# ImpSimple Imperative Programs

(* $Date: 2012-09-15 09:25:36 -0400 (Sat, 15 Sep 2012) $ *)

In this chapter, we begin a new direction that will
continue for the rest of the course. Up to now we've been mostly
studying Coq itself, but from now on we'll mostly be using Coq to
formalize other things.
Our first case study is a
This chapter looks at how to define the

*simple imperative programming language*called Imp. 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

Y ::= 1;

WHILE not (Z = 0) DO

Y ::= Y * Z;

Z ::= Z - 1

END

*syntax*and*semantics*of Imp; the chapters that follow develop a theory of*program equivalence*and introduce*Hoare Logic*, the best-known logic for reasoning about imperative programs.### Sflib

Require Export 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.

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
For comparison, here's a conventional BNF (Backus-Naur Form)
grammar defining the same abstract syntax:
Compared to the Coq version above...
It's good to be comfortable with both sorts of notations: informal
ones for communicating between humans and formal ones for carrying
out implementations and proofs.

*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 want to skim it.
aexp ::= nat

| aexp '+' aexp

| aexp '-' aexp

| aexp '*' aexp

bexp ::= true

| false

| aexp '=' aexp

| aexp '<=' aexp

| bexp 'and' bexp

| 'not' bexp

| aexp '+' aexp

| aexp '-' aexp

| aexp '*' aexp

bexp ::= true

| false

| aexp '=' aexp

| aexp '<=' aexp

| bexp 'and' bexp

| 'not' bexp

- 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
*) unspecified. Some additional information — and human
intelligence — would be required to turn this description
into a formal definition (when implementing a compiler, for
example).
- 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.

Fixpoint aeval (e : aexp) : nat :=

match e with

| ANum n => n

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

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

| AMult a1 a2 => (aeval a1) * (aeval a2)

end.

Example test_aeval1:

aeval (APlus (ANum 2) (ANum 2)) = 4.

Proof. reflexivity. Qed.

Similarly, evaluating a boolean expression yields a boolean.

Fixpoint beval (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: ∀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

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

### The repeat Tactical

Theorem ev100 : ev 100.

Proof.

repeat (apply ev_SS). (* applies ev_SS 50 times,

then apply ev_SS fails *)

apply ev_0.

Qed.

(* Print ev100. *)

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 (it repeats zero times).

Theorem ev100' : ev 100.

Proof.

repeat (apply ev_0). (* doesn't fail, applies ev_0 zero times *)

repeat (apply ev_SS). apply ev_0. (* we can continue the proof *)

Qed.

The repeat T tactic does not have any 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.

### The try Tactical

*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.

Using try in a completely manual proof is a bit silly, but we'll
see below that try is very useful for doing automated proofs
in conjunction with the ; tactical.

### The ; Tactical (Simple Form)

*each subgoal*generated by T.

Lemma foo : ∀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 this proof using the ; tactical:

Lemma foo' : ∀n, ble_nat 0 n = true.

Proof.

intros.

destruct n; (* destruct the current goal *)

simpl; (* then simpl each resulting subgoal *)

reflexivity. (* then do reflexivity on each resulting subgoal *)

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': ∀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).

(* The remaining cases -- ANum and APlus -- are more

interesting... *)

Case "ANum". reflexivity.

Case "APlus".

destruct e1;

(* Again, most cases follow directly by the IH *)

try (simpl; simpl in IHe1; rewrite IHe1;

rewrite IHe2; 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. *)

SCase "e1 = ANum n". destruct n;

simpl; rewrite IHe2; 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:
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'': ∀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.

### The ; Tactical (General Form)

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']

The form T;T' is used most often in practice.
## Defining New Tactic Notations

- The Tactic Notation idiom illustrated below gives a handy
way to define "shorthand tactics" that bundle several tactics
into a single command.
- For more sophisticated programming, Coq offers a 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, which can be used to build tactics that access Coq's internal structures at a lower level, but this is seldom worth the trouble for ordinary Coq users.

Tactic Notation "simpl_and_try" tactic(c) :=

simpl;

try c.

This defines a new tactical called simpl_and_try 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...

### 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 trick (due to Aaron Bohannon) that smoothly achieves this.

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

first;

[ Case_aux c "ANum" | Case_aux c "APlus"

| Case_aux c "AMinus" | Case_aux c "AMult" ].

(Case_aux implements the common functionality of Case,
SCase, SSCase, etc. For example, Case "foo" is defined as
Case_aux Case "foo".)
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''': ∀e,

aeval (optimize_0plus e) = aeval e.

Proof.

intros e.

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_example : ∀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." We'll see
examples of its use below.

## 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. - 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.

# Evaluation as a Relation

*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
vertical bar as the closest approximation in .v files.)

Notation "e '⇓' n" := (aevalR e n) : 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.

Tactic Notation "aevalR_cases" tactic(first) ident(c) :=

first;

[ Case_aux c "E_ANum" | Case_aux c "E_APlus"

| Case_aux c "E_AMinus" | Case_aux c "E_AMult" ].

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

Theorem aeval_iff_aevalR : ∀a n,

(a ⇓ n) ↔ aeval a = n.

Proof.

split.

Case "→".

intros H.

aevalR_cases (induction H) SCase; simpl.

SCase "E_ANum".

reflexivity.

SCase "E_APlus".

rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.

SCase "E_AMinus".

rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.

SCase "E_AMult".

rewrite IHaevalR1. rewrite IHaevalR2. reflexivity.

Case "←".

generalize dependent n.

aexp_cases (induction a) SCase;

simpl; intros; subst.

SCase "ANum".

apply E_ANum.

SCase "APlus".

apply E_APlus.

apply IHa1. reflexivity.

apply IHa2. reflexivity.

SCase "AMinus".

apply E_AMinus.

apply IHa1. reflexivity.

apply IHa2. reflexivity.

SCase "AMult".

apply E_AMult.

apply IHa1. reflexivity.

apply IHa2. reflexivity.

Qed.

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

Theorem aeval_iff_aevalR' : ∀a n,

(a ⇓ n) ↔ aeval a = n.

Proof.

(* WORKED IN CLASS *)

split.

Case "→".

intros H; induction H; subst; reflexivity.

Case "←".

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 *)

*)

☐
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 necessarily 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
computational mechanism.
However, there are circumstances where relational definitions of
evaluation are greatly preferable to functional ones, as we'll see
shortly.

## Inference Rule Notation

*inference rules*, where the premises above the line justify the conclusion below the line. 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:
aevalR e1 n1 →

aevalR e2 n2 →

aevalR (APlus e1 e2) (n1 + n2)

e1 ⇓ n1 | |

e2 ⇓ n2 | (E_APlus) |

APlus e1 e2 ⇓ n1+n2 |

(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 |

End AExp.

# Expressions With Variables

## Identifiers

*identifiers*such as program variables. We could use strings for this — or, in a real compiler, fancier structures like pointers into a symbol table. But for simplicity let's just use natural numbers as identifiers.

Module Id.

We define a new inductive datatype Id so that we won't confuse
identifiers and numbers.

Inductive id : Type :=

Id : nat → id.

Definition beq_id X1 X2 :=

match (X1, X2) with

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

end.

After we "wrap" 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 : ∀X,

true = beq_id X X.

Proof.

intros. destruct X.

apply beq_nat_refl. Qed.

#### Exercise: 1 star, optional (beq_id_eq)

For this and the following exercises, do not use induction, but rather apply similar results already proved for natural numbers. Some of the tactics mentioned above may prove useful.
Theorem beq_id_eq : ∀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 : ∀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 : ∀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.

Theorem beq_id_sym: ∀i1 i2,

beq_id i1 i2 = beq_id i2 i1.

Proof.

(* FILL IN HERE *) Admitted.

beq_id i1 i2 = beq_id i2 i1.

Proof.

(* FILL IN HERE *) Admitted.

☐

End Id.

## States

*state*represents the current values of all the variables at some point in the execution of a program. For simplicity (to avoid dealing with partial functions), we let the state be defined for

*all*variables, even though any given program is only going to mention a finite number of them.

Definition state := id → nat.

Definition empty_state : state :=

fun _ => 0.

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

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

For proofs involving states, we'll need several simple properties
of update.

#### Exercise: 1 star (update_eq)

Theorem update_eq : ∀n X st,

(update st X n) X = n.

Proof.

(* FILL IN HERE *) Admitted.

(update st X n) X = n.

Proof.

(* FILL IN HERE *) Admitted.

Theorem update_neq : ∀X2 X1 n st,

beq_id X2 X1 = false →

(update st X2 n) X1 = (st X1).

Proof.

(* FILL IN HERE *) Admitted.

beq_id X2 X1 = false →

(update st X2 n) X1 = (st X1).

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 1 star (update_example)

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

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

Proof.

(* FILL IN HERE *) Admitted.

Theorem update_shadow : ∀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 : ∀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 : ∀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.

☐

## Syntax

Inductive aexp : Type :=

| ANum : nat → aexp

| AId : id → aexp (* <----- NEW *)

| APlus : aexp → aexp → aexp

| AMinus : aexp → aexp → aexp

| AMult : aexp → aexp → aexp.

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

first;

[ Case_aux c "ANum" | Case_aux c "AId" | Case_aux c "APlus"

| Case_aux c "AMinus" | Case_aux c "AMult" ].

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 the same 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) ident(c) :=

first;

[ Case_aux c "BTrue" | Case_aux c "BFalse" | Case_aux c "BEq"

| Case_aux c "BLe" | Case_aux c "BNot" | Case_aux c "BAnd" ].

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

match e with

| ANum n => n

| AId X => st X (* <----- NEW *)

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

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

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

end.

Fixpoint beval (st : state) (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.

## Syntax

com ::= 'SKIP'

| X '::=' aexp

| com ';' com

| 'WHILE' bexp 'DO' com 'END'

| 'IFB' bexp 'THEN' com 'ELSE' com 'FI'

For example, here's the factorial function in Imp.
| X '::=' aexp

| com ';' com

| 'WHILE' bexp 'DO' com 'END'

| 'IFB' bexp 'THEN' com 'ELSE' com 'FI'

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.
Y ::= 1;

WHILE not (Z = 0) DO

Y ::= Y * Z;

Z ::= Z - 1

END

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) ident(c) :=

first;

[ Case_aux c "SKIP" | Case_aux c "::=" | Case_aux c ";"

| Case_aux c "IFB" | Case_aux c "WHILE" ].

As usual, we can use a few Notation declarations to make things
more readable. We need to be a bit careful to avoid conflicts
with Coq's built-in notations, so we'll keep this light — in
particular, we won't introduce any notations for aexps and
bexps to avoid confusion with the numerical and boolean
operators we've already defined. We use the keyword IFB for
conditionals instead of IF, for similar reasons.

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' e1 'THEN' e2 'ELSE' e3 'FI'" :=

(CIf e1 e2 e3) (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.

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.

Factorial again (broken up into smaller pieces this time, for
convenience when we come back to proving things about it
later).

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 (Failed Attempt)

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

match c with

| SKIP =>

st

| l ::= a1 =>

update st l (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 b1 DO c1 END =>

st (* bogus *)

end.

In a traditional functional programming language like ML or
Haskell we could write the WHILE case as follows:
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, which make everything much more complicated
(see chapter ImpCEvalFun if curious).

Fixpoint ceval_fun (st : state) (c : com) : state := match c with ... | WHILE b1 DO c1 END => if (beval st b1) then ceval_fun st (c1; WHILE b1 DO c1 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, 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.

## Evaluation as a Relation

*relation*rather than a

*function*— i.e., we define it in Prop instead of Type, as we did for aevalR and bevalR above.

(E_Skip) | |

SKIP / st ⇓ st |

aeval st a1 = n | (E_Ass) |

X := a1 / st ⇓ (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 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'' |

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 ⇓ (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' b1 c1 c2,

beval st b1 = true →

c1 / st ⇓ st' →

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

| E_IfFalse : ∀st st' b1 c1 c2,

beval st b1 = false →

c2 / st ⇓ st' →

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

| E_WhileEnd : ∀b1 st c1,

beval st b1 = false →

(WHILE b1 DO c1 END) / st ⇓ st

| E_WhileLoop : ∀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 c1 st st').

Tactic Notation "ceval_cases" tactic(first) ident(c) :=

first;

[ Case_aux c "E_Skip" | Case_aux c "E_Ass" | Case_aux c "E_Seq"

| Case_aux c "E_IfTrue" | Case_aux c "E_IfFalse"

| Case_aux c "E_WhileEnd" | Case_aux 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.

☐

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

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

(* FILL IN HERE *) admit.

Theorem pup_to_2_ceval :

pup_to_n / (update empty_state X 2) ⇓

update (update (update (update (update (update empty_state

X 2) Y 0) Y 2) X 1) Y 3) X 0.

Proof.

(* FILL IN HERE *) Admitted.

☐

## Determinism of Evaluation

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.

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.

# Reasoning About Programs

## Basic 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 update_eq. Qed.

(* FILL IN HERE *)

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.

(* 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: 3 stars, optional (no_whilesR)

The no_whiles property yields true on just those 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, optional (no_whiles_terminating)

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

Print fact_body. Print fact_loop. Print fact_com.

Here is an alternative "mathematical" definition of the factorial
function:

Fixpoint real_fact (n:nat) : nat :=

match n with

| O => 1

| S n' => n * (real_fact n')

end.

We would like to show that they agree — if we start fact_com in
a state where variable X contains some number x, then it will
terminate in a state where variable Y contains the factorial of
x.
To show this, we rely on the critical idea of a

*loop invariant*.Definition fact_invariant (x:nat) (st:state) :=

(st Y) * (real_fact (st Z)) = real_fact x.

Theorem fact_body_preserves_invariant: ∀st st' x,

fact_invariant x st →

st Z <> 0 →

fact_body / st ⇓ st' →

fact_invariant x st'.

Proof.

unfold fact_invariant, fact_body.

intros st st' x Hm HZnz He.

inversion He; subst; clear He.

inversion H1; subst; clear H1.

inversion H4; subst; clear H4.

unfold update. simpl.

(* Show that st Z = S z' for some z' *)

destruct (st Z) as [| z'].

apply ex_falso_quodlibet. apply HZnz. reflexivity.

rewrite ← Hm. rewrite ← mult_assoc.

replace (S z' - 1) with z' by omega.

reflexivity. Qed.

Theorem fact_loop_preserves_invariant : ∀st st' x,

fact_invariant x st →

fact_loop / st ⇓ st' →

fact_invariant x st'.

Proof.

intros st st' x H Hce.

remember fact_loop as c.

ceval_cases (induction Hce) Case;

inversion Heqc; subst; clear Heqc.

Case "E_WhileEnd".

(* trivial when the loop doesn't run... *)

assumption.

Case "E_WhileLoop".

(* if the loop does run, we know that fact_body preserves

fact_invariant -- we just need to assemble the pieces *)

apply IHHce2.

apply fact_body_preserves_invariant with st;

try assumption.

intros Contra. simpl in H0; subst.

rewrite Contra in H0. inversion H0.

reflexivity. Qed.

Theorem guard_false_after_loop: ∀b c st st',

(WHILE b DO c END) / st ⇓ st' →

beval st' b = false.

Proof.

intros b c st st' Hce.

remember (WHILE b DO c END) as cloop.

ceval_cases (induction Hce) Case;

inversion Heqcloop; subst; clear Heqcloop.

Case "E_WhileEnd".

assumption.

Case "E_WhileLoop".

apply IHHce2. reflexivity. Qed.

Patching it all together...

Theorem fact_com_correct : ∀st st' x,

st X = x →

fact_com / st ⇓ st' →

st' Y = real_fact x.

Proof.

intros st st' x HX Hce.

inversion Hce; subst; clear Hce.

inversion H1; subst; clear H1.

inversion H4; subst; clear H4.

inversion H1; subst; clear H1.

rename st' into st''. simpl in H5.

(* The invariant is true before the loop runs... *)

remember (update (update st Z (st X)) Y 1) as st'.

assert (fact_invariant (st X) st').

subst. unfold fact_invariant, update. simpl. omega.

(* ...so when the loop is done running, the invariant

is maintained *)

assert (fact_invariant (st X) st'').

apply fact_loop_preserves_invariant with st'; assumption.

unfold fact_invariant in H0.

(* Finally, if the loop terminated, then Z is 0; so Y must be

factorial of X *)

apply guard_false_after_loop in H5. simpl in H5.

destruct (st'' Z).

Case "st'' Z = 0". simpl in H0. omega.

Case "st'' Z > 0 (impossible)". inversion H5.

Qed.

One might wonder whether all this work with poking at states and
unfolding definitions could be ameliorated with some more powerful
lemmas and/or more uniform reasoning principles... Indeed, this is
exactly the topic of the upcoming Hoare chapter!

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

Prove a specification for subtract_slowly, using the above specification of fact_com and the invariant below as guides.Definition ss_invariant (x:nat) (z:nat) (st:state) :=

minus (st Z) (st X) = minus z x.

(* FILL IN HERE *)

☐

# 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, recommended (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:

[] | 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 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

| 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.

Fixpoint s_compile (e : aexp) : list sinstr :=

(* FILL IN HERE *) admit.

(*

Example s_compile1 :

s_compile (AMinus (AId X) (AMult (ANum 2) (AId Y)))

= SLoad X, SPush 2, SLoad Y, SMult, SMinus.

Proof. reflexivity. Qed.

*)

Finally, prove the following theorem, stating that the compile
function behaves correctly. You will need to start by stating a
more general lemma to get a usable induction hypothesis.

(* FILL IN HERE *)

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

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

Proof.

(* FILL IN HERE *) Admitted.

☐