# Imp: Simple Imperative Programs

(* Version of 4/21/2010 *)

In this chapter, we begin a new direction that we'll continue for
the rest of the course: whereas up to now we've been mostly
studying Coq itself, from now on we'll mostly be using Coq to
formalize other things.
Our first case study is a
A minor technical point: Instead of asking Coq to import our
earlier definitions from Logic.v, we import a small library called
Sflib.v, containing just a few definitions and theorems from
earlier chapters that we'll actually use in the rest of the
course. You won't notice much difference, since most of what's
missing from Sflib has identical definitions in the Coq standard
library. The main reason for doing this is to tidy the global Coq
environment so that, for example, it is easier to search for
relevant theorems.

*simple imperative programming language*called Imp. This chapter looks at how to define the*syntax*and*semantics*of Imp; the chapters that follow will develop a theory of*program equivalence*and introduce*Hoare Logic*, the best known logic for reasoning about imperative programs.### Sflib

# Arithmetic and Boolean Expressions

*arithmetic and boolean expressions*, then an extension of these expressions with

*variables*, and finally a language of

*commands*including assignment, conditions, sequencing, and loops.

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 file ImpParser.v develops a simple implementation of a lexical
analyzer and parser that can perform this translation. You do

*not*need to understand that file to understand this one, but if you haven't taken a course where these techniques are covered (e.g., a compilers course) you may enjoy skimming it.## Evaluation

*Evaluating*an arithmetic expression reduces it to a single number.Fixpoint aeval (e : aexp) : nat :=

match e with

| ANum n => n

| APlus a1 a2 => plus (aeval a1) (aeval a2)

| AMinus a1 a2 => minus (aeval a1) (aeval a2)

| AMult a1 a2 => mult (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 (e : bexp) : bool :=

match e with

| BTrue => true

| BFalse => false

| BEq a1 a2 => beq_nat (aeval a1) (aeval a2)

| BLe a1 a2 => ble_nat (aeval a1) (aeval a2)

| BNot b1 => negb (beval b1)

| BAnd b1 b2 => andb (beval b1) (beval b2)

end.

## Optimization

Fixpoint optimize_0plus (e:aexp) : aexp :=

match e 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 e,

aeval (optimize_0plus e) = aeval e.

Proof.

intros e. induction e.

Case "ANum". reflexivity.

Case "APlus". destruct e1.

SCase "e1 = ANum n". destruct n.

SSCase "n = 0". simpl. apply IHe2.

SSCase "n <> 0". simpl. rewrite IHe2. reflexivity.

SCase "e1 = APlus e1_1 e1_2".

simpl. simpl in IHe1. rewrite IHe1. rewrite IHe2. reflexivity.

SCase "e1 = AMinus e1_1 e1_2".

simpl. simpl in IHe1. rewrite IHe1. rewrite IHe2. reflexivity.

SCase "e1 = AMult e1_1 e1_2".

simpl. simpl in IHe1. rewrite IHe1. rewrite IHe2. reflexivity.

Case "AMinus".

simpl. rewrite IHe1. rewrite IHe2. reflexivity.

Case "AMult".

simpl. rewrite IHe1. rewrite IHe2. reflexivity. Qed.

# Coq Automation

## Tacticals

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

### The try Tactical

### The ; Tactical

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.
T; [T' | T' | ... | T']

That is, if T and T' are tactics, then T;T' is a tactic that
first performs T and then performs T' on *each subgoal*generated by T. This is the form of ; that is used most often in practice.

Lemma foo : forall n, ble_nat 0 n = true.

Proof.

intros.

destruct n.

(* Leaves two subgoals... *)

Case "n=0". simpl. reflexivity.

Case "n=Sn'". simpl. reflexivity.

(* ... which are discharged similarly *)

Qed.

We can simplify the proof above using the ; tactical.

Lemma foo' : forall n, ble_nat 0 n = true.

Proof.

intros.

(* Apply destruct to the current goal *)

destruct n;

(* then apply simpl to each resulting subgoal *)

simpl;

(* then apply reflexivity to each resulting subgoal *)

reflexivity.

Qed.

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

Theorem optimize_0plus_sound': forall e,

aeval (optimize_0plus e) = aeval e.

Proof.

intros e.

induction e;

(* Most cases follow directly by the IH *)

try (simpl; rewrite IHe1; rewrite IHe2; reflexivity).

Case "ANum". reflexivity.

Case "APlus".

destruct e1;

(* Most cases follow directly by the IH *)

try (simpl; simpl in IHe1; rewrite IHe1; rewrite IHe2; reflexivity).

(* The interesting case, on which the above fails, is when e1 =

ANum n. In this case, we have to destruct n (to see whether the

optimization applies) and rewrite with the inductive

hypothesis. *)

SCase "e1 = ANum n". destruct n;

simpl; rewrite IHe2; reflexivity. Qed.

In practice, Coq experts often use try with a tactic like
induction to take care of many similar "straightforward" 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:
This proof can still be improved: the first case (for e = 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*: For all arithmetic expressions e,
aeval (optimize_0plus e) = aeval e.

*Proof*: By induction on e. The AMinus and AMult cases follow directly from the IH. The remaining cases are as follows:- Suppose e = 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 e = APlus e1 e2 for some e1 and e2. We
must show
aeval (optimize_0plus (APlus e1 e2))Consider the possible forms of e1. For most of them, optimize_0plus simply calls itself recursively for the subexpressions and rebuilds a new expression of the same form as e1; in these cases, the result follows directly from the IH.

= aeval (APlus e1 e2).optimize_0plus (APlus e1 e2) = optimize_0plus e2and the IH for e2 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. ☐

Theorem optimize_0plus_sound'': forall e,

aeval (optimize_0plus e) = aeval e.

Proof.

intros e.

induction e;

(* Most cases follow directly by the IH *)

try (simpl; rewrite IHe1; rewrite IHe2; reflexivity);

(* ... or are immediate by definition *)

try reflexivity.

(* The interesting case is when e = APlus e1 e2. *)

Case "APlus".

destruct e1;

try (simpl; simpl in IHe1; rewrite IHe1; rewrite IHe2; reflexivity).

SCase "e1 = ANum n". destruct n;

simpl; rewrite IHe2; reflexivity. Qed.

## Defining New Tactic Notations

- The Tactic Notation command gives a handy way to define
"shorthand tactics" that, when invoked, apply several tactics
at the same time.
- 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
there are many examples of Ltac definitions in the Coq
standard library that you can use as examples.
- There is also an OCaml API that can be used to build new 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 which
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 next subsection gives a more sophisticated use of this
feature...
Being able to deal with most of the cases of an induction or
destruct all at the same time is very convenient, but it can
also be a little confusing. One problem that often comes up is
that

### Bulletproofing Case Analyses

*maintaining*proofs written in this style can be difficult. For example, suppose that, later, we extended the definition of aexp with another constructor that also required a special argument. The above proof might break because Coq generated the subgoals for this constructor before the one for APlus, so that, at the point when we start working on the APlus case, Coq is actually expecting the argument for a completely different constructor. What we'd like is to get a sensible error message saying "I was expecting the AFoo case at this point, but the proof script is talking about APlus." Here's a nice little trick that smoothly achieves this.Tactic Notation "aexp_cases" tactic(first) tactic(c) :=

first;

[ c "ANum" | c "APlus" | c "AMinus" | c "AMult" ].

For example, if e is a variable of type aexp, then doing

aexp_cases (induction e) Case

will perform an induction on e (the same as if we had just typed
induction e) and *also*add a Case tag to each subgoal generated by the induction, labeling which constructor it comes from. For example, here is yet another proof of optimize_0plus_sound, using aexp_cases:Theorem optimize_0plus_sound''': forall e,

aeval (optimize_0plus e) = aeval e.

Proof.

intros e.

(* Note that we must put the entire aexp_cases expression in

parentheses when following it by a semicolon! *)

(aexp_cases (induction e) Case);

try (simpl; rewrite IHe1; rewrite IHe2; reflexivity);

try reflexivity.

(* At this point, there is already an "APlus" case name in the

context. The Case "APlus" here in the proof text has the

effect of a sanity check: if the "Case" string in the context is

anything _other_ than "APlus" (for example, because we added a

clause to the definition of aexp and forgot to change the

proof) we'll get a helpful error at this point telling us that

this is now the wrong case. *)

Case "APlus".

(aexp_cases (destruct e1) SCase);

try (simpl; simpl in IHe1; rewrite IHe1; rewrite IHe2; reflexivity).

SCase "ANum". destruct n;

simpl; rewrite IHe2; reflexivity. Qed.

#### Exercise: 3 stars (optimize_0plus_b)

Since the optimize_0plus tranformation 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.(* FILL IN HERE *)

☐
(* FILL IN HERE *)

☐

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

DESIGN EXERCISE: The optimization implemented by our optimize_0plus function is only one of many imaginable optimizations on arithmetic and boolean expressions. Write a more sophisticated optimizer and prove it correct.☐

## The omega Tactic

*Presburger arithmetic*. It is based on the Omega algorithm invented in 1992 by William Pugh.

- 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 ->,

Example silly_presburger_formula : forall m n o p,

m + n <= n + o /\ o + 3 = p + 3 ->

m <= p.

Proof.

intros. omega.

Qed.

Andrew Appel calls this the "Santa Claus tactic."
Finally, here are some miscellaneous tactics that you may find
convenient.
We'll see many examples of these in the proofs below.
Now 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.
In the rest of the course, we'll often need to talk about
"identifiers," such as program variables. We could use strings
for this, or (as in a real compiler) some kind of fancier
structures like symbols from a symbol table. But for simplicity
let's just use natural numbers as identifiers.
We define a new inductive datatype Id so that we won't
confuse identifiers and numbers.

## A Few More Handy Tactics

- 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. - assumption: Try to find a hypothesis H in the context that exactly matches the goal; if one is found, behave just like apply H.

# Expressions With Variables

## Identifiers

Inductive id : Type :=

Id : nat -> id.

Definition beq_id id1 id2 :=

match (id1, id2) with

(Id n1, Id n2) => beq_nat n1 n2

end.

Now, having "wrapped" numbers as identifiers in this way, it is
convenient to recapitulate a few properties of numbers as
analogous properties of identifiers, so that we can work with
identifiers in definitions and proofs abstractly, without
unwrapping them to expose the underlying numbers. Since all we
need to know about identifiers is whether they are the same or
different, just a few basic facts are all we need.

Theorem beq_id_refl : forall i,

true = beq_id i i.

Proof.

intros. destruct i.

apply beq_nat_refl. Qed.

#### Exercise: 1 star, optional

For this and the following exercises, do not prove by induction, but rather by applying similar results already proved for natural numbers. Some of the tactics mentioned above may prove useful.
Theorem beq_id_eq : forall i1 i2,

true = beq_id i1 i2 -> i1 = i2.

Proof.

(* FILL IN HERE *) Admitted.

true = beq_id i1 i2 -> i1 = i2.

Proof.

(* FILL IN HERE *) Admitted.

Theorem beq_id_false_not_eq : forall i1 i2,

beq_id i1 i2 = false -> i1 <> i2.

Proof.

(* FILL IN HERE *) Admitted.

beq_id i1 i2 = false -> i1 <> i2.

Proof.

(* FILL IN HERE *) Admitted.

Theorem not_eq_beq_id_false : forall i1 i2,

i1 <> i2 -> beq_id i1 i2 = false.

Proof.

(* FILL IN HERE *) Admitted.

i1 <> i2 -> beq_id i1 i2 = false.

Proof.

(* FILL IN HERE *) Admitted.

☐
A

## States

*state*represents the current set of values for all the variables at some point in the execution of a program.Definition state := id -> nat.

Definition empty_state : state := fun _ => 0.

Definition update (st : state) (V:id) (n : nat) : state :=

fun V' => if beq_id V V' then n else st V'.

Theorem update_neq : forall V2 V1 n st,

beq_id V2 V1 = false ->

(update st V2 n) V1 = (st V1).

Proof.

(* FILL IN HERE *) Admitted.

beq_id V2 V1 = false ->

(update st V2 n) V1 = (st V1).

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 2 stars, optional

Before starting to play with tactics, make sure you understand exactly what the theorem is saying!Theorem update_example : forall (n:nat),

(update empty_state (Id 2) n) (Id 3) = 0.

Proof.

(* FILL IN HERE *) Admitted.

Theorem update_shadow : forall x1 x2 k1 k2 (f : state),

(update (update f k2 x1) k2 x2) k1 = (update f k2 x2) k1.

Proof.

(* FILL IN HERE *) Admitted.

(update (update f k2 x1) k2 x2) k1 = (update f k2 x2) k1.

Proof.

(* FILL IN HERE *) Admitted.

Theorem update_same : forall x1 k1 k2 (f : state),

f k1 = x1 ->

(update f k1 x1) k2 = f k2.

Proof.

(* FILL IN HERE *) Admitted.

f k1 = x1 ->

(update f k1 x1) k2 = f k2.

Proof.

(* FILL IN HERE *) Admitted.

Theorem update_permute : forall x1 x2 k1 k2 k3 f,

beq_id k2 k1 = false ->

(update (update f k2 x1) k1 x2) k3 = (update (update f k1 x2) k2 x1) k3.

Proof.

(* FILL IN HERE *) Admitted.

beq_id k2 k1 = false ->

(update (update f k2 x1) k1 x2) k3 = (update (update f k1 x2) k2 x1) k3.

Proof.

(* FILL IN HERE *) Admitted.

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

## Syntax

Inductive aexp : Type :=

| ANum : nat -> aexp

| AId : id -> aexp

| APlus : aexp -> aexp -> aexp

| AMinus : aexp -> aexp -> aexp

| AMult : aexp -> aexp -> aexp.

Tactic Notation "aexp_cases" tactic(first) tactic(c) :=

first;

[ c "ANum" | c "AId" | c "APlus" | c "AMinus" | c "AMult" ].

Shorthands for variables:

(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 will hopefully not cause
confusion.)
Same bexps as before (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.

Tactic Notation "bexp_cases" tactic(first) tactic(c) :=

first;

[ c "BTrue" | c "BFalse" |

c "BEq" | c "BLe" |

c "BNot" | c "BAnd" ].

| BTrue : bexp

| BFalse : bexp

| BEq : aexp -> aexp -> bexp

| BLe : aexp -> aexp -> bexp

| BNot : bexp -> bexp

| BAnd : bexp -> bexp -> bexp.

Tactic Notation "bexp_cases" tactic(first) tactic(c) :=

first;

[ c "BTrue" | c "BFalse" |

c "BEq" | c "BLe" |

c "BNot" | c "BAnd" ].

Fixpoint aeval (st : state) (e : aexp) : nat :=

match e with

| ANum n => n

| AId i => st i

| APlus a1 a2 => plus (aeval st a1) (aeval st a2)

| AMinus a1 a2 => minus (aeval st a1) (aeval st a2)

| AMult a1 a2 => mult (aeval st a1) (aeval st a2)

end.

We update the boolean evaluator with the new aeval.

Fixpoint beval (st : state) (e : bexp) : bool :=

match e with

| BTrue => true

| BFalse => false

| BEq a1 a2 => beq_nat (aeval st a1) (aeval st a2)

| BLe a1 a2 => ble_nat (aeval st a1) (aeval st a2)

| BNot b1 => negb (beval st b1)

| BAnd b1 b2 => andb (beval st b1) (beval st b2)

end.

Example aexp1 :

aeval (update empty_state X 5)

(APlus (ANum 3) (AMult (AId X) (ANum 2)))

= 13.

Proof. reflexivity. Qed.

Example bexp1 :

beval (update empty_state X 5)

(BAnd BTrue (BNot (BLe (AId X) (ANum 4))))

= true.

Proof. reflexivity. Qed.

# Commands

*commands*(or

*statements*).

## Syntax

Inductive com : Type :=

| CSkip : com

| CAss : id -> aexp -> com

| CSeq : com -> com -> com

| CIf : bexp -> com -> com -> com

| CWhile : bexp -> com -> com.

Tactic Notation "com_cases" tactic(first) tactic(c) :=

first;

[ c "SKIP" | c "::=" | c ";" | c "IFB" | c "WHILE" ].

More readable concrete syntax, for examples:

Notation "'SKIP'" :=

CSkip.

Notation "l '::=' a" :=

(CAss l 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' e1 'THEN' e2 'ELSE' e3 'FI'" :=

(CIf e1 e2 e3) (at level 80, right associativity).

Definition plus2 : com :=

X ::= (APlus (AId X) (ANum 2)).

Definition XtimesYinZ : com :=

Z ::= (AMult (AId X) (AId Y)).

Loops:

Definition subtract_slowly_body : com :=

Z ::= AMinus (AId Z) (ANum 1) ;

X ::= AMinus (AId X) (ANum 1).

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:

Factorial:

Definition fact_body : com :=

Y ::= AMult (AId Y) (AId Z) ;

Z ::= AMinus (AId Z) (ANum 1).

Definition fact_loop : com :=

WHILE BNot (BEq (AId Z) (ANum 0)) DO

fact_body

END.

Definition fact_com : com :=

Z ::= AId X ;

Y ::= ANum 1 ;

fact_loop.

# Evaluation

## Evaluation Function

Fixpoint ceval_step1 (st : state) (c : com) : state :=

match c with

| SKIP =>

st

| l ::= a1 =>

update st l (aeval st a1)

| c1 ; c2 =>

let st' := ceval_step1 st c1 in

ceval_step1 st' c2

| IFB b THEN c1 ELSE c2 FI =>

if (beval st b) then ceval_step1 st c1 else ceval_step1 st c2

| WHILE b1 DO c1 END =>

st (* bogus *)

end.

Second try, using an extra numeric argument as a "step index" to
ensure that evaluation always terminates.

Fixpoint ceval_step2 (st : state) (c : com) (i : nat) : state :=

match i with

| O => empty_state

| S i' =>

match c with

| SKIP =>

st

| l ::= a1 =>

update st l (aeval st a1)

| c1 ; c2 =>

let st' := ceval_step2 st c1 i' in

ceval_step2 st' c2 i'

| IFB b THEN c1 ELSE c2 FI =>

if (beval st b) then ceval_step2 st c1 i' else ceval_step2 st c2 i'

| WHILE b1 DO c1 END =>

if (beval st b1)

then let st' := ceval_step2 st c1 i' in

ceval_step2 st' c i'

else st

end

end.

*Note*: It is tempting to think that the index i here is counting the "number of steps of evaluation." But if you look closely you'll see that this is not the case: for example, in the rule for sequencing, the same i is passed to both recursive calls. Understanding the exact way that i is treated will be important in the proof of ceval__ceval_step, which is given as an exercise below.

Fixpoint ceval_step3 (st : state) (c : com) (i : nat)

: option state :=

match i with

| O => None

| S i' =>

match c with

| SKIP =>

Some st

| l ::= a1 =>

Some (update st l (aeval st a1))

| c1 ; c2 =>

match (ceval_step3 st c1 i') with

| Some st' => ceval_step3 st' c2 i'

| None => None

end

| IFB b THEN c1 ELSE c2 FI =>

if (beval st b) then ceval_step3 st c1 i' else ceval_step3 st c2 i'

| WHILE b1 DO c1 END =>

if (beval st b1)

then match (ceval_step3 st c1 i') with

| Some st' => ceval_step3 st' c i'

| None => None

end

else Some st

end

end.

We can improve the readability of this definition by introducing
an auxiliary function bind_option to hide some of the "plumbing"
involved in repeatedly matching against optional states.

Definition bind_option {X Y : Type} (xo : option X) (f : X -> option Y)

: option Y :=

match xo with

| None => None

| Some x => f x

end.

Fixpoint ceval_step (st : state) (c : com) (i : nat)

: option state :=

match i with

| O => None

| S i' =>

match c with

| SKIP =>

Some st

| l ::= a1 =>

Some (update st l (aeval st a1))

| c1 ; c2 =>

bind_option

(ceval_step st c1 i')

(fun st' => ceval_step st' c2 i')

| IFB b THEN c1 ELSE c2 FI =>

if (beval st b) then ceval_step st c1 i' else ceval_step st c2 i'

| WHILE b1 DO c1 END =>

if (beval st b1)

then bind_option

(ceval_step st c1 i')

(fun st' => ceval_step st' c i')

else Some st

end

end.

Definition test_ceval (st:state) (c:com) :=

match ceval_step st c 500 with

| None => None

| Some st => Some (st X, st Y, st Z)

end.

(*

Eval compute in

(test_ceval empty_state

(X ::= ANum 2;

IFB BLe (AId X) (ANum 1)

THEN Y ::= ANum 3

ELSE Z ::= ANum 4

FI)).

====>

Some (2, 0, 4)

*)

#### Exercise: 2 stars

Write an Imp program that sums the numbers from 1 to X (inclusive: 1 + 2 + ... + X) in the variable Y. Make sure your solution satisfies the test that follows.Example pup_to_n_1 : test_ceval (update empty_state X 5) pup_to_n = Some (0, 15, 0). Proof. reflexivity. Qed.☐

#### Exercise: 2 stars, optional

Write a While program that sets Z to 0 if X is even and sets Z to 1 otherwise. Use ceval_test to test your program.(* FILL IN HERE *)

☐
Here's a better way: define ceval as a
This is an important change. Besides freeing us from the
silliness of passing around step indices all over the place, it
gives us a lot more flexibility in the definition. For example,
if we added concurrency features to the language, we'd want the
definition of evaluation to be non-deterministic -- not only not
total, but not even a function.

## Evaluation Relation

*relation*rather than a*function*-- i.e., define it in Prop instead of Type.Module CEvalFirstTry.

Inductive ceval : com -> state -> state -> Prop :=

| E_Skip : forall st,

ceval SKIP st st

| E_Ass : forall st a1 n l,

aeval st a1 = n ->

ceval (l ::= a1) st (update st l n)

| E_Seq : forall c1 c2 st st' st'',

ceval c1 st st' ->

ceval c2 st' st'' ->

ceval (c1 ; c2) st st''

| E_IfTrue : forall st st' b1 c1 c2,

beval st b1 = true ->

ceval c1 st st' ->

ceval (IFB b1 THEN c1 ELSE c2 FI) st st'

| E_IfFalse : forall st st' b1 c1 c2,

beval st b1 = false ->

ceval c2 st st' ->

ceval (IFB b1 THEN c1 ELSE c2 FI) st st'

| E_WhileEnd : forall b1 st c1,

beval st b1 = false ->

ceval (WHILE b1 DO c1 END) st st

| E_WhileLoop : forall st st' st'' b1 c1,

beval st b1 = true ->

ceval c1 st st' ->

ceval (WHILE b1 DO c1 END) st' st'' ->

ceval (WHILE b1 DO c1 END) st st''.

Since we'll be using ceval a lot, let's define some notation.
We'll use the notation c / st ==> st' for our ceval relation,
that is 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'".

Notation "c1 '/' st '==>' st'" := (ceval st c1 st') (at level 40, st at level 39).

End CEvalFirstTry.

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

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

Inductive ceval : state -> com -> state -> Prop :=

| E_Skip : forall st,

SKIP / st ==> st

| E_Ass : forall st a1 n l,

aeval st a1 = n ->

(l ::= a1) / st ==> (update st l n)

| E_Seq : forall c1 c2 st st' st'',

c1 / st ==> st' ->

c2 / st' ==> st'' ->

(c1 ; c2) / st ==> st''

| E_IfTrue : forall st st' b1 c1 c2,

beval st b1 = true ->

c1 / st ==> st' ->

(IFB b1 THEN c1 ELSE c2 FI) / st ==> st'

| E_IfFalse : forall st st' b1 c1 c2,

beval st b1 = false ->

c2 / st ==> st' ->

(IFB b1 THEN c1 ELSE c2 FI) / st ==> st'

| E_WhileEnd : forall b1 st c1,

beval st b1 = false ->

(WHILE b1 DO c1 END) / st ==> st

| E_WhileLoop : forall st st' st'' b1 c1,

beval st b1 = true ->

c1 / st ==> st' ->

(WHILE b1 DO c1 END) / st' ==> st'' ->

(WHILE b1 DO c1 END) / st ==> st''

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

Tactic Notation "ceval_cases" tactic(first) tactic(c) := first; [

c "E_Skip" | c "E_Ass" | c "E_Seq" | c "E_IfTrue" | c "E_IfFalse"

| c "E_WhileEnd" | c "E_WhileLoop" ].

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

==> (update (update empty_state X 2) Z 4).

Proof.

(* We must supply the intermediate state *)

apply E_Seq with (update empty_state X 2).

Case "assignment command".

apply E_Ass. reflexivity.

Case "if command".

apply E_IfFalse.

reflexivity.

apply E_Ass. reflexivity. Qed.

Example ceval_example2:

(X ::= ANum 0; Y ::= ANum 1; Z ::= ANum 2) / empty_state ==>

(update (update (update empty_state X 0) Y 1) Z 2).

Proof.

(* FILL IN HERE *) Admitted.

(X ::= ANum 0; Y ::= ANum 1; Z ::= ANum 2) / empty_state ==>

(update (update (update empty_state X 0) Y 1) Z 2).

Proof.

(* FILL IN HERE *) Admitted.

☐
We will sometimes (especially in informal discussions)
write the rules for ceval in a more "graphical" form called

Formally, there is nothing deep or complex about inference rules:
they are just implications. You can read the rule name on the
right as the name of the constructor and read both the spaces
between premises above the line and the line itself as ->. All
the variables mentioned in the rule (c1, c2, etc.) are
implicitly bound by a universal quantifier at the beginning. The
whole collection of rules is implicitly wrapped in an Inductive
declaration; this is sometimes indicated informally by something
like "Let ceval be the smallest relation closed under the
following rules...".
Here is a complete set of inference rules for ceval:

Naturally, we'd hope that the two alternative definitions of
evaluation actually boil down to the same thing. This section
shows that this is the case. Make sure you understand the
statements of the theorems and can follow the structure of the
proofs.
<<
Inductive aeval_rel :
(* FILL IN HERE *)

Theorem aeval_iff_aeval_rel : forall st a n,
(* FILL IN HERE *)

Inductive beval_rel :
(* FILL IN HERE *)

Lemma beval_iff_beval_rel : forall st b tv,
(* FILL IN HERE *)

(** *)

>>

## Inference Rule Notation

*inference rules*, where the premises above the line allow you to derive the conclusion below the line. For example, the constructor E_Seq would be written like this as an inference rule:c1 / st ==> st' | |

c2 / st' ==> st'' | (E_Seq) |

c1;c2 / st ==> st'' |

(E_Skip) | |

SKIP / st ==> st |

aeval st a1 = n | (E_Ass) |

l := a1 / st ==> (update st l 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 b1 = false | (E_WhileEnd) |

WHILE b1 DO c1 END / st ==> st |

beval st b1 = true | |

c1 / st ==> st' | |

WHILE b1 DO c1 END / st' ==> st'' | (E_WhileLoop) |

WHILE b1 DO c1 END / st ==> st |

## Equivalence of Relational and Step-Indexed Evaluation

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

Write relations aeval_rel and beval_rel in the same style as ceval, and prove that they are equivalent to aeval and beval.(** *)

>>

Theorem ceval_step__ceval: forall c st st',

(exists i, ceval_step st c i = Some st') ->

c / st ==> st'.

Proof.

intros c st st' H.

inversion H as [i E].

clear H.

generalize dependent st'.

generalize dependent st.

generalize dependent c.

induction i as [| i' ].

Case "i = 0 -- contradictory".

intros c st st' H. inversion H.

Case "i = S i'".

intros c st st' H.

(com_cases (destruct c) SCase); simpl in H; inversion H; subst; clear H.

SCase "SKIP". apply E_Skip.

SCase "::=". apply E_Ass. reflexivity.

SCase ";".

remember (ceval_step st c1 i') as r1. destruct r1.

SSCase "Evaluation of r1 terminates normally".

apply E_Seq with s.

apply IHi'. rewrite Heqr1. reflexivity.

apply IHi'. simpl in H1. assumption.

SSCase "Evaluation of r1 terminates abnormally -- contradiction".

inversion H1.

SCase "IFB".

remember (beval st b) as r. destruct r.

SSCase "r = true".

apply E_IfTrue. rewrite Heqr. reflexivity.

apply IHi'. assumption.

SSCase "r = false".

apply E_IfFalse. rewrite Heqr. reflexivity.

apply IHi'. assumption.

SCase "WHILE". remember (beval st b) as r. destruct r.

SSCase "r = true".

remember (ceval_step st c i') as r1. destruct r1.

SSSCase "r1 = Some s".

apply E_WhileLoop with s. rewrite Heqr. reflexivity.

apply IHi'. rewrite Heqr1. reflexivity.

apply IHi'. simpl in H1. assumption.

SSSCase "r1 = None".

inversion H1.

SSCase "r = false".

inversion H1.

apply E_WhileEnd.

rewrite Heqr. subst. reflexivity. Qed.

#### Exercise: 4 stars (ceval_step__ceval_inf)

Write an informal proof of ceval_step__ceval, following the usual template. (The template for case analysis on an inductively defined value should look the same as for induction, except that there is no induction hypothesis.) Make your proof communicate the main ideas to a human reader; do *not* simply transcribe the steps of the formal proof.☐

Theorem ceval_step_more: forall i1 i2 st st' c,

i1 <= i2 -> ceval_step st c i1 = Some st' ->

ceval_step st c i2 = Some st'.

Proof.

induction i1 as [|i1']; intros i2 st st' c Hle Hceval.

Case "i1 = 0".

inversion Hceval.

Case "i1 = S i1'".

destruct i2 as [|i2']. inversion Hle.

assert (Hle': i1' <= i2') by omega.

com_cases (destruct c) SCase.

SCase "SKIP".

simpl in Hceval. inversion Hceval.

reflexivity.

SCase "::=".

simpl in Hceval. inversion Hceval.

reflexivity.

SCase ";".

simpl in Hceval. simpl.

remember (ceval_step st c1 i1') as st1'o.

destruct st1'o.

SSCase "st1'o = Some".

symmetry in Heqst1'o.

apply (IHi1' i2') in Heqst1'o; try assumption.

rewrite Heqst1'o. simpl. simpl in Hceval.

apply (IHi1' i2') in Hceval; try assumption.

SSCase "st1'o = None".

inversion Hceval.

SCase "IFB".

simpl in Hceval. simpl.

remember (beval st b) as bval.

destruct bval; apply (IHi1' i2') in Hceval; assumption.

SCase "WHILE".

simpl in Hceval. simpl.

destruct (beval st b); try assumption.

remember (ceval_step st c i1') as st1'o.

destruct st1'o.

SSCase "st1'o = Some".

symmetry in Heqst1'o.

apply (IHi1' i2') in Heqst1'o; try assumption.

rewrite -> Heqst1'o. simpl. simpl in Hceval.

apply (IHi1' i2') in Hceval; try assumption.

SSCase "i1'o = None".

simpl in Hceval. inversion Hceval. Qed.

#### Exercise: 3 stars

Finish the following proof. You'll need ceval_step_more in a few places, as well as some basic facts about <= and plus.
Theorem ceval__ceval_step: forall c st st',

c / st ==> st' ->

exists i, ceval_step st c i = Some st'.

Proof.

intros c st st' Hce.

ceval_cases (induction Hce) Case.

(* FILL IN HERE *) Admitted.

c / st ==> st' ->

exists i, ceval_step st c i = Some st'.

Proof.

intros c st st' Hce.

ceval_cases (induction Hce) Case.

(* FILL IN HERE *) Admitted.

☐

Theorem ceval_and_ceval_step_coincide: forall c st st',

c / st ==> st'

<-> exists i, ceval_step st c i = Some st'.

Proof.

intros c st st'.

split. apply ceval__ceval_step. apply ceval_step__ceval.

Qed.

## Determinacy of Evaluation

Theorem ceval_deterministic: forall c st st1 st2,

c / st ==> st1 ->

c / st ==> st2 ->

st1 = st2.

Proof.

intros c st st1 st2 E1 E2.

generalize dependent st2.

(ceval_cases (induction E1) Case); intros st2 E2; inversion E2; subst.

Case "E_Skip". reflexivity.

Case "E_Ass". reflexivity.

Case "E_Seq".

assert (st' = st'0) as EQ1.

SCase "Proof of assertion". apply IHE1_1; assumption.

subst st'0.

apply IHE1_2. assumption.

Case "E_IfTrue".

SCase "b1 evaluates to true".

apply IHE1. assumption.

SCase "b1 evaluates to false (contradiction)".

rewrite H in H5. inversion H5.

Case "E_IfFalse".

SCase "b1 evaluates to true (contradiction)".

rewrite H in H5. inversion H5.

SCase "b1 evaluates to false".

apply IHE1. assumption.

Case "E_WhileEnd".

SCase "b1 evaluates to true".

reflexivity.

SCase "b1 evaluates to false (contradiction)".

rewrite H in H2. inversion H2.

Case "E_WhileLoop".

SCase "b1 evaluates to true (contradiction)".

rewrite H in H4. inversion H4.

SCase "b1 evaluates to false".

assert (st' = st'0) as EQ1.

SSCase "Proof of assertion". apply IHE1_1; assumption.

subst st'0.

apply IHE1_2. assumption. Qed.

Here's a slicker proof, using the fact that the relational and
step-indexed definition of evaluation are the same.

Theorem ceval_deterministic' : forall c st st1 st2,

c / st ==> st1 ->

c / st ==> st2 ->

st1 = st2.

Proof.

intros c st st1 st2 He1 He2.

apply ceval__ceval_step in He1.

apply ceval__ceval_step in He2.

inversion He1 as [i1 E1].

inversion He2 as [i2 E2].

apply ceval_step_more with (i2 := i1 + i2) in E1.

apply ceval_step_more with (i2 := i1 + i2) in E2.

rewrite E1 in E2. inversion E2. reflexivity.

omega. omega. Qed.

# Reasoning about programs

Theorem plus2_spec : forall st n st',

st X = n ->

plus2 / st ==> st' ->

st' X = plus 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.

apply update_eq. Qed.

(* FILL IN HERE *)

Theorem loop_never_stops : forall st st',

~(loop / st ==> st').

Proof.

intros st st' contra. unfold loop in contra.

remember (WHILE BTrue DO SKIP END) as loopdef.

(* 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.

~(loop / st ==> st').

Proof.

intros st st' contra. unfold loop in contra.

remember (WHILE BTrue DO SKIP END) as loopdef.

(* 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.

☐

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.

#### Exercise: 2 stars, optional

The no_whiles property yields true on just those programs that have no while loops. Using Inductive, write a property no_Whiles such that no_Whiles c is provable exactly when c is a program with no while loops. Then prove its equivalence with no_whiles.Inductive no_Whiles: com -> Prop :=

(* FILL IN HERE *)

.

Theorem no_whiles_eqv:

forall c, no_whiles c = true <-> no_Whiles c.

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 4 stars, optional

Imp programs that don't involve while loops always terminate. State and prove a theorem that says this. (Use either no_whiles or no_Whiles, as you prefer.)
(* FILL IN HERE *)

☐
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.)

# Additional Exercises

#### 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.
(* FILL IN HERE *)

☐
Write an alternate version of beval that performs short-circuit
evaluation of BAnd in this manner, and prove that it is
equivalent to beval.

#### 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.
(* FILL IN HERE *)

#### Exercise: 4 stars (stack_compiler)

HP Calculators, programming languages like Forth and Postscript, and 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:

[] | 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] |

- SPush n: Push the number n on the stack.
- SLoad i: Load the identifier i from the store and push it on the stack
- SPlus: Pop the two top numbers from the stack, add them, and push the result onto the stack.
- SMinus: Similar, but subtract.
- SMult: Similar, but multiply.

Inductive sinstr : Type :=

| SPush : nat -> sinstr

| SLoad : id -> sinstr

| SPlus : sinstr

| SMinus : sinstr

| SMult : sinstr.

Write a function to evaluate programs in the stack language. It
takes 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 returns 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, since our compiler will never emit such a malformed
program. However, when you do the correctness proof you may find
some choices makes the proof easier than others.

Fixpoint s_execute (st : state) (stack : list nat) (prog : list sinstr)

: list nat :=

(* FILL IN HERE *) admit.

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 (update empty_state X 3) [3,4] [SPush 4, SLoad X, SMult, SPlus]

= [15, 4].

(* FILL IN HERE *) Admitted.

Next, write a function which 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.

Example s_compile1 : s_compile (AMinus (AId X) (AMult (ANum 2) (AId Y))) = [SLoad X, SPush 2, SLoad Y, SMult, SMinus]. Proof. reflexivity. Qed.

(* FILL IN HERE *)

Theorem s_compile_correct : forall (st : state) (e : aexp),

s_execute st [] (s_compile e) = [ aeval st e ].

Proof.

(* FILL IN HERE *) Admitted.

☐