# Hoare2Hoare Logic, Part II

Set Warnings "-notation-overridden,-parsing".
From Coq Require Import Strings.String.
From PLF Require Import Maps.
From Coq Require Import Bool.Bool.
From Coq Require Import Arith.Arith.
From Coq Require Import Arith.EqNat.
From Coq Require Import Arith.PeanoNat. Import Nat.
From Coq Require Import omega.Omega.
From PLF Require Import Hoare.
From PLF Require Import Imp.

# Decorated Programs

The beauty of Hoare Logic is that it is compositional: the structure of proofs exactly follows the structure of programs.
This suggests that we can record the essential ideas of a proof (informally, and leaving out some low-level calculational details) by "decorating" a program with appropriate assertions on each of its commands.
Such a decorated program carries within it an argument for its own correctness.
For example, consider the program:
X ::= m;;
Z ::= p;
WHILE ~(X = 0) DO
Z ::= Z - 1;;
X ::= X - 1
END
(Note the parameters m and p, which stand for fixed-but-arbitrary numbers. Formally, they are simply Coq variables of type nat.) Here is one possible specification for this program:
{True }}
X ::= m;;
Z ::= p;
WHILE ~(X = 0) DO
Z ::= Z - 1;;
X ::= X - 1
END
{Z = p - m }}
Here is a decorated version of the program, embodying a proof of this specification:
{True }->>
{m = m }}
X ::= m;;
{X = m }->>
{X = m ∧ p = p }}
Z ::= p;
{X = m ∧ Z = p }->>
{Z - X = p - m }}
WHILE ~(X = 0) DO
{Z - X = p - m ∧ X ≠ 0 }->>
{{ (Z - 1) - (X - 1) = p - m }}
Z ::= Z - 1;;
{Z - (X - 1) = p - m }}
X ::= X - 1
{Z - X = p - m }}
END
{Z - X = p - m ∧ ¬(X ≠ 0) }->>
{Z = p - m }}
Concretely, a decorated program consists of the program text interleaved with assertions (either a single assertion or possibly two assertions separated by an implication).
To check that a decorated program represents a valid proof, we check that each individual command is locally consistent with its nearby assertions in the following sense:
• SKIP is locally consistent if its precondition and postcondition are the same:
{P }SKIP {P }}
• The sequential composition of c1 and c2 is locally consistent (with respect to assertions P and R) if c1 is locally consistent (with respect to P and Q) and c2 is locally consistent (with respect to Q and R):
{P }c1;; {Q }c2 {R }}
• An assignment is locally consistent if its precondition is the appropriate substitution of its postcondition:
{P [X > a}}
X ::= a
{P }}
• A conditional is locally consistent (with respect to assertions P and Q) if the assertions at the top of its "then" and "else" branches are exactly P b and P ¬b and if its "then" branch is locally consistent (with respect to P b and Q) and its "else" branch is locally consistent (with respect to P ¬b and Q):
{P }}
TEST b THEN
{P ∧ b }}
c1
{Q }}
ELSE
{P ∧ ¬b }}
c2
{Q }}
FI
{Q }}
• A while loop with precondition P is locally consistent if its postcondition is P ¬b, if the pre- and postconditions of its body are exactly P b and P, and if its body is locally consistent:
{P }}
WHILE b DO
{P ∧ b }}
c1
{P }}
END
{P ∧ ¬b }}
• A pair of assertions separated by ->> is locally consistent if the first implies the second:
{P }->>
{P' }}
This corresponds to the application of hoare_consequence, and it is the only place in a decorated program where checking whether decorations are correct is not fully mechanical and syntactic, but rather may involve logical and/or arithmetic reasoning.
These local consistency conditions essentially describe a procedure for verifying the correctness of a given proof. This verification involves checking that every single command is locally consistent with the accompanying assertions.
If we are instead interested in finding a proof for a given specification, we need to discover the right assertions. This can be done in an almost mechanical way, with the exception of finding loop invariants. In the remainder of this section we explain in detail how to construct decorations for several short programs, all of which are loop-free or have simple loop invariants. We defer a full discussion of finding loop invariants to the next section.

## Example: Swapping Using Addition and Subtraction

Here is a program that swaps the values of two variables using addition and subtraction (instead of by assigning to a temporary variable).
X ::= X + Y;;
Y ::= X - Y;;
X ::= X - Y
We can prove (informally) using decorations that this program is correct — i.e., it always swaps the values of variables X and Y.
(1)     {X = m ∧ Y = n }->>
(2)     {{ (X + Y) - ((X + Y) - Y) = n ∧ (X + Y) - Y = m }}
X ::= X + Y;;
(3)     {X - (X - Y) = n ∧ X - Y = m }}
Y ::= X - Y;;
(4)     {X - Y = n ∧ Y = m }}
X ::= X - Y
(5)     {X = n ∧ Y = m }}
These decorations can be constructed as follows:
• We begin with the undecorated program (the unnumbered lines).
• We add the specification — i.e., the outer precondition (1) and postcondition (5). In the precondition, we use parameters m and n to remember the initial values of variables X and Y so that we can refer to them in the postcondition (5).
• We work backwards, mechanically, starting from (5) and proceeding until we get to (2). At each step, we obtain the precondition of the assignment from its postcondition by substituting the assigned variable with the right-hand-side of the assignment. For instance, we obtain (4) by substituting X with X - Y in (5), and we obtain (3) by substituting Y with X - Y in (4).
• Finally, we verify that (1) logically implies (2) — i.e., that the step from (1) to (2) is a valid use of the law of consequence. For this we substitute X by m and Y by n and calculate as follows:
(m + n) - ((m + n) - n) = n ∧ (m + n) - n = m
(m + n) - m = n ∧ m = m
n = n ∧ m = m
Note that, since we are working with natural numbers rather than fixed-width machine integers, we don't need to worry about the possibility of arithmetic overflow anywhere in this argument. This makes life quite a bit simpler!

## Example: Simple Conditionals

Here is a simple decorated program using conditionals:
(1)     {{True}}
TEST X ≤ Y THEN
(2)       {{True ∧ X ≤ Y}->>
(3)       {{(Y - X) + X = Y ∨ (Y - X) + Y = X}}
Z ::= Y - X
(4)       {{Z + X = Y ∨ Z + Y = X}}
ELSE
(5)       {{True ∧ ~(X ≤ Y}->>
(6)       {{(X - Y) + X = Y ∨ (X - Y) + Y = X}}
Z ::= X - Y
(7)       {{Z + X = Y ∨ Z + Y = X}}
FI
(8)     {{Z + X = Y ∨ Z + Y = X}}
These decorations were constructed as follows:
• We follow the format dictated by the hoare_if rule and copy the postcondition (8) to (4) and (7). We conjoin the precondition (1) with the guard of the conditional to obtain (2). We conjoin (1) with the negated guard of the conditional to obtain (5).
• In order to use the assignment rule and obtain (3), we substitute Z by Y - X in (4). To obtain (6) we substitute Z by X - Y in (7).
• Finally, we verify that (2) implies (3) and (5) implies (6). Both of these implications crucially depend on the ordering of X and Y obtained from the guard. For instance, knowing that X Y ensures that subtracting X from Y and then adding back X produces Y, as required by the first disjunct of (3). Similarly, knowing that ¬ (X Y) ensures that subtracting Y from X and then adding back Y produces X, as needed by the second disjunct of (6). Note that n - m + m = n does not hold for arbitrary natural numbers n and m (for example, 3 - 5 + 5 = 5).

#### Exercise: 2 stars, standard (if_minus_plus_reloaded)

Fill in valid decorations for the following program:
{True }}
TEST X ≤ Y THEN
{{                         }->>
{{                         }}
Z ::= Y - X
{{                         }}
ELSE
{{                         }->>
{{                         }}
Y ::= X + Z
{{                         }}
FI
{Y = X + Z }}
(* Do not modify the following line: *)

## Example: Reduce to Zero

Here is a WHILE loop that is so simple it needs no invariant (i.e., the invariant True will do the job).
(1)      {True }}
WHILE ~(X = 0) DO
(2)        {True ∧ X ≠ 0 }->>
(3)        {True }}
X ::= X - 1
(4)        {True }}
END
(5)      {True ∧ ~(X ≠ 0) }->>
(6)      {X = 0 }}
The decorations can be constructed as follows:
• Following the format dictated by the hoare_while rule, we copy (1) to (4). We conjoin (1) with the guard to obtain (2). The guard is a Boolean expression ~(X = 0), which for simplicity we express in assertion (2) as X 0. We also conjoin (1) with the negation of the guard to obtain (5). Note that, because the outer postcondition (6) does not syntactically match (5), we need a trivial use of the consequence rule from (5) to (6).
• Assertion (3) is the same as (4), because X does not appear in 4, so the substitution in the assignment rule is trivial.
• Finally, the implication between (2) and (3) is also trivial.
From an informal proof in the form of a decorated program, it is easy to read off a formal proof using the Coq versions of the Hoare rules. Note that we do not unfold the definition of hoare_triple anywhere in this proof — the idea is to use the Hoare rules as a self-contained logic for reasoning about programs.
Definition reduce_to_zero' : com :=
WHILE ~(X = 0) DO
X ::= X - 1
END.

Theorem reduce_to_zero_correct' :
{{fun stTrue}}
reduce_to_zero'
{{fun stst X = 0}}.
Proof.
unfold reduce_to_zero'.
(* First we need to transform the postcondition so
that hoare_while will apply. *)

eapply hoare_consequence_post.
apply hoare_while.
- (* Loop body preserves invariant *)
(* Need to massage precondition before hoare_asgn applies *)
eapply hoare_consequence_pre. apply hoare_asgn.
(* Proving trivial implication (2) ->> (3) *)
intros st [HT Hbp]. unfold assn_sub. constructor.
- (* Invariant and negated guard imply postcondition *)
intros st [Inv GuardFalse].
unfold bassn in GuardFalse. simpl in GuardFalse.
rewrite not_true_iff_false in GuardFalse.
rewrite negb_false_iff in GuardFalse.
apply eqb_eq in GuardFalse.
apply GuardFalse. Qed.

## Example: Division

The following Imp program calculates the integer quotient and remainder of two numbers m and n that are arbitrary constants in the program.
X ::= m;;
Y ::= 0;;
WHILE n ≤ X DO
X ::= X - n;;
Y ::= Y + 1
END;
In we replace m and n by concrete numbers and execute the program, it will terminate with the variable X set to the remainder when m is divided by n and Y set to the quotient.
In order to give a specification to this program we need to remember that dividing m by n produces a remainder X and a quotient Y such that n * Y + X = m X < n.
It turns out that we get lucky with this program and don't have to think very hard about the loop invariant: the invariant is just the first conjunct n * Y + X = m, and we can use this to decorate the program.
(1)    {True }->>
(2)    {n * 0 + m = m }}
X ::= m;;
(3)    {n * 0 + X = m }}
Y ::= 0;;
(4)    {n * Y + X = m }}
WHILE n ≤ X DO
(5)      {n * Y + X = m ∧ n ≤ X }->>
(6)      {n * (Y + 1) + (X - n) = m }}
X ::= X - n;;
(7)      {n * (Y + 1) + X = m }}
Y ::= Y + 1
(8)      {n * Y + X = m }}
END
(9)    {n * Y + X = m ∧ ¬(n ≤ X}->>
(10)    {n * Y + X = m ∧ X < n }}
Assertions (4), (5), (8), and (9) are derived mechanically from the invariant and the loop's guard. Assertions (8), (7), and (6) are derived using the assignment rule going backwards from (8) to (6). Assertions (4), (3), and (2) are again backwards applications of the assignment rule.
Now that we've decorated the program it only remains to check that the uses of the consequence rule are correct — i.e., that (1) implies (2), that (5) implies (6), and that (9) implies (10). This is indeed the case, so we have a valid decorated program.

# Finding Loop Invariants

Once the outermost precondition and postcondition are chosen, the only creative part in verifying programs using Hoare Logic is finding the right loop invariants. The reason this is difficult is the same as the reason that inductive mathematical proofs are: strengthening the loop invariant (or the induction hypothesis) means that you have a stronger assumption to work with when trying to establish the postcondition of the loop body (or complete the induction step of the proof), but it also means that the loop body's postcondition (or the statement being proved inductively) is stronger and thus harder to prove!
This section explains how to approach the challenge of finding loop invariants through a series of examples and exercises.

## Example: Slow Subtraction

The following program subtracts the value of X from the value of Y by repeatedly decrementing both X and Y. We want to verify its correctness with respect to the pre- and postconditions shown:
{X = m ∧ Y = n }}
WHILE ~(X = 0) DO
Y ::= Y - 1;;
X ::= X - 1
END
{Y = n - m }}
To verify this program, we need to find an invariant Inv for the loop. As a first step we can leave Inv as an unknown and build a skeleton for the proof by applying the rules for local consistency (working from the end of the program to the beginning, as usual, and without any thinking at all yet).
This leads to the following skeleton:
(1)      {X = m ∧ Y = n }}  ->>             (a)
(2)      {Inv }}
WHILE ~(X = 0) DO
(3)        {Inv ∧ X ≠ 0 }}  ->>              (c)
(4)        {Inv [X > X-1] [Y > Y-1] }}
Y ::= Y - 1;;
(5)        {Inv [X > X-1] }}
X ::= X - 1
(6)        {Inv }}
END
(7)      {Inv ∧ ¬(X ≠ 0) }}  ->>            (b)
(8)      {Y = n - m }}
By examining this skeleton, we can see that any valid Inv will have to respect three conditions:
• (a) it must be weak enough to be implied by the loop's precondition, i.e., (1) must imply (2);
• (b) it must be strong enough to imply the program's postcondition, i.e., (7) must imply (8);
• (c) it must be preserved by each iteration of the loop (given that the loop guard evaluates to true), i.e., (3) must imply (4).
These conditions are actually independent of the particular program and specification we are considering. Indeed, every loop invariant has to satisfy them. One way to find an invariant that simultaneously satisfies these three conditions is by using an iterative process: start with a "candidate" invariant (e.g., a guess or a heuristic choice) and check the three conditions above; if any of the checks fails, try to use the information that we get from the failure to produce another — hopefully better — candidate invariant, and repeat.
For instance, in the reduce-to-zero example above, we saw that, for a very simple loop, choosing True as an invariant did the job. So let's try instantiating Inv with True in the skeleton above and see what we get...
(1)      {X = m ∧ Y = n }->>       (a - OK)
(2)      {True }}
WHILE ~(X = 0) DO
(3)        {True ∧ X ≠ 0 }}  ->>    (c - OK)
(4)        {True }}
Y ::= Y - 1;;
(5)        {True }}
X ::= X - 1
(6)        {True }}
END
(7)      {True ∧ ~(X ≠ 0) }}  ->>       (b - WRONG!)
(8)      {Y = n - m }}
While conditions (a) and (c) are trivially satisfied, condition (b) is wrong, i.e., it is not the case that True X = 0 (7) implies Y = n - m (8). In fact, the two assertions are completely unrelated, so it is very easy to find a counterexample to the implication (say, Y = X = m = 0 and n = 1).
If we want (b) to hold, we need to strengthen the invariant so that it implies the postcondition (8). One simple way to do this is to let the invariant be the postcondition. So let's return to our skeleton, instantiate Inv with Y = n - m, and check conditions (a) to (c) again.
(1)      {X = m ∧ Y = n }}  ->>          (a - WRONG!)
(2)      {Y = n - m }}
WHILE ~(X = 0) DO
(3)        {Y = n - m ∧ X ≠ 0 }}  ->>   (c - WRONG!)
(4)        {Y - 1 = n - m }}
Y ::= Y - 1;;
(5)        {Y = n - m }}
X ::= X - 1
(6)        {Y = n - m }}
END
(7)      {Y = n - m ∧ ~(X ≠ 0) }}  ->>      (b - OK)
(8)      {Y = n - m }}
This time, condition (b) holds trivially, but (a) and (c) are broken. Condition (a) requires that (1) X = m Y = n implies (2) Y = n - m. If we substitute Y by n we have to show that n = n - m for arbitrary m and n, which is not the case (for instance, when m = n = 1). Condition (c) requires that n - m - 1 = n - m, which fails, for instance, for n = 1 and m = 0. So, although Y = n - m holds at the end of the loop, it does not hold from the start, and it doesn't hold on each iteration; it is not a correct invariant.
This failure is not very surprising: the variable Y changes during the loop, while m and n are constant, so the assertion we chose didn't have much chance of being an invariant!
To do better, we need to generalize (8) to some statement that is equivalent to (8) when X is 0, since this will be the case when the loop terminates, and that "fills the gap" in some appropriate way when X is nonzero. Looking at how the loop works, we can observe that X and Y are decremented together until X reaches 0. So, if X = 2 and Y = 5 initially, after one iteration of the loop we obtain X = 1 and Y = 4; after two iterations X = 0 and Y = 3; and then the loop stops. Notice that the difference between Y and X stays constant between iterations: initially, Y = n and X = m, and the difference is always n - m. So let's try instantiating Inv in the skeleton above with Y - X = n - m.
(1)      {X = m ∧ Y = n }}  ->>               (a - OK)
(2)      {Y - X = n - m }}
WHILE ~(X = 0) DO
(3)        {Y - X = n - m ∧ X ≠ 0 }}  ->>    (c - OK)
(4)        {{ (Y - 1) - (X - 1) = n - m }}
Y ::= Y - 1;;
(5)        {Y - (X - 1) = n - m }}
X ::= X - 1
(6)        {Y - X = n - m }}
END
(7)      {Y - X = n - m ∧ ~(X ≠ 0) }}  ->>       (b - OK)
(8)      {Y = n - m }}
Success! Conditions (a), (b) and (c) all hold now. (To verify (c), we need to check that, under the assumption that X 0, we have Y - X = (Y - 1) - (X - 1); this holds for all natural numbers X and Y.)

## Exercise: Slow Assignment

#### Exercise: 2 stars, standard (slow_assignment)

A roundabout way of assigning a number currently stored in X to the variable Y is to start Y at 0, then decrement X until it hits 0, incrementing Y at each step. Here is a program that implements this idea:
{X = m }}
Y ::= 0;;
WHILE ~(X = 0) DO
X ::= X - 1;;
Y ::= Y + 1
END
{Y = m }}
Write an informal decorated program showing that this procedure is correct.
(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_decorations_in_slow_assignment : option (nat*string) := None.

#### Exercise: 3 stars, standard, optional (add_slowly_decoration)

The following program adds the variable X into the variable Z by repeatedly decrementing X and incrementing Z.
WHILE ~(X = 0) DO
Z ::= Z + 1;;
X ::= X - 1
END
Following the pattern of the subtract_slowly example above, pick a precondition and postcondition that give an appropriate specification of add_slowly; then (informally) decorate the program accordingly.
(* FILL IN HERE *)

## Example: Parity

Here is a cute little program for computing the parity of the value initially stored in X (due to Daniel Cristofani).
{X = m }}
WHILE 2 ≤ X DO
X ::= X - 2
END
{X = parity m }}
The mathematical parity function used in the specification is defined in Coq as follows:
Fixpoint parity x :=
match x with
| 0 ⇒ 0
| 1 ⇒ 1
| S (S x') ⇒ parity x'
end.
The postcondition does not hold at the beginning of the loop, since m = parity m does not hold for an arbitrary m, so we cannot use that as an invariant. To find an invariant that works, let's think a bit about what this loop does. On each iteration it decrements X by 2, which preserves the parity of X. So the parity of X does not change, i.e., it is invariant. The initial value of X is m, so the parity of X is always equal to the parity of m. Using parity X = parity m as an invariant we obtain the following decorated program:
{X = m }->>                               (a - OK)
{parity X = parity m }}
WHILE 2 ≤ X DO
{parity X = parity m ∧ 2 ≤ X }}  ->>    (c - OK)
{parity (X-2) = parity m }}
X ::= X - 2
{parity X = parity m }}
END
{parity X = parity m ∧ ~(2 ≤ X}}  ->>       (b - OK)
{X = parity m }}
With this invariant, conditions (a), (b), and (c) are all satisfied. For verifying (b), we observe that, when X < 2, we have parity X = X (we can easily see this in the definition of parity). For verifying (c), we observe that, when 2 X, we have parity X = parity (X-2).

#### Exercise: 3 stars, standard (parity_formal)

Translate the above informal decorated program into a formal proof in Coq. Your proof should use the Hoare logic rules and should not unfold hoare_triple. Refer to reduce_to_zero for an example. You might find the following lemmas to be useful:
Lemma parity_ge_2 : x,
2 ≤ x
parity (x - 2) = parity x.
Proof.
induction x; intro. reflexivity.
destruct x. inversion H. inversion H1.
simpl. rewrite <- minus_n_O. reflexivity.
Qed.

Lemma parity_lt_2 : x,
¬2 ≤ x
parity x = x.
Proof.
intros. induction x. reflexivity. destruct x. reflexivity.
exfalso. apply H. omega.
Qed.

Theorem parity_correct : m,
{{ fun stst X = m }}
WHILE 2 ≤ X DO
X ::= X - 2
END
{{ fun stst X = parity m }}.
Proof.
(* FILL IN HERE *) Admitted.

## Example: Finding Square Roots

The following program computes the (integer) square root of X by naive iteration:
{X=m }}
Z ::= 0;;
WHILE (Z+1)*(Z+1) ≤ X DO
Z ::= Z+1
END
{Z*Zm ∧ m<(Z+1)*(Z+1) }}
As above, we can try to use the postcondition as a candidate invariant, obtaining the following decorated program:
(1)  {X=m }}  ->>           (a - second conjunct of (2) WRONG!)
(2)  {{ 0*0 ≤ m ∧ m<(0+1)*(0+1) }}
Z ::= 0;;
(3)  {Z*Z ≤ m ∧ m<(Z+1)*(Z+1) }}
WHILE (Z+1)*(Z+1) ≤ X DO
(4)    {Z*Zm ∧ (Z+1)*(Z+1)≤X }}  ->>             (c - WRONG!)
(5)    {{ (Z+1)*(Z+1)≤m ∧ m<((Z+1)+1)*((Z+1)+1) }}
Z ::= Z+1
(6)    {Z*Zm ∧ m<(Z+1)*(Z+1) }}
END
(7)  {Z*Zm ∧ m<(Z+1)*(Z+1) ∧ ~((Z+1)*(Z+1)≤X}}  ->> (b - OK)
(8)  {Z*Zm ∧ m<(Z+1)*(Z+1) }}
This didn't work very well: conditions (a) and (c) both failed. Looking at condition (c), we see that the second conjunct of (4) is almost the same as the first conjunct of (5), except that (4) mentions X while (5) mentions m. But note that X is never assigned in this program, so we should always have X=m; we didn't propagate this information from (1) into the loop invariant, but we could!
Also, we don't need the second conjunct of (8), since we can obtain it from the negation of the guard — the third conjunct in (7) — again under the assumption that X=m. This allows us to simplify a bit.
So we now try X=m Z*Z m as the loop invariant:
{X=m }}  ->>                                      (a - OK)
{X=m ∧ 0*0 ≤ m }}
Z ::= 0;
{X=m ∧ Z*Z ≤ m }}
WHILE (Z+1)*(Z+1) ≤ X DO
{X=m ∧ Z*Zm ∧ (Z+1)*(Z+1)≤X }}  ->>        (c - OK)
{X=m ∧ (Z+1)*(Z+1)≤m }}
Z ::= Z + 1
{X=m ∧ Z*Zm }}
END
{X=m ∧ Z*Zm ∧ ~((Z+1)*(Z+1)≤X}}  ->>           (b - OK)
{Z*Zm ∧ m<(Z+1)*(Z+1) }}
This works, since conditions (a), (b), and (c) are now all trivially satisfied.
Very often, if a variable is used in a loop in a read-only fashion (i.e., it is referred to by the program or by the specification and it is not changed by the loop), it is necessary to add the fact that it doesn't change to the loop invariant.

## Example: Squaring

Here is a program that squares X by repeated addition:
{X = m }}
Y ::= 0;;
Z ::= 0;;
WHILE ~(Y = X)  DO
Z ::= Z + X;;
Y ::= Y + 1
END
{Z = m*m }}
The first thing to note is that the loop reads X but doesn't change its value. As we saw in the previous example, it is a good idea in such cases to add X = m to the invariant. The other thing that we know is often useful in the invariant is the postcondition, so let's add that too, leading to the invariant candidate Z = m * m X = m.
{X = m }->>                            (a - WRONG)
{{ 0 = m*m ∧ X = m }}
Y ::= 0;;
{{ 0 = m*m ∧ X = m }}
Z ::= 0;;
{Z = m*m ∧ X = m }}
WHILE ~(Y = XDO
{Z = m*m ∧ X = m ∧ Y ≠ X }->>     (c - WRONG)
{Z+X = m*m ∧ X = m }}
Z ::= Z + X;;
{Z = m*m ∧ X = m }}
Y ::= Y + 1
{Z = m*m ∧ X = m }}
END
{Z = m*m ∧ X = m ∧ ~(Y ≠ X}->>         (b - OK)
{Z = m*m }}
Conditions (a) and (c) fail because of the Z = m*m part. While Z starts at 0 and works itself up to m*m, we can't expect Z to be m*m from the start. If we look at how Z progresses in the loop, after the 1st iteration Z = m, after the 2nd iteration Z = 2*m, and at the end Z = m*m. Since the variable Y tracks how many times we go through the loop, this leads us to derive a new invariant candidate: Z = Y*m X = m.
{X = m }->>                               (a - OK)
{{ 0 = 0*m ∧ X = m }}
Y ::= 0;;
{{ 0 = Y*m ∧ X = m }}
Z ::= 0;;
{Z = Y*m ∧ X = m }}
WHILE ~(Y = XDO
{Z = Y*m ∧ X = m ∧ Y ≠ X }->>        (c - OK)
{Z+X = (Y+1)*m ∧ X = m }}
Z ::= Z + X;
{Z = (Y+1)*m ∧ X = m }}
Y ::= Y + 1
{Z = Y*m ∧ X = m }}
END
{Z = Y*m ∧ X = m ∧ ~(Y ≠ X}->>           (b - OK)
{Z = m*m }}
This new invariant makes the proof go through: all three conditions are easy to check.
It is worth comparing the postcondition Z = m*m and the Z = Y*m conjunct of the invariant. It is often the case that one has to replace parameters with variables — or with expressions involving both variables and parameters, like m - Y — when going from postconditions to invariants.

## Exercise: Factorial

#### Exercise: 3 stars, standard (factorial)

Recall that n! denotes the factorial of n (i.e., n! = 1*2*...*n). Here is an Imp program that calculates the factorial of the number initially stored in the variable X and puts it in the variable Y:
{X = m }}
Y ::= 1 ;;
WHILE ~(X = 0)
DO
Y ::= Y * X ;;
X ::= X - 1
END
{Y = m}}
Fill in the blanks in following decorated program. For full credit, make sure all the arithmetic operations used in the assertions are well-defined on natural numbers.
{X = m }->>
{{                                      }}
Y ::= 1;;
{{                                      }}
WHILE ~(X = 0)
DO   {{                                      }->>
{{                                      }}
Y ::= Y * X;;
{{                                      }}
X ::= X - 1
{{                                      }}
END
{{                                      }->>
{Y = m}}
(* Do not modify the following line: *)
Definition manual_grade_for_decorations_in_factorial : option (nat*string) := None.

## Exercise: Min

#### Exercise: 3 stars, standard (Min_Hoare)

Fill in valid decorations for the following program. For the ->> steps in your annotations, you may rely (silently) on the following facts about min
Lemma lemma1 : x y,
(x≠0 ∧ y≠0) → min x y ≠ 0.
Lemma lemma2 : x y,
min (x-1) (y-1) = (min x y) - 1.
plus standard high-school algebra, as always.
{True }->>
{{                    }}
X ::= a;;
{{                       }}
Y ::= b;;
{{                       }}
Z ::= 0;;
{{                       }}
WHILE ~(X = 0) && ~(Y = 0) DO
{{                                     }->>
{{                                }}
X := X - 1;;
{{                            }}
Y := Y - 1;;
{{                        }}
Z := Z + 1
{{                       }}
END
{{                            }->>
{Z = min a b }}
(* Do not modify the following line: *)
Definition manual_grade_for_decorations_in_Min_Hoare : option (nat*string) := None.

#### Exercise: 3 stars, standard (two_loops)

Here is a very inefficient way of adding 3 numbers:
X ::= 0;;
Y ::= 0;;
Z ::= c;;
X ::= X + 1;;
Z ::= Z + 1
END;;
WHILE ~(Y = bDO
Y ::= Y + 1;;
Z ::= Z + 1
END
Show that it does what it should by filling in the blanks in the following decorated program.
{True }->>
{{                                        }}
X ::= 0;;
{{                                        }}
Y ::= 0;;
{{                                        }}
Z ::= c;;
{{                                        }}
{{                                        }->>
{{                                        }}
X ::= X + 1;;
{{                                        }}
Z ::= Z + 1
{{                                        }}
END;;
{{                                        }->>
{{                                        }}
WHILE ~(Y = bDO
{{                                        }->>
{{                                        }}
Y ::= Y + 1;;
{{                                        }}
Z ::= Z + 1
{{                                        }}
END
{{                                        }->>
{Z = a + b + c }}
(* Do not modify the following line: *)
Definition manual_grade_for_decorations_in_two_loops : option (nat*string) := None.

## Exercise: Power Series

#### Exercise: 4 stars, standard, optional (dpow2_down)

Here is a program that computes the series: 1 + 2 + 2^2 + ... + 2^m = 2^(m+1) - 1
X ::= 0;;
Y ::= 1;;
Z ::= 1;;
WHILE ~(X = mDO
Z ::= 2 * Z;;
Y ::= Y + Z;;
X ::= X + 1
END
Write a decorated program for this.
(* FILL IN HERE *)

# Weakest Preconditions

Some Hoare triples are more interesting than others. For example,
{False }}  X ::= Y + 1  {X ≤ 5 }}
is not very interesting: although it is perfectly valid, it tells us nothing useful. Since the precondition isn't satisfied by any state, it doesn't describe any situations where we can use the command X ::= Y + 1 to achieve the postcondition X 5.
By contrast,
{Y ≤ 4 ∧ Z = 0 }}  X ::= Y + 1 {X ≤ 5 }}
is useful: it tells us that, if we can somehow create a situation in which we know that Y 4 Z = 0, then running this command will produce a state satisfying the postcondition. However, this triple is still not as useful as it could be, because the Z = 0 clause in the precondition actually has nothing to do with the postcondition X 5. The most useful triple (for this command and postcondition) is this one:
{Y ≤ 4 }}  X ::= Y + 1  {X ≤ 5 }}
In other words, Y 4 is the weakest valid precondition of the command X ::= Y + 1 for the postcondition X 5.
In general, we say that "P is the weakest precondition of command c for postcondition Q" if {{P}} c {{Q}} and if, whenever P' is an assertion such that {{P'}} c {{Q}}, it is the case that P' st implies P st for all states st.
Definition is_wp P c Q :=
{{P}} c {{Q}} ∧
P', {{P'}} c {{Q}} → (P' ->> P).
That is, P is the weakest precondition of c for Q if (a) P is a precondition for Q and c, and (b) P is the weakest (easiest to satisfy) assertion that guarantees that Q will hold after executing c.

#### Exercise: 1 star, standard, optional (wp)

What are the weakest preconditions of the following commands for the following postconditions?
1) {{ ? }}  SKIP  {X = 5 }}

2) {{ ? }}  X ::= Y + Z {X = 5 }}

3) {{ ? }}  X ::= Y  {X = Y }}

4) {{ ? }}
TEST X = 0 THEN Y ::= Z + 1 ELSE Y ::= W + 2 FI
{Y = 5 }}

5) {{ ? }}
X ::= 5
{X = 0 }}

6) {{ ? }}
WHILE true DO X ::= 0 END
{X = 0 }}
(* FILL IN HERE *)

#### Exercise: 3 stars, standard (is_wp_formal)

Prove formally, using the definition of hoare_triple, that Y 4 is indeed the weakest precondition of X ::= Y + 1 with respect to postcondition X 5.
Theorem is_wp_example :
is_wp (fun stst Y ≤ 4)
(X ::= Y + 1) (fun stst X ≤ 5).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars, standard (hoare_asgn_weakest)

Show that the precondition in the rule hoare_asgn is in fact the weakest precondition.
Theorem hoare_asgn_weakest : Q X a,
is_wp (Q [X > a]) (X ::= a) Q.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars, standard, optional (hoare_havoc_weakest)

Show that your havoc_pre rule from the himp_hoare exercise in the Hoare chapter returns the weakest precondition.
Module Himp2.
Import Himp.

Lemma hoare_havoc_weakest : (P Q : Assertion) (X : string),
{{ P }} HAVOC X {{ Q }} →
P ->> havoc_pre X Q.
Proof.
(* FILL IN HERE *) Admitted.

# Formal Decorated Programs

Our informal conventions for decorated programs amount to a way of displaying Hoare triples, in which commands are annotated with enough embedded assertions that checking the validity of a triple is reduced to simple logical and algebraic calculations showing that some assertions imply others. In this section, we show that this informal presentation style can actually be made completely formal and indeed that checking the validity of decorated programs can mostly be automated.

## Syntax

The first thing we need to do is to formalize a variant of the syntax of commands with embedded assertions. We call the new commands decorated commands, or dcoms.
The choice of exactly where to put assertions in the definition of dcom is a bit subtle. The simplest thing to do would be to annotate every dcom with a precondition and postcondition. But this would result in very verbose programs with a lot of repeated annotations: for example, a program like SKIP;SKIP would have to be annotated as
{{P}} ({{P}SKIP {{P}}) ;; ({{P}SKIP {{P}}) {{P}},
with pre- and post-conditions on each SKIP, plus identical pre- and post-conditions on the semicolon!
We don't want both preconditions and postconditions on each command, because a sequence of two commands would contain redundant decorations—the postcondition of the first likely being the same as the precondition of the second.
Instead, we decorate commands as follows:
• The post-condition expected by each dcom d is embedded in d.
• The pre-condition is supplied by the context.
Inductive dcom : Type :=
| DCSkip : Assertiondcom
| DCSeq : dcomdcomdcom
| DCAsgn : stringaexpAssertiondcom
| DCIf : bexpAssertiondcomAssertiondcom
→ Assertiondcom
| DCWhile : bexpAssertiondcomAssertiondcom
| DCPre : Assertiondcomdcom
| DCPost : dcomAssertiondcom.
DCPre is used to provide the weakened precondition from the rule of consequence. To provide the initial precondition at the very top of the program, we use Decorated:
Inductive decorated : Type :=
| Decorated : Assertiondcomdecorated.
To avoid clashing with the existing Notation definitions for ordinary commands, we introduce these notations in a special scope called dcom_scope, and we Open this scope for the remainder of the file.
Delimit Scope default with default.

Notation "'SKIP' {{ P }}"
:= (DCSkip P)
(at level 10) : dcom_scope.
Notation "l '::=' a {{ P }}"
:= (DCAsgn l a P)
(at level 60, a at next level) : dcom_scope.
Notation "'WHILE' b 'DO' {{ Pbody }} d 'END' {{ Ppost }}"
:= (DCWhile b Pbody d Ppost)
(at level 80, right associativity) : dcom_scope.
Notation "'TEST' b 'THEN' {{ P }} d 'ELSE' {{ P' }} d' 'FI' {{ Q }}"
:= (DCIf b P d P' d' Q)
(at level 80, right associativity) : dcom_scope.
Notation "'->>' {{ P }} d"
:= (DCPre P d)
(at level 90, right associativity) : dcom_scope.
Notation "d '->>' {{ P }}"
:= (DCPost d P)
(at level 80, right associativity) : dcom_scope.
Notation " d ;; d' "
:= (DCSeq d d')
(at level 80, right associativity) : dcom_scope.
Notation "{{ P }} d"
:= (Decorated P d)
(at level 90) : dcom_scope.

Delimit Scope dcom_scope with dcom.
Open Scope dcom_scope.

Example dec0 :=
SKIP {{ fun stTrue }}.
Example dec1 :=
WHILE true DO {{ fun stTrue }} SKIP {{ fun stTrue }} END
{{ fun stTrue }}.

Example dec_while : decorated :=
{{ fun stTrue }}
WHILE ~(X = 0)
DO
{{ fun stTruest X ≠ 0}}
X ::= X - 1
{{ fun _True }}
END
{{ fun stTruest X = 0}} ->>
{{ fun stst X = 0 }}.
It is easy to go from a dcom to a com by erasing all annotations.
Fixpoint extract (d : dcom) : com :=
match d with
| DCSkip _SKIP
| DCSeq d1 d2 ⇒ (extract d1 ;; extract d2)
| DCAsgn X a _X ::= a
| DCIf b _ d1 _ d2 _TEST b THEN extract d1 ELSE extract d2 FI
| DCWhile b _ d _WHILE b DO extract d END
| DCPre _ dextract d
| DCPost d _extract d
end.

Definition extract_dec (dec : decorated) : com :=
match dec with
| Decorated P dextract d
end.

Print dec_while.
(*
dec_while =
{{fun _ : state => True}} (WHILE ~ X = 0
DO {{fun st : state => True /\ st X <> 0}
X ::= X - 1 {{fun _ : state => True}}
END {{fun st : state => True /\ st X = 0}}) ->>
{{fun st : state => st X = 0}}
: decorated
*)

Compute (extract_dec dec_while).
(*
= (WHILE ~ "X"%string = 0 DO "X" ::= "X"%string - 1 END)%imp
: com
*)

It is straightforward to extract the precondition and postcondition from a decorated program.
Fixpoint post (d : dcom) : Assertion :=
match d with
| DCSkip PP
| DCSeq d1 d2post d2
| DCAsgn X a QQ
| DCIf _ _ d1 _ d2 QQ
| DCWhile b Pbody c PpostPpost
| DCPre _ dpost d
| DCPost c QQ
end.

Definition pre_dec (dec : decorated) : Assertion :=
match dec with
| Decorated P dP
end.

Definition post_dec (dec : decorated) : Assertion :=
match dec with
| Decorated P dpost d
end.

Print dec_while.
(*
dec_while =
{{fun _ : state => True}} (WHILE ~ X = 0
DO {{fun st : state => True /\ st X <> 0}
X ::= X - 1 {{fun _ : state => True}}
END {{fun st : state => True /\ st X = 0}}) ->>
{{fun st : state => st X = 0}}
: decorated
*)

Compute pre_dec dec_while.
(*
= fun _ : state => True
: Assertion
*)

Compute post_dec dec_while.
(*
= fun x : state => x "X"%string = 0
: Assertion
*)

We can express what it means for a decorated program to be correct as follows:
Definition dec_correct (dec : decorated) :=
{{pre_dec dec}} (extract_dec dec) {{post_dec dec}}.

Example dec_while_triple_correct :
dec_correct dec_while
= {{ fun stTrue }}
(WHILE ~(X = 0) DO X ::= X - 1 END)%imp
{{ fun stst X = 0 }}.
Proof. reflexivity. Qed.
To check whether this Hoare triple is valid, we need a way to extract the "proof obligations" from a decorated program. These obligations are often called verification conditions, because they are the facts that must be verified to see that the decorations are logically consistent and thus constitute a proof of correctness.

## Extracting Verification Conditions

The function verification_conditions takes a dcom d together with a precondition P and returns a proposition that, if it can be proved, implies that the triple {{P}} (extract d) {{post d}} is valid.
It does this by walking over d and generating a big conjunction including all the "local checks" that we listed when we described the informal rules for decorated programs. (Strictly speaking, we need to massage the informal rules a little bit to add some uses of the rule of consequence, but the correspondence should be clear.)
Fixpoint verification_conditions (P : Assertion) (d : dcom) : Prop :=
match d with
| DCSkip Q
(P ->> Q)
| DCSeq d1 d2
verification_conditions P d1
∧ verification_conditions (post d1) d2
| DCAsgn X a Q
(P ->> Q [X > a])
| DCIf b P1 d1 P2 d2 Q
((fun stP stbassn b st) ->> P1)
∧ ((fun stP st ∧ ¬(bassn b st)) ->> P2)
∧ (post d1 ->> Q) ∧ (post d2 ->> Q)
∧ verification_conditions P1 d1
∧ verification_conditions P2 d2
| DCWhile b Pbody d Ppost
(* post d is the loop invariant and the initial
precondition *)

(P ->> post d)
∧ ((fun stpost d stbassn b st) ->> Pbody)
∧ ((fun stpost d st ∧ ~(bassn b st)) ->> Ppost)
∧ verification_conditions Pbody d
| DCPre P' d
(P ->> P') ∧ verification_conditions P' d
| DCPost d Q
verification_conditions P d ∧ (post d ->> Q)
end.
And now the key theorem, stating that verification_conditions does its job correctly. Not surprisingly, we need to use each of the Hoare Logic rules at some point in the proof.
Theorem verification_correct : d P,
verification_conditions P d{{P}} (extract d) {{post d}}.
Proof.
induction d; intros P H; simpl in *.
- (* Skip *)
eapply hoare_consequence_pre.
apply hoare_skip.
assumption.
- (* Seq *)
destruct H as [H1 H2].
eapply hoare_seq.
apply IHd2. apply H2.
apply IHd1. apply H1.
- (* Asgn *)
eapply hoare_consequence_pre.
apply hoare_asgn.
assumption.
- (* If *)
destruct H as [HPre1 [HPre2 [Hd1 [Hd2 [HThen HElse]]]]].
apply IHd1 in HThen. clear IHd1.
apply IHd2 in HElse. clear IHd2.
apply hoare_if.
+ eapply hoare_consequence_post with (Q':=post d1); eauto.
eapply hoare_consequence_pre; eauto.
+ eapply hoare_consequence_post with (Q':=post d2); eauto.
eapply hoare_consequence_pre; eauto.
- (* While *)
destruct H as [Hpre [Hbody1 [Hpost1 Hd]]].
eapply hoare_consequence_pre; eauto.
eapply hoare_consequence_post; eauto.
apply hoare_while.
eapply hoare_consequence_pre; eauto.
- (* Pre *)
destruct H as [HP Hd].
eapply hoare_consequence_pre. apply IHd. apply Hd. assumption.
- (* Post *)
destruct H as [Hd HQ].
eapply hoare_consequence_post. apply IHd. apply Hd. assumption.
Qed.
Now that all the pieces are in place, we can verify an entire program.
Definition verification_conditions_dec (dec : decorated) : Prop :=
match dec with
| Decorated P dverification_conditions P d
end.

Corollary verification_correct_dec : dec,
verification_conditions_dec decdec_correct dec.
Proof.
intros [P d]. apply verification_correct.
Qed.
The propositions generated by verification_conditions are fairly big, and they contain many conjuncts that are essentially trivial.
Eval simpl in (verification_conditions_dec dec_while).
===>
(((fun _ : state ⇒ True->> (fun _ : state ⇒ True)) ∧
((fun st : state ⇒ True ∧ bassn (~(X = 0)) st->>
(fun st : state ⇒ True ∧ st X ≠ 0)) ∧
((fun st : state ⇒ True ∧ ¬bassn (~(X = 0)) st->>
(fun st : state ⇒ True ∧ st X = 0)) ∧
(fun st : state ⇒ True ∧ st X ≠ 0) ->>
(fun _ : state ⇒ True) [X > X - 1]) ∧
(fun st : state ⇒ True ∧ st X = 0) ->>
(fun st : state ⇒ st X = 0)

## Automation

In principle, we could work with such propositions using just the tactics we have so far, but we can make things much smoother with a bit of automation. We first define a custom verify tactic that uses split repeatedly to turn all the conjunctions into separate subgoals and then uses omega and eauto (described in chapter Auto in Logical Foundations) to deal with as many of them as possible.
Tactic Notation "verify" :=
apply verification_correct;
repeat split;
simpl; unfold assert_implies;
unfold bassn in *; unfold beval in *; unfold aeval in *;
unfold assn_sub; intros;
repeat rewrite t_update_eq;
repeat (rewrite t_update_neq; [| (intro X; inversion X)]);
simpl in *;
repeat match goal with [H : ___] ⇒ destruct H end;
repeat rewrite not_true_iff_false in *;
repeat rewrite not_false_iff_true in *;
repeat rewrite negb_true_iff in *;
repeat rewrite negb_false_iff in *;
repeat rewrite eqb_eq in *;
repeat rewrite eqb_neq in *;
repeat rewrite leb_iff in *;
repeat rewrite leb_iff_conv in *;
try subst;
repeat
match goal with
[st : state_] ⇒
match goal with
[H : st _ = __] ⇒ rewriteH in *; clear H
| [H : _ = st __] ⇒ rewrite <- H in *; clear H
end
end;
try eauto; try omega.
What's left after verify does its thing is "just the interesting parts" of checking that the decorations are correct. For very simple examples, verify sometimes even immediately solves the goal (provided that the annotations are correct!).
Theorem dec_while_correct :
dec_correct dec_while.
Proof. verify. Qed.

## Examples

In this section, we use the automation developed above to verify formal decorated programs corresponding to most of the informal ones we have seen.

### Slow Subtraction

Example subtract_slowly_dec (m : nat) (p : nat) : decorated :=
{{ fun stst X = mst Z = p }} ->>
{{ fun stst Z - st X = p - m }}
WHILE ~(X = 0)
DO {{ fun stst Z - st X = p - mst X ≠ 0 }} ->>
{{ fun st ⇒ (st Z - 1) - (st X - 1) = p - m }}
Z ::= Z - 1
{{ fun stst Z - (st X - 1) = p - m }} ;;
X ::= X - 1
{{ fun stst Z - st X = p - m }}
END
{{ fun stst Z - st X = p - mst X = 0 }} ->>
{{ fun stst Z = p - m }}.

Theorem subtract_slowly_dec_correct : m p,
dec_correct (subtract_slowly_dec m p).
Proof. intros m p. verify. (* this grinds for a bit! *) Qed.

### Swapping Using Addition and Subtraction

Definition swap : com :=
X ::= X + Y;;
Y ::= X - Y;;
X ::= X - Y.

Definition swap_dec m n : decorated :=
{{ fun stst X = mst Y = n}} ->>
{{ fun st ⇒ (st X + st Y) - ((st X + st Y) - st Y) = n
∧ (st X + st Y) - st Y = m }}
X ::= X + Y
{{ fun stst X - (st X - st Y) = nst X - st Y = m }};;
Y ::= X - Y
{{ fun stst X - st Y = nst Y = m }};;
X ::= X - Y
{{ fun stst X = nst Y = m}}.

Theorem swap_correct : m n,
dec_correct (swap_dec m n).
Proof. intros; verify. Qed.

### Division

Definition div_mod_dec (a b : nat) : decorated :=
{{ fun stTrue }} ->>
{{ fun stb * 0 + a = a }}
X ::= a
{{ fun stb * 0 + st X = a }};;
Y ::= 0
{{ fun stb * st Y + st X = a }};;
WHILE bX DO
{{ fun stb * st Y + st X = abst X }} ->>
{{ fun stb * (st Y + 1) + (st X - b) = a }}
X ::= X - b
{{ fun stb * (st Y + 1) + st X = a }};;
Y ::= Y + 1
{{ fun stb * st Y + st X = a }}
END
{{ fun stb * st Y + st X = a ∧ ~(bst X) }} ->>
{{ fun stb * st Y + st X = a ∧ (st X < b) }}.

Theorem div_mod_dec_correct : a b,
dec_correct (div_mod_dec a b).
Proof. intros a b. verify.
rewrite mult_plus_distr_l. omega.
Qed.

### Parity

Definition find_parity : com :=
WHILE 2 ≤ X DO
X ::= X - 2
END.
There are actually several ways to phrase the loop invariant for this program. Here is one natural one, which leads to a rather long proof:
Inductive ev : natProp :=
| ev_0 : ev O
| ev_SS : n : nat, ev nev (S (S n)).

Definition find_parity_dec m : decorated :=
{{ fun stst X = m}} ->>
{{ fun stst Xmev (m - st X) }}
WHILE 2 ≤ X DO
{{ fun st ⇒ (st Xmev (m - st X)) ∧ 2 ≤ st X }} ->>
{{ fun stst X - 2 ≤ m ∧ (ev (m - (st X - 2))) }}
X ::= X - 2
{{ fun stst Xmev (m - st X) }}
END
{{ fun st ⇒ (st Xmev (m - st X)) ∧ st X < 2 }} ->>
{{ fun stst X=0 ↔ ev m }}.

Lemma l1 : m n p,
pn
nm
m - (n - p) = m - n + p.
Proof. intros. omega. Qed.

Lemma l2 : m,
ev m
ev (m + 2).
Proof. intros. rewrite plus_comm. simpl. constructor. assumption. Qed.

Lemma l3' : m,
ev m
¬ev (S m).
Proof. induction m; intros H1 H2. inversion H2. apply IHm.
inversion H2; subst; assumption. assumption. Qed.

Lemma l3 : m,
1 ≤ m
ev m
ev (m - 1) →
False.
Proof. intros. apply l2 in H1.
assert (G : m - 1 + 2 = S m). clear H0 H1. omega.
rewrite G in H1. apply l3' in H0. apply H0. assumption. Qed.

Theorem find_parity_correct : m,
dec_correct (find_parity_dec m).
Proof.
intro m. verify;
(* simplification too aggressive ... reverting a bit *)
fold (2 <=? (st X)) in *;
try rewrite leb_iff in *;
try rewrite leb_iff_conv in *; eauto; try omega.
- (* invariant holds initially *)
rewrite minus_diag. constructor.
- (* invariant preserved *)
rewrite l1; try assumption.
apply l2; assumption.
- (* invariant strong enough to imply conclusion
(-> direction) *)

rewrite <- minus_n_O in H2. assumption.
- (* invariant strong enough to imply conclusion
(<- direction) *)

destruct (st X) as [| [| n]].
(* by H1 X can only be 0 or 1 *)
+ (* st X = 0 *)
reflexivity.
+ (* st X = 1 *)
apply l3 in H; try assumption. inversion H.
+ (* st X = 2 *)
clear H0 H2. (* omega confused otherwise *)
omega.
Qed.
Here is a more intuitive way of writing the invariant:
Definition find_parity_dec' m : decorated :=
{{ fun stst X = m}} ->>
{{ fun stev (st X) ↔ ev m }}
WHILE 2 ≤ X DO
{{ fun st ⇒ (ev (st X) ↔ ev m) ∧ 2 ≤ st X }} ->>
{{ fun st ⇒ (ev (st X - 2) ↔ ev m) }}
X ::= X - 2
{{ fun st ⇒ (ev (st X) ↔ ev m) }}
END
{{ fun st ⇒ (ev (st X) ↔ ev m) ∧ ~(2 ≤ st X) }} ->>
{{ fun stst X=0 ↔ ev m }}.

Lemma l4 : m,
2 ≤ m
(ev (m - 2) ↔ ev m).
Proof.
induction m; intros. split; intro; constructor.
destruct m. inversion H. inversion H1. simpl in *.
rewrite <- minus_n_O in *. split; intro.
constructor. assumption.
inversion H0. assumption.
Qed.

Theorem find_parity_correct' : m,
dec_correct (find_parity_dec' m).
Proof.
intros m. verify;
(* simplification too aggressive ... reverting a bit *)
fold (2 <=? (st X)) in *;
try rewrite leb_iff in *;
try rewrite leb_iff_conv in *; intuition; eauto; try omega.
- (* invariant preserved (part 1) *)
rewrite l4 in H0; eauto.
- (* invariant preserved (part 2) *)
rewrite l4; eauto.
- (* invariant strong enough to imply conclusion
(-> direction) *)

apply H0. constructor.
- (* invariant strong enough to imply conclusion
(<- direction) *)

destruct (st X) as [| [| n]]. (* by H1 X can only be 0 or 1 *)
+ (* st X = 0 *)
reflexivity.
+ (* st X = 1 *)
inversion H.
+ (* st X = 2 *)
clear H0 H H3. (* omega confused otherwise *)
omega.
Qed.

### Square Roots

Definition sqrt_dec m : decorated :=
{{ fun stst X = m }} ->>
{{ fun stst X = m ∧ 0*0 ≤ m }}
Z ::= 0
{{ fun stst X = mst Z*st Zm }};;
WHILE (Z+1)*(Z+1) ≤ X DO
{{ fun st ⇒ (st X = mst Z*st Zm)
∧ (st Z + 1)*(st Z + 1) ≤ st X }} ->>
{{ fun stst X = m ∧ (st Z+1)*(st Z+1)≤m }}
Z ::= Z + 1
{{ fun stst X = mst Z*st Zm }}
END
{{ fun st ⇒ (st X = mst Z*st Zm)
∧ ~((st Z + 1)*(st Z + 1) ≤ st X) }} ->>
{{ fun stst Z*st Zmm<(st Z+1)*(st Z+1) }}.

Theorem sqrt_correct : m,
dec_correct (sqrt_dec m).
Proof. intro m. verify. Qed.

### Squaring

Again, there are several ways of annotating the squaring program. The simplest variant we've found, square_simpler_dec, is given last.
Definition square_dec (m : nat) : decorated :=
{{ fun stst X = m }}
Y ::= X
{{ fun stst X = mst Y = m }};;
Z ::= 0
{{ fun stst X = mst Y = mst Z = 0}} ->>
{{ fun stst Z + st X * st Y = m * m }};;
WHILE ~(Y = 0) DO
{{ fun stst Z + st X * st Y = m * mst Y ≠ 0 }} ->>
{{ fun st ⇒ (st Z + st X) + st X * (st Y - 1) = m * m }}
Z ::= Z + X
{{ fun stst Z + st X * (st Y - 1) = m * m }};;
Y ::= Y - 1
{{ fun stst Z + st X * st Y = m * m }}
END
{{ fun stst Z + st X * st Y = m * mst Y = 0 }} ->>
{{ fun stst Z = m * m }}.

Theorem square_dec_correct : m,
dec_correct (square_dec m).
Proof.
intro n. verify.
- (* invariant preserved *)
destruct (st Y) as [| y']. apply False_ind. apply H0.
reflexivity.
simpl. rewrite <- minus_n_O.
assert (G : n m, n * S m = n + n * m). {
clear. intros. induction n. reflexivity. simpl.
rewrite IHn. omega. }
rewrite <- H. rewrite G. rewrite plus_assoc. reflexivity.
Qed.

Definition square_dec' (n : nat) : decorated :=
{{ fun stTrue }}
X ::= n
{{ fun stst X = n }};;
Y ::= X
{{ fun stst X = nst Y = n }};;
Z ::= 0
{{ fun stst X = nst Y = nst Z = 0 }} ->>
{{ fun stst Z = st X * (st X - st Y)
∧ st X = nst Yst X }};;
WHILE ~(Y = 0) DO
{{ fun st ⇒ (st Z = st X * (st X - st Y)
∧ st X = nst Yst X)
∧ st Y ≠ 0 }}
Z ::= Z + X
{{ fun stst Z = st X * (st X - (st Y - 1))
∧ st X = nst Yst X }};;
Y ::= Y - 1
{{ fun stst Z = st X * (st X - st Y)
∧ st X = nst Yst X }}
END
{{ fun st ⇒ (st Z = st X * (st X - st Y)
∧ st X = nst Yst X)
∧ st Y = 0 }} ->>
{{ fun stst Z = n * n }}.

Theorem square_dec'_correct : n,
dec_correct (square_dec' n).
Proof.
intro n. verify.
- (* invariant holds initially *)
rewrite minus_diag. omega.
- (* invariant preserved *) subst.
rewrite mult_minus_distr_l.
repeat rewrite mult_minus_distr_l. rewrite mult_1_r.
assert (G : n m p,
mnpmn - (m - p) = n - m + p).
intros. omega.
rewrite G. reflexivity. apply mult_le_compat_l. assumption.
destruct (st Y). apply False_ind. apply H0. reflexivity.
clear. rewrite mult_succ_r. rewrite plus_comm.
apply le_plus_l.
- (* invariant + negation of guard imply
desired postcondition *)

rewrite <- minus_n_O. reflexivity.
Qed.

Definition square_simpler_dec (m : nat) : decorated :=
{{ fun stst X = m }} ->>
{{ fun st ⇒ 0 = 0*mst X = m }}
Y ::= 0
{{ fun st ⇒ 0 = (st Y)*mst X = m }};;
Z ::= 0
{{ fun stst Z = (st Y)*mst X = m }}->>
{{ fun stst Z = (st Y)*mst X = m }};;
WHILE ~(Y = X) DO
{{ fun st ⇒ (st Z = (st Y)*mst X = m)
∧ st Yst X }} ->>
{{ fun stst Z + st X = ((st Y) + 1)*mst X = m }}
Z ::= Z + X
{{ fun stst Z = ((st Y) + 1)*mst X = m }};;
Y ::= Y + 1
{{ fun stst Z = (st Y)*mst X = m }}
END
{{ fun st ⇒ (st Z = (st Y)*mst X = m) ∧ st Y = st X }} ->>
{{ fun stst Z = m*m }}.

Theorem square_simpler_dec_correct : m,
dec_correct (square_simpler_dec m).
Proof.
intro m. verify.
rewrite mult_plus_distr_r. simpl. rewrite <- plus_n_O.
reflexivity.
Qed.

### Power Series

Fixpoint pow2 n :=
match n with
| 0 ⇒ 1
| S n' ⇒ 2 * (pow2 n')
end.

Definition dpow2_down (n : nat) :=
{{ fun stTrue }} ->>
{{ fun st ⇒ 1 = (pow2 (0 + 1))-1 ∧ 1 = pow2 0 }}
X ::= 0
{{ fun st ⇒ 1 = (pow2 (0 + 1))-1 ∧ 1 = pow2 (st X) }};;
Y ::= 1
{{ fun stst Y = (pow2 (st X + 1))-1 ∧ 1 = pow2 (st X) }};;
Z ::= 1
{{ fun stst Y = (pow2 (st X + 1))-1 ∧ st Z = pow2 (st X) }};;
WHILE ~(X = n) DO
{{ fun st ⇒ (st Y = (pow2 (st X + 1))-1 ∧ st Z = pow2 (st X))
∧ st Xn }} ->>
{{ fun stst Y + 2 * st Z = (pow2 (st X + 2))-1
∧ 2 * st Z = pow2 (st X + 1) }}
Z ::= 2 * Z
{{ fun stst Y + st Z = (pow2 (st X + 2))-1
∧ st Z = pow2 (st X + 1) }};;
Y ::= Y + Z
{{ fun stst Y = (pow2 (st X + 2))-1
∧ st Z = pow2 (st X + 1) }};;
X ::= X + 1
{{ fun stst Y = (pow2 (st X + 1))-1
∧ st Z = pow2 (st X) }}
END
{{ fun st ⇒ (st Y = (pow2 (st X + 1))-1 ∧ st Z = pow2 (st X))
∧ st X = n }} ->>
{{ fun stst Y = pow2 (n+1) - 1 }}.

Lemma pow2_plus_1 : n,
pow2 (n+1) = pow2 n + pow2 n.
Proof. induction n; simpl. reflexivity. omega. Qed.

Lemma pow2_le_1 : n, pow2 n ≥ 1.
Proof. induction n. simpl. constructor. simpl. omega. Qed.

Theorem dpow2_down_correct : n,
dec_correct (dpow2_down n).
Proof.
intro m. verify.
- (* 1 *)
rewrite pow2_plus_1. rewrite <- H0. reflexivity.
- (* 2 *)
rewrite <- plus_n_O.
rewrite <- pow2_plus_1. remember (st X) as n.
replace (pow2 (n + 1) - 1 + pow2 (n + 1))
with (pow2 (n + 1) + pow2 (n + 1) - 1) by omega.
rewrite <- pow2_plus_1.
replace (n + 1 + 1) with (n + 2) by omega.
reflexivity.
- (* 3 *)
rewrite <- plus_n_O. rewrite <- pow2_plus_1.
reflexivity.
- (* 4 *)
replace (st X + 1 + 1) with (st X + 2) by omega.
reflexivity.
Qed.

## Further Exercises

#### Exercise: 3 stars, standard (slow_assignment_dec)

In the slow_assignment exercise above, we saw a roundabout way of assigning a number currently stored in X to the variable Y: start Y at 0, then decrement X until it hits 0, incrementing Y at each step. Tranform the informal decorated program your wrote for slow_assignment into a formal decorated program and prove its correctness. If all goes well, your correctness proof will need only intros and verify.
Hint: most of transformation will involve replacing informal assertions with formal assertions. For example, you would replace {{ X = m Y = 0}} with {{ fun st st X = m st Y = 0 }}. Also, semicolons go after the postcondition of an assignment in a formal decorated program, as in this example:
{X = m ∧ 0 = 0 }}
Y ::= 0;;
{X = m ∧ Y = 0 }}

becomes

{fun st ⇒ st X = m ∧ 0 = 0 }}
Y ::= 0
{fun st ⇒ st X = m ∧ st Y = 0 }} ;;
Example slow_assignment_dec (m : nat) : decorated
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Theorem slow_assignment_dec_correct : m,
dec_correct (slow_assignment_dec m).
Proof. (* FILL IN HERE *) Admitted.

(* Do not modify the following line: *)
Definition manual_grade_for_check_defn_of_slow_assignment_dec : option (nat*string) := None.

#### Exercise: 4 stars, standard (factorial_dec)

The factorial function is defined recursively in the Coq standard library in a way that is equivalent to the following:
Fixpoint fact (n : nat) : nat :=
match n with
| O ⇒ 1
| S n' ⇒ n * (fact n')
end.
Compute fact 5. (* ==> 120 *)
Using your solutions to factorial and slow_assignment_dec as a guide, write a formal decorated program factorial_dec that implements the factorial function, state a theorem named factorial_dec_correct that says factorial_dec is correct, and prove the theorem. If all goes well, verify will leave you with just two subgoals, each of which requires establishing some mathematical property of fact rather than proving anything about your program.
Hint: when transforming a loop guard into a formal assertion, make sure to express it with bassn.
(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_factorial_dec : option (nat*string) := None.

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

The Fibonacci function is usually written like this:
Fixpoint fib n :=
match n with
| 0 ⇒ 1
| 1 ⇒ 1
| _ ⇒ fib (pred n) + fib (pred (pred n))
end.
This doesn't pass Coq's termination checker, but here is a slightly clunkier definition that does:
Fixpoint fib n :=
match n with
| 0 ⇒ 1
| S n'match n' with
| 0 ⇒ 1
| S n''fib n' + fib n''
end
end.
Prove that fib satisfies the following equation:
Lemma fib_eqn : n,
n > 0 →
fib n + fib (Init.Nat.pred n) = fib (n + 1).
Proof.
(* FILL IN HERE *) Admitted.

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

The following Imp program leaves the value of fib n in the variable Y when it terminates:
X ::= 1;;
Y ::= 1;;
Z ::= 1;;
WHILE ~(X = n + 1) DO
T ::= Z;;
Z ::= Z + Y;;
Y ::= T;;
X ::= X + 1
END
Fill in the following definition of dfib and prove that it satisfies this specification:
{True }dfib {Y = fib n }}
Definition T : string := "T".

Definition dfib (n : nat) : decorated
(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Theorem dfib_correct : n,
dec_correct (dfib n).
(* FILL IN HERE *) Admitted.

#### Exercise: 5 stars, advanced, optional (improve_dcom)

The formal decorated programs defined in this section are intended to look as similar as possible to the informal ones defined earlier in the chapter. If we drop this requirement, we can eliminate almost all annotations, just requiring final postconditions and loop invariants to be provided explicitly. Do this — i.e., define a new version of dcom with as few annotations as possible and adapt the rest of the formal development leading up to the verification_correct theorem.
(* FILL IN HERE *)

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

Adapt the parser for Imp presented in chapter ImpParser of Logical Foundations to parse decorated commands (either ours or, even better, the ones you defined in the previous exercise).
(* FILL IN HERE *)
(* Mon Mar 25 14:39:37 EDT 2019 *)