ImpSimple Imperative Programs
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
Require Export SfLib.
Arithmetic and Boolean Expressions
Abstract syntax trees for 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.
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
| b and b
| not b
| a + a
| a - a
| a × a
b ::= true
| false
| a = a
| a ≤ a
| b and b
| not b
Fixpoint aeval (a : aexp) : nat :=
match a with
| ANum n ⇒ n
| APlus a1 a2 ⇒ (aeval a1) + (aeval a2)
| AMinus a1 a2 ⇒ (aeval a1) - (aeval a2)
| AMult a1 a2 ⇒ (aeval a1) × (aeval a2)
end.
Example test_aeval1:
aeval (APlus (ANum 2) (ANum 2)) = 4.
Proof. reflexivity. Qed.
Similarly, evaluating a boolean expression yields a boolean.
Fixpoint beval (b : bexp) : bool :=
match b with
| BTrue ⇒ true
| BFalse ⇒ false
| BEq a1 a2 ⇒ 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.
What does the following expression evaluate to?
(1) true
(2) false
(3) 0
(4) 3
(5) 6
aeval (APlus (ANum 3) (AMinus (ANum 4) (ANum 1)))
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.
Example test_optimize_0plus:
optimize_0plus (APlus (ANum 2)
(APlus (ANum 0)
(APlus (ANum 0) (ANum 1))))
= APlus (ANum 2) (ANum 1).
Proof. reflexivity. Qed.
Theorem optimize_0plus_sound: ∀a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a. induction a.
Case "ANum". reflexivity.
Case "APlus". destruct a1.
SCase "a1 = ANum n". destruct n.
SSCase "n = 0". simpl. apply IHa2.
SSCase "n ≠ 0". simpl. rewrite IHa2. reflexivity.
SCase "a1 = APlus a1_1 a1_2".
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
SCase "a1 = AMinus a1_1 a1_2".
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
SCase "a1 = AMult a1_1 a1_2".
simpl. simpl in IHa1. rewrite IHa1.
rewrite IHa2. reflexivity.
Case "AMinus".
simpl. rewrite IHa1. rewrite IHa2. reflexivity.
Case "AMult".
simpl. rewrite IHa1. rewrite IHa2. reflexivity. Qed.
Coq Automation
Tacticals
The repeat Tactical
Theorem ev100 : ev 100.
Proof.
repeat (apply ev_SS). (* applies ev_SS 50 times,
until apply ev_SS fails *)
apply ev_0.
Qed.
(* Print ev100. *)
The try Tactical
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.
The ; Tactical (Simple Form)
Lemma foo : ∀n, ble_nat 0 n = true.
Proof.
intros.
destruct n.
(* Leaves two subgoals, which are discharged identically... *)
Case "n=0". simpl. reflexivity.
Case "n=Sn'". simpl. reflexivity.
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. (* and 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': ∀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).
(* The remaining cases -- ANum and APlus -- are different *)
Case "ANum". reflexivity.
Case "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. *)
SCase "a1 = ANum n". destruct n;
simpl; rewrite IHa2; reflexivity. Qed.
This proof can be further improved by removing the trivial
first case (for e = ANum n).
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. *)
Case "APlus".
destruct a1; try (simpl; simpl in IHa1; rewrite IHa1;
rewrite IHa2; reflexivity).
SCase "a1 = ANum n". destruct n;
simpl; rewrite IHa2; reflexivity. Qed.
Defining New Tactic Notations
- Ltac: scripting language for tactics
- Tactic Notation: syntax extension for tactics
- OCaml tactic scripting API (only for wizards)
Tactic Notation "simpl_and_try" tactic(c) :=
simpl;
try c.
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 a is a variable of type aexp, then doing
aexp_cases (induction a) Case
will perform an induction on a (the same as if we had just typed
induction a) 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''': ∀a,
aeval (optimize_0plus a) = aeval a.
Proof.
intros a.
aexp_cases (induction a) Case;
try (simpl; rewrite IHa1; rewrite IHa2; 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 a1) SCase;
try (simpl; simpl in IHa1;
rewrite IHa1; rewrite IHa2; reflexivity).
SCase "ANum". destruct n;
simpl; rewrite IHa2; reflexivity. Qed.
The omega Tactic
- numeric constants, addition (+ and S), subtraction (-
and pred), and multiplication by constants (this is what
makes it Presburger arithmetic),
- equality (= and ≠) and 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.
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
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).
A standard notation for "evaluates to":
Notation "e '⇓' n" := (aevalR e n) : type_scope.
If we "reserve" the notation in advance, we can also use it
in the definition:
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" ].
Write down a term with the following type:
(AMinus (ANum 3) (AMinus (ANum 2) (ANum 1))) ⇓ 2
Inference Rule Notation
| 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 |
- rule name corresponds to a constructor name
- above the line are premises
- below the line is conclusion
- metavariables like e1 and n1 are implicitly universally quantified
- the whole collection of rules is implicitly wrapped in an Inductive (sometimes we write this slightly more explicitly, as "...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 |
Here, again, is the Coq definition of the beval function:
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.
Write out a corresponding definition of boolean evaluation as a
relation (in inference rule notation).
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.
Answer to the above:
(E_BTrue) | |
BTrue ⇓ true |
(E_BFalse) | |
BFalse ⇓ false |
e1 ⇓ n1 | |
e2 ⇓ n2 | |
beq_nat n1 n2 = b | (E_BEq) |
BEq e1 e2 ⇓ b |
e1 ⇓ n1 | |
e2 ⇓ n2 | (E_BLe) |
BEq e1 e2 ⇓ ble_nat n1 n2 |
e1 ⇓ b1 | (E_BNot) |
BNot e1 ⇓ negb b1 |
e1 ⇓ b1 | |
e2 ⇓ b2 | (E_BAnd) |
BAnd e1 e2 ⇓ andb b1 b2 |
Equivalence of the Definitions
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.
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
use of tacticals...
Theorem aeval_iff_aevalR' : ∀a n,
(a ⇓ n) ↔ aeval a = n.
Proof.
(* WORK IN CLASS *) Admitted.
Computational vs. Relational Definitions
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.
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) → (mult n2 n3 = n1) → (ADiv a1 a2) ⇓ n3
where "a '⇓' n" := (aevalR a n) : type_scope.
Suppose, instead, that we want to extend the arithmetic operations
by a nondeterministic number generator any:
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 (because 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.
Expressions With Variables
Identifiers
We define a new inductive datatype Id so that we won't confuse
identifiers and numbers. We use sumbool to define a computable
equality operator on Id.
Inductive id : Type :=
Id : nat → id.
Theorem eq_id_dec : ∀id1 id2 : id, {id1 = id2} + {id1 ≠ id2}.
Proof.
intros id1 id2.
destruct id1 as [n1]. destruct id2 as [n2].
destruct (eq_nat_dec n1 n2) as [Heq | Hneq].
Case "n1 = n2".
left. rewrite Heq. reflexivity.
Case "n1 ≠ n2".
right. intros contra. inversion contra. apply Hneq. apply H0.
Defined.
The following lemmas will be useful for rewriting terms involving eq_id_dec.
Lemma eq_id : ∀(T:Type) x (p q:T),
(if eq_id_dec x x then p else q) = p.
Proof.
intros.
destruct (eq_id_dec x x).
Case "x = x".
reflexivity.
Case "x ≠ x (impossible)".
apply ex_falso_quodlibet; apply n; reflexivity. Qed.
Lemma neq_id : ∀(T:Type) x y (p q:T), x ≠ y →
(if eq_id_dec x y then p else q) = q.
Proof.
(* FILL IN HERE *) Admitted.
(if eq_id_dec x y then p else q) = q.
Proof.
(* FILL IN HERE *) Admitted.
☐
States
Definition state := id → nat.
Definition empty_state : state :=
fun _ ⇒ 0.
Definition update (st : state) (x : id) (n : nat) : state :=
fun x' ⇒ if eq_id_dec 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,
x2 ≠ x1 →
(update st x2 n) x1 = (st x1).
Proof.
(* FILL IN HERE *) Admitted.
x2 ≠ x1 →
(update st x2 n) x1 = (st x1).
Proof.
(* FILL IN HERE *) Admitted.
Theorem update_shadow : ∀n1 n2 x1 x2 (st : state),
(update (update st x2 n1) x2 n2) x1 = (update st x2 n2) x1.
Proof.
(* FILL IN HERE *) Admitted.
(update (update st x2 n1) x2 n2) x1 = (update st x2 n2) x1.
Proof.
(* FILL IN HERE *) Admitted.
Theorem update_same : ∀n1 x1 x2 (st : state),
st x1 = n1 →
(update st x1 n1) x2 = st x2.
Proof.
(* FILL IN HERE *) Admitted.
st x1 = n1 →
(update st x1 n1) x2 = st x2.
Proof.
(* FILL IN HERE *) Admitted.
Theorem update_permute : ∀n1 n2 x1 x2 x3 st,
x2 ≠ x1 →
(update (update st x2 n1) x1 n2) x3 = (update (update st x1 n2) x2 n1) x3.
Proof.
(* FILL IN HERE *) Admitted.
x2 ≠ x1 →
(update (update st x2 n1) x1 n2) x3 = (update (update st x1 n2) x2 n1) x3.
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.
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) (a : aexp) : nat :=
match a 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) (b : bexp) : bool :=
match b 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.
Syntax
c ::= SKIP
| x ::= a
| c ;; c
| WHILE b DO c END
| IFB b THEN c ELSE c FI
For example, here's the factorial function in Imp.
| x ::= a
| c ;; c
| WHILE b DO c END
| IFB b THEN c ELSE c 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' 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.
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
Evaluation as a Function (Failed Attempt)
Fixpoint ceval_fun_no_while (st : state) (c : com) : state :=
match c with
| SKIP ⇒
st
| x ::= a1 ⇒
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 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 (see chapter ImpCEvalFun if curious).
Fixpoint ceval_fun (st : state) (c : com) : state := match c with ... | WHILE b DO c END => if (beval st b1) then ceval_fun st (c1; 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.
Evaluation as a Relation
(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 b DO c END / st ⇓ st |
beval st b1 = true | |
c / st ⇓ st' | |
WHILE b DO c END / st' ⇓ st'' | (E_WhileLoop) |
WHILE b DO c 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' 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').
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.
Is the following proposition provable?
(2) No
(3) Not sure
∀c st st',
(SKIP ; c) / st ⇓ st' →
c / st ⇓ st'
(1) Yes
(SKIP ; c) / st ⇓ st' →
c / st ⇓ st'
Is the following proposition provable?
(2) No
(3) Not sure
∀c1 c2 st st',
(c1;c2) / st ⇓ st' →
c1 / st ⇓ st →
c2 / st ⇓ st'
(1) Yes
(c1;c2) / st ⇓ st' →
c1 / st ⇓ st →
c2 / st ⇓ st'
Is the following proposition provable?
(2) No
(3) Not sure
∀b c st st',
(IFB b THEN c ELSE c FI) / st ⇓ st' →
c / st ⇓ st'
(1) Yes
(IFB b THEN c ELSE c FI) / st ⇓ st' →
c / st ⇓ st'
Is the following proposition provable?
(2) No
(3) Not sure
∀b,
(∀st, beval st b = true) →
∀c st,
~(∃st', (WHILE b DO c END) / st ⇓ st')
(1) Yes
(∀st, beval st b = true) →
∀c st,
~(∃st', (WHILE b DO c END) / st ⇓ st')
Is the following proposition provable?
(2) No
(3) Not sure
∀b c st,
~(∃st', (WHILE b DO c END) / st ⇓ st') →
∀st'', beval st'' b = true
(1) Yes
~(∃st', (WHILE b DO c END) / st ⇓ st') →
∀st'', beval st'' b = true
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 false".
reflexivity.
SCase "b1 evaluates to true (contradiction)".
rewrite H in H2. inversion H2.
Case "E_WhileLoop".
SCase "b1 evaluates to false (contradiction)".
rewrite H in H4. inversion H4.
SCase "b1 evaluates to true".
assert (st' = st'0) as EQ1.
SSCase "Proof of assertion". apply IHE1_1; assumption.
subst st'0.
apply IHE1_2. assumption. Qed.
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 false".
reflexivity.
SCase "b1 evaluates to true (contradiction)".
rewrite H in H2. inversion H2.
Case "E_WhileLoop".
SCase "b1 evaluates to false (contradiction)".
rewrite H in H4. inversion H4.
SCase "b1 evaluates to true".
assert (st' = st'0) as EQ1.
SSCase "Proof of assertion". apply IHE1_1; assumption.
subst st'0.
apply IHE1_2. assumption. Qed.
Reasoning About Imp Programs
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.
Additional Exercises
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:
[] | 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 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 :=
(* 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.
After you've defined s_compile, uncomment the following to test
that it works.
(*
Example s_compile1 :
s_compile (AMinus (AId X) (AMult (ANum 2) (AId Y)))
= SLoad X; SPush 2; SLoad Y; SMult; SMinus.
Proof. reflexivity. Qed.
*)
☐
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; the main
theorem will then be a simple corollary of this lemma.
Exercise: 3 stars, advanced (stack_compiler_correct)
The task of this exercise is to prove the correctness of the calculator 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 may find it useful to go back and change your implementation!)Theorem s_compile_correct : ∀(st : state) (e : aexp),
s_execute st [] (s_compile e) = [ aeval st e ].
Proof.
(* FILL IN HERE *) Admitted.
☐