(** * ImpCEvalFun: Evaluation Function for Imp *)
(** We saw in the [Imp] chapter how a naive approach to defining a
function representing evaluation for Imp runs into difficulties.
There, we adopted the solution of changing from a functional to a
relational definition of evaluation. In this optional chapter, we
consider strategies for getting the functional approach to
work. *)
(* ################################################################# *)
(** * A Broken Evaluator *)
(* IMPORTS *)
Require Import Coq.omega.Omega.
Require Import Coq.Arith.Arith.
Require Import Imp.
Require Import Maps.
(* /IMPORTS *)
(** Here was our first try at an evaluation function for commands,
omitting [WHILE]. *)
Fixpoint ceval_step1 (st : state) (c : com) : state :=
match c with
| SKIP =>
st
| l ::= a1 =>
t_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.
(** As we remarked in chapter [Imp], in a traditional functional
programming language like ML or Haskell we could write the WHILE
case as follows:
| WHILE b1 DO c1 END => if (beval st b1) then ceval_step1 st (c1;;
WHILE b1 DO c1 END) else st
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 changed
[ceval_step1] function applied to the [loop] program from [Imp.v]
would never terminate. Since Coq is not just a functional
programming language, but also a consistent logic, any potentially
non-terminating function needs to be rejected. Here is an
invalid(!) Coq program showing what would go wrong if Coq allowed
non-terminating recursive functions:
Fixpoint loop_false (n : nat) : False := loop_false n.
That is, propositions like [False] would become
provable (e.g., [loop_false 0] would be a proof of [False]), which
would be a disaster for Coq's logical consistency.
Thus, because it doesn't terminate on all inputs, the full version
of [ceval_step1] cannot be written in Coq -- at least not without
one additional trick... *)
(* ################################################################# *)
(** * A Step-Indexed Evaluator *)
(** The trick we need is to pass an _additional_ parameter to the
evaluation function that tells it how long to run. Informally, we
start the evaluator with a certain amount of "gas" in its tank,
and we allow it to run until either it terminates in the usual way
_or_ it runs out of gas, at which point we simply stop evaluating
and say that the final result is the empty memory. (We could also
say that the result is the current state at the point where the
evaluator runs out fo gas -- it doesn't really matter because the
result is going to be wrong in either case!) *)
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 =>
t_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.
One thing that is not so nice about this evaluator is that we
can't tell, from its result, whether it stopped because the
program terminated normally or because it ran out of gas. Our
next version returns an [option state] instead of just a [state],
so that we can distinguish between normal and abnormal
termination. *)
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 (t_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 version by introducing a
bit of auxiliary notation to hide the plumbing involved in
repeatedly matching against optional states. *)
Notation "'LETOPT' x <== e1 'IN' e2"
:= (match e1 with
| Some x => e2
| None => None
end)
(right associativity, at level 60).
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 (t_update st l (aeval st a1))
| c1 ;; c2 =>
LETOPT st' <== ceval_step st c1 i' IN
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 LETOPT st' <== ceval_step st c1 i' IN
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.
(* Compute
(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, recommended (pup_to_n) *)
(** 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. *)
Definition pup_to_n : com
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
(*
Example pup_to_n_1 :
test_ceval (t_update empty_state X 5) pup_to_n
= Some (0, 15, 0).
Proof. reflexivity. Qed.
*)
(** [] *)
(** **** Exercise: 2 stars, optional (peven) *)
(** 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 *)
(** [] *)
(* ################################################################# *)
(** * Relational vs. Step-Indexed Evaluation *)
(** As for arithmetic and boolean expressions, we'd hope that
the two alternative definitions of evaluation would actually
amount to the same thing in the end. This section shows that this
is the case. *)
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' ].
- (* i = 0 -- contradictory *)
intros c st st' H. inversion H.
- (* i = S i' *)
intros c st st' H.
destruct c;
simpl in H; inversion H; subst; clear H.
+ (* SKIP *) apply E_Skip.
+ (* ::= *) apply E_Ass. reflexivity.
+ (* ;; *)
destruct (ceval_step st c1 i') eqn:Heqr1.
* (* Evaluation of r1 terminates normally *)
apply E_Seq with s.
apply IHi'. rewrite Heqr1. reflexivity.
apply IHi'. simpl in H1. assumption.
* (* Otherwise -- contradiction *)
inversion H1.
+ (* IFB *)
destruct (beval st b) eqn:Heqr.
* (* r = true *)
apply E_IfTrue. rewrite Heqr. reflexivity.
apply IHi'. assumption.
* (* r = false *)
apply E_IfFalse. rewrite Heqr. reflexivity.
apply IHi'. assumption.
+ (* WHILE *) destruct (beval st b) eqn :Heqr.
* (* r = true *)
destruct (ceval_step st c i') eqn:Heqr1.
{ (* r1 = Some s *)
apply E_WhileLoop with s. rewrite Heqr.
reflexivity.
apply IHi'. rewrite Heqr1. reflexivity.
apply IHi'. simpl in H1. assumption. }
{ (* r1 = None *) inversion H1. }
* (* 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.
(* FILL IN HERE *)
[]
*)
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.
- (* i1 = 0 *)
simpl in Hceval. inversion Hceval.
- (* i1 = S i1' *)
destruct i2 as [|i2']. inversion Hle.
assert (Hle': i1' <= i2') by omega.
destruct c.
+ (* SKIP *)
simpl in Hceval. inversion Hceval.
reflexivity.
+ (* ::= *)
simpl in Hceval. inversion Hceval.
reflexivity.
+ (* ;; *)
simpl in Hceval. simpl.
destruct (ceval_step st c1 i1') eqn:Heqst1'o.
* (* st1'o = Some *)
apply (IHi1' i2') in Heqst1'o; try assumption.
rewrite Heqst1'o. simpl. simpl in Hceval.
apply (IHi1' i2') in Hceval; try assumption.
* (* st1'o = None *)
inversion Hceval.
+ (* IFB *)
simpl in Hceval. simpl.
destruct (beval st b); apply (IHi1' i2') in Hceval;
assumption.
+ (* WHILE *)
simpl in Hceval. simpl.
destruct (beval st b); try assumption.
destruct (ceval_step st c i1') eqn: Heqst1'o.
* (* st1'o = Some *)
apply (IHi1' i2') in Heqst1'o; try assumption.
rewrite -> Heqst1'o. simpl. simpl in Hceval.
apply (IHi1' i2') in Hceval; try assumption.
* (* i1'o = None *)
simpl in Hceval. inversion Hceval. Qed.
(** **** Exercise: 3 stars, recommended (ceval__ceval_step) *)
(** 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.
induction Hce.
(* 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.
(* ################################################################# *)
(** * Determinism of Evaluation Again *)
(** Using the fact that the relational and step-indexed definition of
evaluation are the same, we can give a slicker proof that the
evaluation _relation_ is deterministic. *)
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.
(** $Date: 2016-10-22 20:19:37 -0400 (Sat, 22 Oct 2016) $ *)