_0plus a) = aeval a.
Proof: By induction on a. Most cases follow directly from the IH. The remaining cases are as follows:
• Suppose a = ANum n for some n. We must show
aeval (optimize_0plus (ANum n)) = aeval (ANum n).
This is immediate from the definition of optimize_0plus.
• Suppose a = APlus a1 a2 for some a1 and a2. We must show
aeval (optimize_0plus (APlus a1 a2))
aeval (APlus a1 a2).
Consider the possible forms of a1. For most of them, optimize_0plus simply calls itself recursively for the subexpressions and rebuilds a new expression of the same form as a1; in these cases, the result follows directly from the IH.
The interesting case is when a1 = ANum n for some n. If n = ANum 0, then
optimize_0plus (APlus a1 a2) = optimize_0plus a2
and the IH for a2 is exactly what we need. On the other hand, if n = S n' for some n', then again optimize_0plus simply calls itself recursively, and the result follows from the IH.
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'': 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.
(* Print In10. *)

The repeat T tactic never fails: if the tactic T doesn't apply to the original goal, then repeat still succeeds without changing the original goal (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 repeat T tactic also does not have any upper bound on the number of times it applies T. If T is a tactic that always succeeds, then repeat T will loop forever (e.g., repeat simpl loops forever, since simpl always succeeds). While Coq's term language is guaranteed to terminate, Coq's tactic language is not!

#### Exercise: 3 stars (optimize_0plus_b)

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 which performs that transformation on bexps, 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 : b,
beval (optimize_0plus_b b) = beval b.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 4 stars, optional (optimizer)

Design exercise: The optimization implemented by our optimize_0plus function is only one of many possible optimizations on arithmetic and boolean expressions. Write a more sophisticated optimizer and prove it correct.
(* 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 small built-in programming language called Ltac with primitives that can examine and modify the proof state. The details are a bit too complicated to get into here (and it is generally agreed that Ltac is not the most beautiful part of Coq's design!), but they can be found in the reference manual and 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. For example, writing "simpl_and_try reflexivity." in a proof would be the same as writing "simpl; try reflexivity."

## The omega Tactic

The omega tactic implements a decision procedure for a subset of first-order logic called Presburger arithmetic. It is based on the Omega algorithm invented in 1991 by William Pugh [Pugh 1991].
If the goal is a universally quantified formula made out of
• numeric constants, addition (+ and S), subtraction (- and pred), and multiplication by constants (this is what makes it Presburger arithmetic),
• equality (= and ) and inequality (), and
• the logical connectives , , ¬, and ,
then invoking omega will either solve the goal or tell you that it is actually false.

Require Import Coq.omega.Omega.

Example silly_presburger_example : m n o p,
m + nn + o o + 3 = p + 3
mp.
Proof.
intros. omega.
Qed.

Leibniz wrote, "It is unworthy of excellent men to lose hours like slaves in the labor of calculation which could be relegated to anyone else if machines were used." We recommend that excellent people of all genders use the omega tactic whenever possible.

## A Few More Handy Tactics

Finally, here are some miscellaneous tactics that you may find convenient.
• clear H: Delete hypothesis H from the context.
• subst x: Find an assumption x = e or e = x in the context, replace x with e throughout the context and current goal, and clear the assumption.
• subst: Substitute away all assumptions of the form x = e or e = x.
• rename... into...: Change the name of a hypothesis in the proof context. For example, if the context includes a variable named x, then rename x into y will change all occurrences of x to y.
• assumption: Try to find a hypothesis H in the context that exactly matches the goal; if one is found, behave just like apply H.
• contradiction: Try to find a hypothesis H in the current context that is logically equivalent to False. If one is found, solve the goal.
• constructor: Try to find a constructor c (from some Inductive definition in the current environment) that can be applied to solve the current goal. If one is found, behave like apply c.
We'll see many examples of these in the proofs below.

# Evaluation as a Relation

We have presented aeval and beval as functions defined by 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 one for arithmetic expressions...

Module aevalR_first_try.

Inductive aevalR : aexp nat Prop :=
| 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).

As is often the case with relations, we'll find it convenient to define infix notation for aevalR. We'll write e n to mean that arithmetic expression e evaluates to value n. (This notation is one place where the limitation to ASCII symbols becomes a little bothersome. The standard notation for the evaluation relation is a double down-arrow. We'll typeset it like this in the HTML version of the notes and use a double slash as the closest approximation in .v files.)

Notation "e '' n"
:= (aevalR e n)
(at level 50, left associativity)
: type_scope.

End aevalR_first_try.

In fact, Coq provides a way to use this notation in the definition of aevalR itself. This avoids situations where we're working on a proof involving statements in the form e n but we have to refer back to a definition written using the form aevalR e n.
We do this by first "reserving" the notation, then giving the definition together with a declaration of what the notation means.

Reserved Notation "e '' n" (at level 50, left associativity).

Inductive aevalR : aexp nat Prop :=
| E_ANum : (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 Prop chapter).
For example, the constructor E_APlus...
| E_APlus : (e1 e2aexp) (n1 n2nat),
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 and the line itself as . All the variables mentioned in the rule (e1, n1, etc.) are implicitly bound by universal quantifiers at the beginning. (Such variables are often called metavariables to distinguish them from the variables of the language we are defining. At the moment, our arithmetic expressions don't include variables, but we'll soon be adding them.) The whole collection of rules is understood as being wrapped in an Inductive declaration. In informal prose, this is either elided or else indicated by saying something like "Let aevalR be the smallest relation closed under the following rules...".
For example, is the smallest relation closed under these rules:
 (E_ANum) ANum n ⇓ n
 e1 ⇓ n1 e2 ⇓ n2 (E_APlus) APlus e1 e2 ⇓ n1+n2
 e1 ⇓ n1 e2 ⇓ n2 (E_AMinus) AMinus e1 e2 ⇓ n1-n2
 e1 ⇓ n1 e2 ⇓ n2 (E_AMult) AMult e1 e2 ⇓ n1*n2

## Equivalence of the Definitions

It is straightforward to prove that the relational and functional definitions of evaluation agree, for all arithmetic expressions...

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.

We can make the proof quite a bit shorter by making more use of tacticals...

Theorem aeval_iff_aevalR' : a n,
(a n) aeval a = n.
Proof.
(* WORKED IN CLASS *)
split.
- (* -> *)
intros H; induction H; subst; reflexivity.
- (* <- *)
generalize dependent n.
induction a; simpl; intros; subst; constructor;
try apply IHa1; try apply IHa2; reflexivity.
Qed.

#### Exercise: 3 stars (bevalR)

Write a relation bevalR in the same style as aevalR, and prove that it is equivalent to beval.

(*
Inductive bevalR:
(* FILL IN HERE *)
*)

End AExp.

## Computational vs. Relational Definitions

For the definitions of evaluation for arithmetic and boolean expressions, the choice of whether to use functional or relational definitions is mainly a matter of taste. In general, Coq has somewhat better support for working with relations. On the other hand, in some sense function definitions carry more information, because functions are by definition deterministic and defined on all arguments; for a relation we have to show these properties explicitly if we need them. Functions also take advantage of Coq's computation mechanism.
However, there are circumstances where relational definitions of evaluation are preferable to functional ones.

Module aevalR_division.

For example, suppose that we wanted to extend the arithmetic operations by considering also a division operation:

Inductive aexp : Type :=
| ANum : nat aexp
| APlus : aexp aexp aexp
| AMinus : aexp aexp aexp
| AMult : aexp aexp aexp
| ADiv : aexp aexp aexp. (* <--- new *)

Extending the definition of aeval to handle this new operation would not be straightforward (what should we return as the result of ADiv (ANum 5) (ANum 0)?). But extending aevalR is straightforward.

Reserved Notation "e '' n"
(at level 50, left associativity).

Inductive aevalR : aexp nat Prop :=
| E_ANum : (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.

Suppose, instead, that we want to extend the arithmetic operations by a nondeterministic number generator any that, when evaluated, may yield any number. (Note that this is not the same as making a probabilistic choice among all possible numbers — we're not specifying any particular distribution of results, but just saying what results are possible.)

Reserved Notation "e '' n" (at level 50, left associativity).

Inductive aexp : Type :=
| AAny : aexp (* <--- NEW *)
| ANum : nat aexp
| APlus : aexp aexp aexp
| AMinus : aexp aexp aexp
| AMult : aexp aexp aexp.

Again, extending aeval would be tricky, since now evaluation is not a deterministic function from expressions to numbers, but extending aevalR is no problem:

Inductive aevalR : aexp nat Prop :=
| E_Any : (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.

# Expressions With Variables

Let's turn our attention back to defining Imp. The next thing we need to do is to enrich our arithmetic and boolean expressions with variables. To keep things simple, we'll assume that all variables are global and that they only hold numbers.

## States

Since we'll want to look variables up to find out their current values, we'll reuse the type id from the Maps chapter for the type of variables in Imp.
A machine state (or just state) represents the current values of all variables at some point in the execution of a program.
For simplicity, we assume that the state is defined for all variables, even though any given program is only going to mention a finite number of them. The state captures all of the information stored in memory. For Imp programs, because each variable stores a natural number, we can represent the state as a mapping from identifiers to nat. For more complex programming languages, the state might have more structure.

Definition state := total_map nat.

Definition empty_state : state :=
t_empty 0.

## Syntax

We can add variables to the arithmetic expressions we had before by simply adding one more constructor:

Inductive aexp : Type :=
| ANum : nat aexp
| AId : id aexp (* <----- NEW *)
| APlus : aexp aexp aexp
| AMinus : aexp aexp aexp
| AMult : aexp aexp aexp.

Defining a few variable names as notational shorthands will make examples easier to read:

Definition X : id := Id 0.
Definition Y : id := Id 1.
Definition Z : id := Id 2.

(This convention for naming program variables (X, Y, Z) clashes a bit with our earlier use of uppercase letters for types. Since we're not using polymorphism heavily in this part of the course, this overloading should not cause confusion.)
The definition of bexps is unchanged (except for using the new aexps):

Inductive bexp : Type :=
| BTrue : bexp
| BFalse : bexp
| BEq : aexp aexp bexp
| BLe : aexp aexp bexp
| BNot : bexp bexp
| BAnd : bexp bexp bexp.

## Evaluation

The arith and boolean evaluators are extended to handle variables in the obvious way, taking a state as an extra argument:

Fixpoint aeval (st : state) (a : aexp) : nat :=
match a with
| ANum nn
| AId xst 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
| BTruetrue
| BFalsefalse
| BEq a1 a2beq_nat (aeval st a1) (aeval st a2)
| BLe a1 a2leb (aeval st a1) (aeval st a2)
| BNot b1negb (beval st b1)
| BAnd b1 b2andb (beval st b1) (beval st b2)
end.

Example aexp1 :
aeval (t_update empty_state X 5)
(APlus (ANum 3) (AMult (AId X) (ANum 2)))
= 13.
Proof. reflexivity. Qed.

Example bexp1 :
beval (t_update empty_state X 5)
(BAnd BTrue (BNot (BLe (AId X) (ANum 4))))
= true.
Proof. reflexivity. Qed.

# Commands

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

## Syntax

Informally, commands c are described by the following BNF grammar. (We choose this slightly awkward concrete syntax for the sake of being able to define Imp syntax using Coq's Notation mechanism. In particular, we use IFB to avoid conflicting with the if notation from the standard library.)
c ::= SKIP | x ::= a | c ;; c | IFB b THEN c ELSE c FI
| WHILE b DO c END
For example, here's factorial in Imp:
Z ::= X;;
Y ::= 1;;
WHILE not (Z = 0) DO
Y ::= Y * Z;;
Z ::= Z - 1
END
When this command terminates, the variable Y will contain the factorial of the initial value of X.
Here is the formal definition of the abstract syntax of commands:

Inductive com : Type :=
| CSkip : com
| CAss : id aexp com
| CSeq : com com com
| CIf : bexp com com com
| CWhile : bexp com com.

As usual, we can use a few Notation declarations to make things more readable. To avoid conflicts with Coq's built-in notations, we keep this light — in particular, we don't introduce any notations for aexps and bexps to avoid confusion with the numeric and boolean operators we've already defined.

Notation "'SKIP'" :=
CSkip.
Notation "x '::=' a" :=
(CAss x a) (at level 60).
Notation "c1 ;; c2" :=
(CSeq c1 c2) (at level 80, right associativity).
Notation "'WHILE' b 'DO' c 'END'" :=
(CWhile b c) (at level 80, right associativity).
Notation "'IFB' c1 'THEN' c2 'ELSE' c3 'FI'" :=
(CIf c1 c2 c3) (at level 80, right associativity).

For example, here is the factorial function again, written as a formal definition to Coq:

Definition fact_in_coq : com :=
Z ::= AId X;;
Y ::= ANum 1;;
WHILE BNot (BEq (AId Z) (ANum 0)) DO
Y ::= AMult (AId Y) (AId Z);;
Z ::= AMinus (AId Z) (ANum 1)
END.

## Examples

Assignment:

Definition plus2 : com :=
X ::= (APlus (AId X) (ANum 2)).

Definition XtimesYinZ : com :=
Z ::= (AMult (AId X) (AId Y)).

Definition subtract_slowly_body : com :=
Z ::= AMinus (AId Z) (ANum 1) ;;
X ::= AMinus (AId X) (ANum 1).

### Loops

Definition subtract_slowly : com :=
WHILE BNot (BEq (AId X) (ANum 0)) DO
subtract_slowly_body
END.

Definition subtract_3_from_5_slowly : com :=
X ::= ANum 3 ;;
Z ::= ANum 5 ;;
subtract_slowly.

### An infinite loop:

Definition loop : com :=
WHILE BTrue DO
SKIP
END.

# Evaluation

Next we need to define what it means to evaluate an Imp command. The fact that WHILE loops don't necessarily terminate makes defining an evaluation function tricky...

## Evaluation as a Function (Failed Attempt)

Here's an attempt at defining an evaluation function for commands, omitting the WHILE case.

Fixpoint ceval_fun_no_while (st : state) (c : com)
: state :=
match c with
| SKIP
st
| x ::= a1
t_update st x (aeval st a1)
| c1 ;; c2
let st' := ceval_fun_no_while st c1 in
ceval_fun_no_while st' c2
| IFB b THEN c1 ELSE c2 FI
if (beval st b)
then ceval_fun_no_while st c1
else ceval_fun_no_while st c2
| WHILE b DO c END
st (* bogus *)
end.

In a traditional functional programming language like 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!) Coq program showing what would go wrong if Coq allowed non-terminating recursive functions:
         Fixpoint loop_false (n : nat) : False := loop_false n.

That is, propositions like False would become provable (e.g., loop_false 0 would be a proof of False), which would be a disaster for Coq's logical consistency.
Thus, because it doesn't terminate on all inputs, the full version of ceval_fun cannot be written in Coq — at least not without additional tricks (see chapter ImpCEvalFun if 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 the awkward workarounds that would be needed to define evaluation as a function, it gives us a lot more flexibility in the definition. For example, if we add nondeterministic features like any to the language, we want the definition of evaluation to be nondeterministic — i.e., not only will it not be total, it will not even be a function! We'll use the notation c / st st' for our ceval relation: c / st st' means that executing program c in a starting state st results in an ending state st'. This can be pronounced "c takes state st to st'".

### Operational Semantics

Here is an informal definition of evaluation, presented as inference rules for the sake of readability:
 (E_Skip) SKIP / st ⇓ st
 aeval st a1 = n (E_Ass) x := a1 / st ⇓ (t_update st x n)
 c1 / st ⇓ st' c2 / st' ⇓ st'' (E_Seq) c1;;c2 / st ⇓ st''
 beval st b1 = true c1 / st ⇓ st' (E_IfTrue) IF b1 THEN c1 ELSE c2 FI / st ⇓ st'
 beval st b1 = false c2 / st ⇓ st' (E_IfFalse) IF b1 THEN c1 ELSE c2 FI / st ⇓ st'
 beval st b = false (E_WhileEnd) WHILE b DO c END / st ⇓ st
 beval st b = true c / st ⇓ st' WHILE b DO c END / st' ⇓ st'' (E_WhileLoop) WHILE b DO c END / st ⇓ st''
Here is the formal definition. Make sure you understand how it corresponds to the inference rules.

Reserved Notation "c1 '/' st '' st'"
(at level 40, st at level 39).

Inductive ceval : com state state Prop :=
| E_Skip : st,
SKIP / st st
| E_Ass : st a1 n x,
aeval st a1 = n
(x ::= a1) / st (t_update st x n)
| E_Seq : c1 c2 st st' st'',
c1 / st st'
c2 / st' st''
(c1 ;; c2) / st st''
| E_IfTrue : st st' b c1 c2,
beval st b = true
c1 / st st'
(IFB b THEN c1 ELSE c2 FI) / st st'
| E_IfFalse : st st' b c1 c2,
beval st b = false
c2 / st st'
(IFB b THEN c1 ELSE c2 FI) / st st'
| E_WhileEnd : b st c,
beval st b = false
(WHILE b DO c END) / st st
| E_WhileLoop : st st' st'' b c,
beval st b = true
c / st st'
(WHILE b DO c END) / st' st''
(WHILE b DO c END) / st st''

where "c1 '/' st '' st'" := (ceval c1 st st').

The cost of defining evaluation as a relation instead of a function is that we now need to construct proofs that some program evaluates to some result state, rather than just letting Coq's computation mechanism do it for us.

Example ceval_example1:
(X ::= ANum 2;;
IFB BLe (AId X) (ANum 1)
THEN Y ::= ANum 3
ELSE Z ::= ANum 4
FI)
/ empty_state
(t_update (t_update empty_state X 2) Z 4).
Proof.
(* We must supply the intermediate state *)
apply E_Seq with (t_update empty_state X 2).
- (* assignment command *)
apply E_Ass. reflexivity.
- (* if command *)
apply E_IfFalse.
reflexivity.
apply E_Ass. reflexivity. Qed.

#### Exercise: 2 stars (ceval_example2)

Example ceval_example2:
(X ::= ANum 0;; Y ::= ANum 1;; Z ::= ANum 2) / empty_state
(t_update (t_update (t_update empty_state X 0) Y 1) Z 2).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, advanced (pup_to_n)

Write an Imp program that sums the numbers from 1 to X (inclusive: 1 + 2 + ... + X) in the variable Y. Prove that this program executes as intended for X = 2 (the latter is trickier than you might expect).

Definition pup_to_n : com
(* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Theorem pup_to_2_ceval :
pup_to_n / (t_update empty_state X 2)
t_update (t_update (t_update (t_update (t_update (t_update empty_state
X 2) Y 0) Y 2) X 1) Y 3) X 0.
Proof.
(* FILL IN HERE *) Admitted.

## Determinism of Evaluation

Changing from a computational to a relational definition of evaluation is a good move because it allows us to escape 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: c st st1 st2,
c / st st1
c / st st2
st1 = st2.
Proof.
intros c st st1 st2 E1 E2.
generalize dependent st2.
induction E1;
intros st2 E2; inversion E2; subst.
- (* E_Skip *) reflexivity.
- (* E_Ass *) reflexivity.
- (* E_Seq *)
assert (st' = st'0) as EQ1.
{ (* Proof of assertion *) apply IHE1_1; assumption. }
subst st'0.
apply IHE1_2. assumption.
- (* E_IfTrue, b1 evaluates to true *)
apply IHE1. assumption.
- (* E_IfTrue,  b1 evaluates to false (contradiction) *)
rewrite H in H5. inversion H5.
- (* E_IfFalse, b1 evaluates to true (contradiction) *)
rewrite H in H5. inversion H5.
- (* E_IfFalse, b1 evaluates to false *)
apply IHE1. assumption.
- (* E_WhileEnd, b1 evaluates to false *)
reflexivity.
- (* E_WhileEnd, b1 evaluates to true (contradiction) *)
rewrite H in H2. inversion H2.
- (* E_WhileLoop, b1 evaluates to false (contradiction) *)
rewrite H in H4. inversion H4.
- (* E_WhileLoop, b1 evaluates to true *)
assert (st' = st'0) as EQ1.
{ (* Proof of assertion *) apply IHE1_1; assumption. }
subst st'0.
apply IHE1_2. assumption. Qed.

We'll get much deeper into systematic techniques for reasoning about Imp programs in the following chapters, but we can do quite a bit just working with the bare definitions.
This section explores some examples.

Theorem plus2_spec : st n st',
st X = n
plus2 / st st'
st' X = n + 2.
Proof.
intros st n st' HX Heval.

Inverting Heval essentially forces Coq to expand one step of the ceval computation — in this case revealing that st' must be st extended with the new value of X, since plus2 is an assignment

inversion Heval. subst. clear Heval. simpl.
apply t_update_eq. Qed.

#### Exercise: 3 stars, recommended (XtimesYinZ_spec)

State and prove a specification of XtimesYinZ.

(* FILL IN HERE *)

#### Exercise: 3 stars, recommended (loop_never_stops)

Theorem loop_never_stops : st st',
~(loop / st st').
Proof.
intros st st' contra. unfold loop in contra.
remember (WHILE BTrue DO SKIP END) as loopdef
eqn:Heqloopdef.

Proceed by induction on the assumed derivation showing that loopdef terminates. Most of the cases are immediately contradictory (and so can be solved in one step with inversion).

(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars (no_whilesR)

Consider the definition of the no_whiles boolean predicate below:

Fixpoint no_whiles (c : com) : bool :=
match c with
| SKIP
true
| _ ::= _
true
| c1 ;; c2
andb (no_whiles c1) (no_whiles c2)
| IFB _ THEN ct ELSE cf FI
andb (no_whiles ct) (no_whiles cf)
| WHILE _ DO _ END
false
end.

This predicate yields true just on programs that have no while loops. Using Inductive, write a property no_whilesR such that no_whilesR c is provable exactly when c is a program with no while loops. Then prove its equivalence with no_whiles.

Inductive no_whilesR: com Prop :=
(* FILL IN HERE *)
.

Theorem no_whiles_eqv:
c, no_whiles c = true no_whilesR c.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 4 stars (no_whiles_terminating)

Imp programs that don't involve while loops always terminate. State and prove a theorem no_whiles_terminating that says this. Use either no_whiles or no_whilesR, as you prefer.

(* FILL IN HERE *)

#### Exercise: 3 stars (stack_compiler)

HP Calculators, programming languages like Forth and Postscript, and abstract machines like the Java Virtual Machine all evaluate arithmetic expressions using a stack. For instance, the expression
      (2*3)+(3*(4-2))

would be entered as
      2 3 * 3 4 2 - * +

and evaluated like this (where we show the program being evaluated on the right and the contents of the stack on the left):
      []            |    2 3 * 3 4 2 - * +
[2]           |    3 * 3 4 2 - * +
[3, 2]        |    * 3 4 2 - * +
[6]           |    3 4 2 - * +
[3, 6]        |    4 2 - * +
[4, 3, 6]     |    2 - * +
[2, 4, 3, 6]  |    - * +
[2, 3, 6]     |    * +
[6, 6]        |    +
[12]          |

The task of this exercise is to write a small compiler that translates 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.
• SMult: Similar, but multiply.

Inductive sinstr : Type :=
| SPush : nat sinstr
| SPlus : sinstr
| SMinus : sinstr
| SMult : sinstr.

Write a function to evaluate programs in the stack language. It should take as input a state, a stack represented as a list of numbers (top stack item is the head of the list), and a program represented as a list of instructions, and it should return the stack after executing the program. Test your function on the examples below.
Note that the specification leaves unspecified what to do when encountering an SPlus, SMinus, or SMult instruction if the stack contains less than two elements. In a sense, it is immaterial what we do, since our compiler will never emit such a malformed program.

Fixpoint s_execute (st : state) (stack : list nat)
(prog : list sinstr)
: list nat
(* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

Example s_execute1 :
s_execute empty_state []
[SPush 5; SPush 3; SPush 1; SMinus]
= [2; 5].
(* FILL IN HERE *) Admitted.

Example s_execute2 :
s_execute (t_update empty_state X 3) [3;4]
[SPush 4; SLoad X; SMult; SPlus]
= [15; 4].
(* FILL IN HERE *) Admitted.

Next, write a function that compiles an aexp into a stack machine program. The effect of running the program should be the same as pushing the value of the expression on the stack.

Fixpoint s_compile (e : aexp) : list sinstr
(* REPLACE THIS LINE WITH   := _your_definition_ . *) . Admitted.

After you've defined s_compile, prove the following to test that it works.

Example s_compile1 :
s_compile (AMinus (AId X) (AMult (ANum 2) (AId Y)))
(* 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 : (st : state) (e : aexp),
s_execute st [] (s_compile e) = [ aeval st e ].
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 5 stars, advanced (break_imp)

Module BreakImp.

Imperative languages like C and Java often include a break or similar statement for interrupting the execution of loops. In this exercise we consider how to add break to Imp. First, we need to enrich the language of commands with an additional case.

Inductive com : Type :=
| CSkip : com
| CBreak : com (* <-- new *)
| CAss : id aexp com
| CSeq : com com com
| CIf : bexp com com com
| CWhile : bexp com com.

Notation "'SKIP'" :=
CSkip.
Notation "'BREAK'" :=
CBreak.
Notation "x '::=' a" :=
(CAss x a) (at level 60).
Notation "c1 ;; c2" :=
(CSeq c1 c2) (at level 80, right associativity).
Notation "'WHILE' b 'DO' c 'END'" :=
(CWhile b c) (at level 80, right associativity).
Notation "'IFB' c1 'THEN' c2 'ELSE' c3 'FI'" :=
(CIf c1 c2 c3) (at level 80, right associativity).

Next, we need to define the behavior of BREAK. Informally, whenever BREAK is executed in a sequence of commands, it stops the execution of that sequence and signals that the innermost enclosing loop should terminate. (If there aren't any enclosing loops, then the whole program simply terminates.) The final state should be the same as the one in which the BREAK statement was executed.
One important point is what to do when there are multiple loops enclosing a given BREAK. In those cases, BREAK should only terminate the innermost loop. Thus, after executing the following...
X ::= 0;;
Y ::= 1;;
WHILE 0 ≠ Y DO
WHILE TRUE DO
BREAK
END;;
X ::= 1;;
Y ::= Y - 1
END
... the value of X should be 1, and not 0.
One way of expressing this behavior is to add another parameter to the evaluation relation that specifies whether evaluation of a command executes a BREAK statement:

Inductive status : Type :=
| SContinue : status
| SBreak : status.

Reserved Notation "c1 '/' st '' s '/' st'"
(at level 40, st, s at level 39).

Intuitively, c / st s / st' means that, if c is started in state st, then it terminates in state st' and either signals that the innermost surrounding loop (or the whole program) should exit immediately (s = SBreak) or that execution should continue normally (s = SContinue).
The definition of the "c / st s / st'" relation is very similar to the one we gave above for the regular evaluation relation (c / st st') — we just need to handle the termination signals appropriately:
• If the command is SKIP, then the state doesn't change, and execution of any enclosing loop can continue normally.
• If the command is BREAK, the state stays unchanged, but we signal a SBreak.
• If the command is an assignment, then we update the binding for that variable in the state accordingly and signal that execution can continue normally.
• If the command is of the form IFB b THEN c1 ELSE c2 FI, then the state is updated as in the original semantics of Imp, except that we also propagate the signal from the execution of whichever branch was taken.
• If the command is a sequence c1 ;; c2, we first execute c1. If this yields a SBreak, we skip the execution of c2 and propagate the SBreak signal to the surrounding context; the resulting state is the same as the one obtained by executing c1 alone. Otherwise, we execute c2 on the state obtained after executing c1, and propagate the signal generated there.
• Finally, for a loop of the form WHILE b DO c END, the semantics is almost the same as before. The only difference is that, when b evaluates to true, we execute c and check the signal that it raises. If that signal is SContinue, then the execution proceeds as in the original semantics. Otherwise, we stop the execution of the loop, and the resulting state is the same as the one resulting from the execution of the current iteration. In either case, since BREAK only terminates the innermost loop, WHILE signals SContinue.
Based on the above description, complete the definition of the ceval relation.

Inductive ceval : com state status state Prop :=
| E_Skip : st,
CSkip / st SContinue / st
(* FILL IN HERE *)

where "c1 '/' st '' s '/' st'" := (ceval c1 st s st').

Now prove the following properties of your definition of ceval:

Theorem break_ignore : c st st' s,
(BREAK;; c) / st s / st'
st = st'.
Proof.
(* FILL IN HERE *) Admitted.

Theorem while_continue : b c st st' s,
(WHILE b DO c END) / st s / st'
s = SContinue.
Proof.
(* FILL IN HERE *) Admitted.

Theorem while_stops_on_break : b c st st',
beval st b = true
c / st SBreak / st'
(WHILE b DO c END) / st SContinue / st'.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, advanced, optional (while_break_true)

Theorem while_break_true : b c st st',
(WHILE b DO c END) / st SContinue / st'
beval st' b = true
st'', c / st'' SBreak / st'.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 4 stars, advanced, optional (ceval_deterministic)

Theorem ceval_deterministic: (c:com) st st1 st2 s1 s2,
c / st s1 / st1
c / st s2 / st2
st1 = st2 s1 = s2.
Proof.
(* FILL IN HERE *) Admitted.

End BreakImp.

#### Exercise: 3 stars, optional (short_circuit)

Most modern programming languages use a "short-circuit" evaluation rule for boolean and: to evaluate BAnd b1 b2, first evaluate b1. If it evaluates to false, then the entire BAnd expression evaluates to false immediately, without evaluating b2. Otherwise, b2 is evaluated to determine the result of the BAnd expression.
Write an alternate version of beval that performs short-circuit evaluation of BAnd in this manner, and prove that it is equivalent to beval.

(* FILL IN HERE *)

#### Exercise: 4 stars, optional (add_for_loop)

Add C-style for loops to the language of commands, update the ceval definition to define the semantics of for loops, and add cases for for loops as needed so that all the proofs in this file are accepted by Coq.
A for loop should be parameterized by (a) a statement executed initially, (b) a test that is run on each iteration of the loop to determine whether the loop should continue, (c) a statement executed at the end of each loop iteration, and (d) a statement that makes up the body of the loop. (You don't need to worry about making up a concrete Notation for for loops, but feel free to play with this too if you like.)

(* FILL IN HERE *)