Hoare2Hoare Logic, Part II
Require Import Coq.Bool.Bool.
Require Import Coq.Arith.Arith.
Require Import Coq.Arith.EqNat.
Require Import Coq.omega.Omega.
Require Import SfLib.
Require Import Maps.
Require Import Imp.
Require Import Hoare.
Decorated Programs
{{ 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 }}
{{ 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 }}
- 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 }}
IFB 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 (in all states):
{{ P }} ⇾This corresponds to the application of hoare_consequence and is the only place in a decorated program where checking if decorations are correct is not fully mechanical and syntactic, but rather may involve logical and/or arithmetic reasoning.
{{ P' }}
Example: Swapping Using Addition and Subtraction
X ::= X + Y;;
Y ::= X - Y;;
X ::= X - Y
We can prove using decorations that this program is correct —
i.e., it always swaps the values of variables X and Y.
Y ::= X - Y;;
X ::= X - 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:
(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 }}
- We begin with the undecorated program (the unnumbered lines).
- We then add the specification — i.e., the outer precondition (1) and postcondition (5). In the precondition we use auxiliary variables (parameters) m and n to remember the initial values of variables X and respectively 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 (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
Example: Simple Conditionals
(1) {{True}}
IFB 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:
IFB 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}}
- We start with the outer precondition (1) and postcondition (8).
- 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 (if_minus_plus_reloaded)
Fill in valid decorations for the following program:
{{ True }}
IFB X ≤ Y THEN
{{ }} ⇾
{{ }}
Z ::= Y - X
{{ }}
ELSE
{{ }} ⇾
{{ }}
Y ::= X + Z
{{ }}
FI
{{ Y = X + Z }}
☐
IFB X ≤ Y THEN
{{ }} ⇾
{{ }}
Z ::= Y - X
{{ }}
ELSE
{{ }} ⇾
{{ }}
Y ::= X + Z
{{ }}
FI
{{ Y = X + Z }}
Example: Reduce to Zero
(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:
WHILE X ≠ 0 DO
(2) {{ True ∧ X ≠ 0 }} ⇾
(3) {{ True }}
X ::= X - 1
(4) {{ True }}
END
(5) {{ True ∧ X = 0 }} ⇾
(6) {{ X = 0 }}
- Start with the outer precondition (1) and postcondition (6).
- Following the format dictated by the hoare_while rule, we copy (1) to (4). We conjoin (1) with the guard to obtain (2) and 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.
Definition reduce_to_zero' : com :=
WHILE BNot (BEq (AId X) (ANum 0)) DO
X ::= AMinus (AId X) (ANum 1)
END.
Theorem reduce_to_zero_correct' :
{{fun st ⇒ True}}
reduce_to_zero'
{{fun st ⇒ st 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. apply I.
- (* Invariant and negated guard imply postcondition *)
intros st [Inv GuardFalse].
unfold bassn in GuardFalse. simpl in GuardFalse.
(* SearchAbout helps to find the right lemmas *)
SearchAbout [not true].
rewrite not_true_iff_false in GuardFalse.
SearchAbout [negb false].
rewrite negb_false_iff in GuardFalse.
SearchAbout [beq_nat true].
apply beq_nat_true in GuardFalse.
apply GuardFalse. Qed.
Example: Division
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.
Y ::= 0;;
WHILE n ≤ X DO
X ::= X - n;;
Y ::= Y + 1
END;
(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 ∧ 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.
(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 ∧ X < n }}
Finding Loop Invariants
Example: Slow Subtraction
{{ 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 I for the
loop. As a first step we can leave I as an unknown and build a
skeleton for the proof by applying (backward) the rules for local
consistency. This process leads to the following skeleton:
WHILE X ≠ 0 DO
Y ::= Y - 1;;
X ::= X - 1
END
{{ Y = n - m }}
(1) {{ X = m ∧ Y = n }} ⇾ (a)
(2) {{ I }}
WHILE X ≠ 0 DO
(3) {{ I ∧ X ≠ 0 }} ⇾ (c)
(4) {{ I [X ↦ X-1] [Y ↦ Y-1] }}
Y ::= Y - 1;;
(5) {{ I [X ↦ X-1] }}
X ::= X - 1
(6) {{ I }}
END
(7) {{ I ∧ ¬ (X ≠ 0) }} ⇾ (b)
(8) {{ Y = n - m }}
By examining this skeleton, we can see that any valid I will
have to respect three conditions:
(2) {{ I }}
WHILE X ≠ 0 DO
(3) {{ I ∧ X ≠ 0 }} ⇾ (c)
(4) {{ I [X ↦ X-1] [Y ↦ Y-1] }}
Y ::= Y - 1;;
(5) {{ I [X ↦ X-1] }}
X ::= X - 1
(6) {{ I }}
END
(7) {{ I ∧ ¬ (X ≠ 0) }} ⇾ (b)
(8) {{ Y = n - m }}
- (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 loop's postcondition, i.e., (7) must imply (8);
- (c) it must be preserved by one iteration of the loop, i.e., (3) must imply (4).
(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 (7) True ∧
X = 0 implies (8) Y = n - m. 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).
(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 }}
(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.
(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 }}
(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.)
(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 }}
Exercise: Slow Assignment
Exercise: 2 stars (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.
Y ::= 0;;
WHILE X ≠ 0 DO
X ::= X - 1;;
Y ::= Y + 1
END
{{ Y = m }}
(* FILL IN HERE *)
☐
Exercise: Slow Addition
Exercise: 3 stars, 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.
Z ::= Z + 1;;
X ::= X - 1
END
(* FILL IN HERE *)
☐
Example: Parity
{{ 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:
WHILE 2 ≤ X DO
X ::= X - 2
END
{{ X = parity m }}
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:
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).
{{ 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 ∧ X < 2 }} ⇾ (b - OK)
{{ X = parity m }}
{{ 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 ∧ X < 2 }} ⇾ (b - OK)
{{ X = parity m }}
Exercise: 3 stars, optional (parity_formal)
Translate this proof to Coq. Refer to the reduce-to-zero example for ideas. You may find the following two lemmas 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.
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.
intros. induction x. reflexivity. destruct x. reflexivity.
exfalso. apply H. omega.
Qed.
Theorem parity_correct : ∀m,
{{ fun st ⇒ st X = m }}
WHILE BLe (ANum 2) (AId X) DO
X ::= AMinus (AId X) (ANum 2)
END
{{ fun st ⇒ st X = parity m }}.
Proof.
(* FILL IN HERE *) Admitted.
☐
Example: Finding Square Roots
{{ X=m }}
Z ::= 0;;
WHILE (Z+1)*(Z+1) ≤ X DO
Z ::= Z+1
END
{{ Z*Z≤m ∧ m<(Z+1)*(Z+1) }}
Z ::= 0;;
WHILE (Z+1)*(Z+1) ≤ X DO
Z ::= Z+1
END
{{ Z*Z≤m ∧ m<(Z+1)*(Z+1) }}
(1) {{ X=m }} ⇾ (a - second conjunct of (2) WRONG!)
(2) {{ 0*0 ≤ m ∧ m<1*1 }}
Z ::= 0;;
(3) {{ Z*Z ≤ m ∧ m<(Z+1)*(Z+1) }}
WHILE (Z+1)*(Z+1) ≤ X DO
(4) {{ Z*Z≤m ∧ (Z+1)*(Z+1)≤X }} ⇾ (c - WRONG!)
(5) {{ (Z+1)*(Z+1)≤m ∧ m<(Z+2)*(Z+2) }}
Z ::= Z+1
(6) {{ Z*Z≤m ∧ m<(Z+1)*(Z+1) }}
END
(7) {{ Z*Z≤m ∧ m<(Z+1)*(Z+1) ∧ X<(Z+1)*(Z+1) }} ⇾ (b - OK)
(8) {{ Z*Z≤m ∧ m<(Z+1)*(Z+1) }}
(2) {{ 0*0 ≤ m ∧ m<1*1 }}
Z ::= 0;;
(3) {{ Z*Z ≤ m ∧ m<(Z+1)*(Z+1) }}
WHILE (Z+1)*(Z+1) ≤ X DO
(4) {{ Z*Z≤m ∧ (Z+1)*(Z+1)≤X }} ⇾ (c - WRONG!)
(5) {{ (Z+1)*(Z+1)≤m ∧ m<(Z+2)*(Z+2) }}
Z ::= Z+1
(6) {{ Z*Z≤m ∧ m<(Z+1)*(Z+1) }}
END
(7) {{ Z*Z≤m ∧ m<(Z+1)*(Z+1) ∧ X<(Z+1)*(Z+1) }} ⇾ (b - OK)
(8) {{ Z*Z≤m ∧ m<(Z+1)*(Z+1) }}
{{ 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*Z≤m ∧ (Z+1)*(Z+1)≤X }} ⇾ (c - OK)
{{ X=m ∧ (Z+1)*(Z+1)≤m }}
Z ::= Z+1
{{ X=m ∧ Z*Z≤m }}
END
{{ X=m ∧ Z*Z≤m ∧ X<(Z+1)*(Z+1) }} ⇾ (b - OK)
{{ Z*Z≤m ∧ m<(Z+1)*(Z+1) }}
{{ X=m ∧ 0*0 ≤ m }}
Z ::= 0;
{{ X=m ∧ Z*Z ≤ m }}
WHILE (Z+1)*(Z+1) ≤ X DO
{{ X=m ∧ Z*Z≤m ∧ (Z+1)*(Z+1)≤X }} ⇾ (c - OK)
{{ X=m ∧ (Z+1)*(Z+1)≤m }}
Z ::= Z+1
{{ X=m ∧ Z*Z≤m }}
END
{{ X=m ∧ Z*Z≤m ∧ X<(Z+1)*(Z+1) }} ⇾ (b - OK)
{{ Z*Z≤m ∧ m<(Z+1)*(Z+1) }}
Example: Squaring
{{ X = m }}
Y ::= 0;;
Z ::= 0;;
WHILE Y ≠ X DO
Z ::= Z + X;;
Y ::= Y + 1
END
{{ Z = m*m }}
Y ::= 0;;
Z ::= 0;;
WHILE Y ≠ X DO
Z ::= Z + X;;
Y ::= Y + 1
END
{{ Z = m*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 ≠ X DO
{{ Z = Y*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 }}
{{ 0 = m*m ∧ X = m }}
Y ::= 0;;
{{ 0 = m*m ∧ X = m }}
Z ::= 0;;
{{ Z = m*m ∧ X = m }}
WHILE Y ≠ X DO
{{ Z = Y*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 }}
{{ 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 ≠ X DO
{{ 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 }}
{{ 0 = 0*m ∧ X = m }}
Y ::= 0;;
{{ 0 = Y*m ∧ X = m }}
Z ::= 0;;
{{ Z = Y*m ∧ X = m }}
WHILE Y ≠ X DO
{{ 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 }}
Exercise: Factorial
Exercise: 3 stars (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! }}
Y ::= 1 ;;
WHILE X ≠ 0
DO
Y ::= Y * X ;;
X ::= X - 1
END
{{ Y = m! }}
{{ X = m }} ⇾
{{ }}
Y ::= 1;;
{{ }}
WHILE X ≠ 0
DO {{ }} ⇾
{{ }}
Y ::= Y * X;;
{{ }}
X ::= X - 1
{{ }}
END
{{ }} ⇾
{{ Y = m! }}
{{ }}
Y ::= 1;;
{{ }}
WHILE X ≠ 0
DO {{ }} ⇾
{{ }}
Y ::= Y * X;;
{{ }}
X ::= X - 1
{{ }}
END
{{ }} ⇾
{{ Y = m! }}
Exercise: Min
Exercise: 3 stars (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.
(x=0 ∨ y=0) → min x y = 0.
Lemma lemma2 : ∀x y,
min (x-1) (y-1) = (min x y) - 1.
{{ 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 }}
{{ }}
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 }}
Exercise: 3 stars (two_loops)
Here is a very inefficient way of adding 3 numbers:
X ::= 0;;
Y ::= 0;;
Z ::= c;;
WHILE X ≠ a DO
X ::= X + 1;;
Z ::= Z + 1
END;;
WHILE Y ≠ b DO
Y ::= Y + 1;;
Z ::= Z + 1
END
Y ::= 0;;
Z ::= c;;
WHILE X ≠ a DO
X ::= X + 1;;
Z ::= Z + 1
END;;
WHILE Y ≠ b DO
Y ::= Y + 1;;
Z ::= Z + 1
END
{{ True }} ⇾
{{ }}
X ::= 0;;
{{ }}
Y ::= 0;;
{{ }}
Z ::= c;;
{{ }}
WHILE X ≠ a DO
{{ }} ⇾
{{ }}
X ::= X + 1;;
{{ }}
Z ::= Z + 1
{{ }}
END;;
{{ }} ⇾
{{ }}
WHILE Y ≠ b DO
{{ }} ⇾
{{ }}
Y ::= Y + 1;;
{{ }}
Z ::= Z + 1
{{ }}
END
{{ }} ⇾
{{ Z = a + b + c }}
{{ }}
X ::= 0;;
{{ }}
Y ::= 0;;
{{ }}
Z ::= c;;
{{ }}
WHILE X ≠ a DO
{{ }} ⇾
{{ }}
X ::= X + 1;;
{{ }}
Z ::= Z + 1
{{ }}
END;;
{{ }} ⇾
{{ }}
WHILE Y ≠ b DO
{{ }} ⇾
{{ }}
Y ::= Y + 1;;
{{ }}
Z ::= Z + 1
{{ }}
END
{{ }} ⇾
{{ Z = a + b + c }}
Exercise: Power Series
Exercise: 4 stars, 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 ≠ m DO
Z ::= 2 * Z;;
Y ::= Y + Z;;
X ::= X + 1
END
Write a decorated program for this.
Y ::= 1;;
Z ::= 1;;
WHILE X ≠ m DO
Z ::= 2 * Z;;
Y ::= Y + Z;;
X ::= X + 1
END
(* FILL IN HERE *)
Weakest Preconditions (Optional)
{{ 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.
{{ 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.
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, 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) {{ ? }}
IFB 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 }}
2) {{ ? }} X ::= Y + Z {{ X = 5 }}
3) {{ ? }} X ::= Y {{ X = Y }}
4) {{ ? }}
IFB 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, advanced, optional (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 st ⇒ st Y ≤ 4)
(X ::= APlus (AId Y) (ANum 1)) (fun st ⇒ st X ≤ 5).
Proof.
(* FILL IN HERE *) Admitted.
☐
Exercise: 2 stars, advanced, optional (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, advanced, 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 : id),
{{ P }} HAVOC X {{ Q }} →
P ⇾ havoc_pre X Q.
Proof.
(* FILL IN HERE *) Admitted.
Import Himp.
Lemma hoare_havoc_weakest : ∀(P Q : Assertion) (X : id),
{{ P }} HAVOC X {{ Q }} →
P ⇾ havoc_pre X Q.
Proof.
(* FILL IN HERE *) Admitted.
☐
Formal Decorated Programs (Optional)
Syntax
Inductive dcom : Type :=
| DCSkip : Assertion → dcom
| DCSeq : dcom → dcom → dcom
| DCAsgn : id → aexp → Assertion → dcom
| DCIf : bexp → Assertion → dcom → Assertion → dcom
→ Assertion→ dcom
| DCWhile : bexp → Assertion → dcom → Assertion → dcom
| DCPre : Assertion → dcom → dcom
| DCPost : dcom → Assertion → dcom.
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 "'IFB' 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 "{{ P }} d"
:= (DCPre P d)
(at level 90) : 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.
Delimit Scope dcom_scope with dcom.
To avoid clashing with the existing Notation definitions
for ordinary commands, we introduce these notations in a special
scope called dcom_scope, and we wrap examples with the
declaration % dcom to signal that we want the notations to be
interpreted in this scope.
Careful readers will note that we've defined two notations for the
DCPre constructor, one with and one without a ⇾. The
"without" version is intended to be used to supply the initial
precondition at the very top of the program.
Example dec_while : dcom := (
{{ fun st ⇒ True }}
WHILE (BNot (BEq (AId X) (ANum 0)))
DO
{{ fun st ⇒ True ∧ st X ≠ 0}}
X ::= (AMinus (AId X) (ANum 1))
{{ fun _ ⇒ True }}
END
{{ fun st ⇒ True ∧ st X = 0}} ⇾
{{ fun st ⇒ st X = 0 }}
) % dcom.
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 _ ⇒ IFB b THEN extract d1 ELSE extract d2 FI
| DCWhile b _ d _ ⇒ WHILE b DO extract d END
| DCPre _ d ⇒ extract d
| DCPost d _ ⇒ extract d
end.
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
Instead, the rule we've followed is this:
In other words, the invariant of the representation is that a
dcom d together with a precondition P determines a Hoare
triple {{P}} (extract d) {{post d}}, where post is defined as
follows:
{{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!
- The post-condition expected by each dcom d is embedded
in d.
- The pre-condition is supplied by the context.
Fixpoint post (d:dcom) : Assertion :=
match d with
| DCSkip P ⇒ P
| DCSeq d1 d2 ⇒ post d2
| DCAsgn X a Q ⇒ Q
| DCIf _ _ d1 _ d2 Q ⇒ Q
| DCWhile b Pbody c Ppost ⇒ Ppost
| DCPre _ d ⇒ post d
| DCPost c Q ⇒ Q
end.
Similarly, we can extract the "initial precondition" from a
decorated program.
Fixpoint pre (d:dcom) : Assertion :=
match d with
| DCSkip P ⇒ fun st ⇒ True
| DCSeq c1 c2 ⇒ pre c1
| DCAsgn X a Q ⇒ fun st ⇒ True
| DCIf _ _ t _ e _ ⇒ fun st ⇒ True
| DCWhile b Pbody c Ppost ⇒ fun st ⇒ True
| DCPre P c ⇒ P
| DCPost c Q ⇒ pre c
end.
This function is not doing anything sophisticated like calculating
a weakest precondition; it just recursively searches for an
explicit annotation at the very beginning of the program,
returning default answers for programs that lack an explicit
precondition (like a bare assignment or SKIP).
Using pre and post, and assuming that we adopt the convention
of always supplying an explicit precondition annotation at the
very beginning of our decorated programs, we can express what it
means for a decorated program to be correct as follows:
Definition dec_correct (d:dcom) :=
{{pre d}} (extract d) {{post d}}.
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 add up to a complete
proof of correctness.
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.)
Extracting Verification Conditions
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 st ⇒ P st ∧ bassn b st) ⇾ P1)
∧ ((fun st ⇒ P st ∧ ¬ (bassn b st)) ⇾ P2)
∧ (Q ⇿ post d1) ∧ (Q ⇿ post d2)
∧ 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)
∧ (Pbody ⇿ (fun st ⇒ post d st ∧ bassn b st))
∧ (Ppost ⇿ (fun st ⇒ post d st ∧ ~(bassn b st)))
∧ 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 *)
inversion H as [H1 H2]. clear H.
eapply hoare_seq.
apply IHd2. apply H2.
apply IHd1. apply H1.
- (* Asgn *)
eapply hoare_consequence_pre.
apply hoare_asgn.
assumption.
- (* If *)
inversion H as [HPre1 [HPre2 [[Hd11 Hd12]
[[Hd21 Hd22] [HThen HElse]]]]].
clear H.
apply IHd1 in HThen. clear IHd1.
apply IHd2 in HElse. clear IHd2.
apply hoare_if.
eapply hoare_consequence_pre; eauto.
eapply hoare_consequence_post; eauto.
eapply hoare_consequence_pre; eauto.
eapply hoare_consequence_post; eauto.
- (* While *)
inversion H as [Hpre [[Hbody1 Hbody2] [[Hpost1 Hpost2] Hd]]];
subst; clear H.
eapply hoare_consequence_pre; eauto.
eapply hoare_consequence_post; eauto.
apply hoare_while.
eapply hoare_consequence_pre; eauto.
- (* Pre *)
inversion H as [HP Hd]; clear H.
eapply hoare_consequence_pre. apply IHd. apply Hd. assumption.
- (* Post *)
inversion H as [Hd HQ]; clear H.
eapply hoare_consequence_post. apply IHd. apply Hd. assumption.
Qed.
induction d; intros P H; simpl in *.
- (* Skip *)
eapply hoare_consequence_pre.
apply hoare_skip.
assumption.
- (* Seq *)
inversion H as [H1 H2]. clear H.
eapply hoare_seq.
apply IHd2. apply H2.
apply IHd1. apply H1.
- (* Asgn *)
eapply hoare_consequence_pre.
apply hoare_asgn.
assumption.
- (* If *)
inversion H as [HPre1 [HPre2 [[Hd11 Hd12]
[[Hd21 Hd22] [HThen HElse]]]]].
clear H.
apply IHd1 in HThen. clear IHd1.
apply IHd2 in HElse. clear IHd2.
apply hoare_if.
eapply hoare_consequence_pre; eauto.
eapply hoare_consequence_post; eauto.
eapply hoare_consequence_pre; eauto.
eapply hoare_consequence_post; eauto.
- (* While *)
inversion H as [Hpre [[Hbody1 Hbody2] [[Hpost1 Hpost2] Hd]]];
subst; clear H.
eapply hoare_consequence_pre; eauto.
eapply hoare_consequence_post; eauto.
apply hoare_while.
eapply hoare_consequence_pre; eauto.
- (* Pre *)
inversion H as [HP Hd]; clear H.
eapply hoare_consequence_pre. apply IHd. apply Hd. assumption.
- (* Post *)
inversion H as [Hd HQ]; clear H.
eapply hoare_consequence_post. apply IHd. apply Hd. assumption.
Qed.
(If you expand the proof, you'll see that it uses an
unfamiliar idiom: simpl in *. We have used ...in... variants
of several tactics before, to apply them to values in the context
rather than the goal. The syntax tactic in * extends this idea,
applying tactic in the goal and every hypothesis in the
context.)
The propositions generated by verification_conditions are fairly
big, and they contain many conjuncts that are essentially trivial.
Automation
Eval simpl in (verification_conditions (fun st ⇒ True) dec_while).
⇒
(((fun _ : state ⇒ True) ⇾ (fun _ : state ⇒ True)) ∧
((fun _ : state ⇒ True) ⇾ (fun _ : state ⇒ True)) ∧
(fun st : state ⇒ True ∧ bassn (BNot (BEq (AId X) (ANum 0))) st) =
(fun st : state ⇒ True ∧ bassn (BNot (BEq (AId X) (ANum 0))) st) ∧
(fun st : state ⇒ True ∧ ¬ bassn (BNot (BEq (AId X) (ANum 0))) st) =
(fun st : state ⇒ True ∧ ¬ bassn (BNot (BEq (AId X) (ANum 0))) st) ∧
(fun st : state ⇒ True ∧ bassn (BNot (BEq (AId X) (ANum 0))) st) ⇾
(fun _ : state ⇒ True) [X ↦ AMinus (AId X) (ANum 1)]) ∧
(fun st : state ⇒ True ∧ ¬ bassn (BNot (BEq (AId X) (ANum 0))) st) ⇾
(fun st : state ⇒ st X = 0)
(((fun _ : state ⇒ True) ⇾ (fun _ : state ⇒ True)) ∧
((fun _ : state ⇒ True) ⇾ (fun _ : state ⇒ True)) ∧
(fun st : state ⇒ True ∧ bassn (BNot (BEq (AId X) (ANum 0))) st) =
(fun st : state ⇒ True ∧ bassn (BNot (BEq (AId X) (ANum 0))) st) ∧
(fun st : state ⇒ True ∧ ¬ bassn (BNot (BEq (AId X) (ANum 0))) st) =
(fun st : state ⇒ True ∧ ¬ bassn (BNot (BEq (AId X) (ANum 0))) st) ∧
(fun st : state ⇒ True ∧ bassn (BNot (BEq (AId X) (ANum 0))) st) ⇾
(fun _ : state ⇒ True) [X ↦ AMinus (AId X) (ANum 1)]) ∧
(fun st : state ⇒ True ∧ ¬ bassn (BNot (BEq (AId X) (ANum 0))) st) ⇾
(fun st : state ⇒ st X = 0)
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 beq_nat_true_iff in *;
repeat rewrite beq_nat_false_iff 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 _ = _ ⊢ _] ⇒ rewrite → H 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 immediately solves the goal (provided
that the annotations are correct).
Theorem dec_while_correct :
dec_correct dec_while.
Proof. verify. Qed.
Another example (formalizing a decorated program we've seen
before):
Example subtract_slowly_dec (m:nat) (p:nat) : dcom := (
{{ fun st ⇒ st X = m ∧ st Z = p }} ⇾
{{ fun st ⇒ st Z - st X = p - m }}
WHILE BNot (BEq (AId X) (ANum 0))
DO {{ fun st ⇒ st Z - st X = p - m ∧ st X ≠ 0 }} ⇾
{{ fun st ⇒ (st Z - 1) - (st X - 1) = p - m }}
Z ::= AMinus (AId Z) (ANum 1)
{{ fun st ⇒ st Z - (st X - 1) = p - m }} ;;
X ::= AMinus (AId X) (ANum 1)
{{ fun st ⇒ st Z - st X = p - m }}
END
{{ fun st ⇒ st Z - st X = p - m ∧ st X = 0 }} ⇾
{{ fun st ⇒ st Z = p - m }}
) % dcom.
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.
Exercise: 3 stars, advanced (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. Write a formal version of this decorated program and prove it correct.Example slow_assignment_dec (m:nat) : dcom :=
(* FILL IN HERE *) admit.
Theorem slow_assignment_dec_correct : ∀m,
dec_correct (slow_assignment_dec m).
Proof. (* FILL IN HERE *) Admitted.
☐
Exercise: 4 stars, advanced (factorial_dec)
Remember the factorial function we worked with before:Fixpoint real_fact (n:nat) : nat :=
match n with
| O ⇒ 1
| S n' ⇒ n * (real_fact n')
end.
Following the pattern of subtract_slowly_dec, write a decorated
program factorial_dec that implements the factorial function and
prove it correct as factorial_dec_correct.
(* FILL IN HERE *)
☐
In this section, we use the automation developed above to verify
formal decorated programs corresponding to most of the informal
ones we have seen.
Examples
Swapping Using Addition and Subtraction
Definition swap : com :=
X ::= APlus (AId X) (AId Y);;
Y ::= AMinus (AId X) (AId Y);;
X ::= AMinus (AId X) (AId Y).
Definition swap_dec m n : dcom :=
({{ fun st ⇒ st X = m ∧ st Y = n}} ⇾
{{ fun st ⇒ (st X + st Y) - ((st X + st Y) - st Y) = n
∧ (st X + st Y) - st Y = m }}
X ::= APlus (AId X) (AId Y)
{{ fun st ⇒ st X - (st X - st Y) = n ∧ st X - st Y = m }};;
Y ::= AMinus (AId X) (AId Y)
{{ fun st ⇒ st X - st Y = n ∧ st Y = m }};;
X ::= AMinus (AId X) (AId Y)
{{ fun st ⇒ st X = n ∧ st Y = m}})%dcom.
Theorem swap_correct : ∀m n,
dec_correct (swap_dec m n).
Proof. intros; verify. Qed.
Definition if_minus_plus_com :=
IFB (BLe (AId X) (AId Y))
THEN (Z ::= AMinus (AId Y) (AId X))
ELSE (Y ::= APlus (AId X) (AId Z))
FI.
Definition if_minus_plus_dec :=
({{fun st ⇒ True}}
IFB (BLe (AId X) (AId Y)) THEN
{{ fun st ⇒ True ∧ st X ≤ st Y }} ⇾
{{ fun st ⇒ st Y = st X + (st Y - st X) }}
Z ::= AMinus (AId Y) (AId X)
{{ fun st ⇒ st Y = st X + st Z }}
ELSE
{{ fun st ⇒ True ∧ ~(st X ≤ st Y) }} ⇾
{{ fun st ⇒ st X + st Z = st X + st Z }}
Y ::= APlus (AId X) (AId Z)
{{ fun st ⇒ st Y = st X + st Z }}
FI
{{fun st ⇒ st Y = st X + st Z}})%dcom.
Theorem if_minus_plus_correct :
dec_correct if_minus_plus_dec.
Proof. intros; verify. Qed.
Definition if_minus_dec :=
( {{fun st ⇒ True}}
IFB (BLe (AId X) (AId Y)) THEN
{{fun st ⇒ True ∧ st X ≤ st Y }} ⇾
{{fun st ⇒ (st Y - st X) + st X = st Y
∨ (st Y - st X) + st Y = st X}}
Z ::= AMinus (AId Y) (AId X)
{{fun st ⇒ st Z + st X = st Y ∨ st Z + st Y = st X}}
ELSE
{{fun st ⇒ True ∧ ~(st X ≤ st Y) }} ⇾
{{fun st ⇒ (st X - st Y) + st X = st Y
∨ (st X - st Y) + st Y = st X}}
Z ::= AMinus (AId X) (AId Y)
{{fun st ⇒ st Z + st X = st Y ∨ st Z + st Y = st X}}
FI
{{fun st ⇒ st Z + st X = st Y ∨ st Z + st Y = st X}})%dcom.
Theorem if_minus_correct :
dec_correct if_minus_dec.
Proof. verify. Qed.
Definition div_mod_dec (a b : nat) : dcom := (
{{ fun st ⇒ True }} ⇾
{{ fun st ⇒ b * 0 + a = a }}
X ::= ANum a
{{ fun st ⇒ b * 0 + st X = a }};;
Y ::= ANum 0
{{ fun st ⇒ b * st Y + st X = a }};;
WHILE (BLe (ANum b) (AId X)) DO
{{ fun st ⇒ b * st Y + st X = a ∧ b ≤ st X }} ⇾
{{ fun st ⇒ b * (st Y + 1) + (st X - b) = a }}
X ::= AMinus (AId X) (ANum b)
{{ fun st ⇒ b * (st Y + 1) + st X = a }};;
Y ::= APlus (AId Y) (ANum 1)
{{ fun st ⇒ b * st Y + st X = a }}
END
{{ fun st ⇒ b * st Y + st X = a ∧ ~(b ≤ st X) }} ⇾
{{ fun st ⇒ b * st Y + st X = a ∧ (st X < b) }}
)%dcom.
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.
Definition find_parity : com :=
WHILE (BLe (ANum 2) (AId X)) DO
X ::= AMinus (AId X) (ANum 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 : nat → Prop :=
| ev_0 : ev O
| ev_SS : ∀n:nat, ev n → ev (S (S n)).
Definition find_parity_dec m : dcom :=
({{ fun st ⇒ st X = m}} ⇾
{{ fun st ⇒ st X ≤ m ∧ ev (m - st X) }}
WHILE (BLe (ANum 2) (AId X)) DO
{{ fun st ⇒ (st X ≤ m ∧ ev (m - st X)) ∧ 2 ≤ st X }} ⇾
{{ fun st ⇒ st X - 2 ≤ m ∧ (ev (m - (st X - 2))) }}
X ::= AMinus (AId X) (ANum 2)
{{ fun st ⇒ st X ≤ m ∧ ev (m - st X) }}
END
{{ fun st ⇒ (st X ≤ m ∧ ev (m - st X)) ∧ st X < 2 }} ⇾
{{ fun st ⇒ st X=0 ↔ ev m }})%dcom.
Lemma l1 : ∀m n p,
p ≤ n →
n ≤ m →
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 (leb 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 : dcom :=
({{ fun st ⇒ st X = m}} ⇾
{{ fun st ⇒ ev (st X) ↔ ev m }}
WHILE (BLe (ANum 2) (AId X)) DO
{{ fun st ⇒ (ev (st X) ↔ ev m) ∧ 2 ≤ st X }} ⇾
{{ fun st ⇒ (ev (st X - 2) ↔ ev m) }}
X ::= AMinus (AId X) (ANum 2)
{{ fun st ⇒ (ev (st X) ↔ ev m) }}
END
{{ fun st ⇒ (ev (st X) ↔ ev m) ∧ ~(2 ≤ st X) }} ⇾
{{ fun st ⇒ st X=0 ↔ ev m }})%dcom.
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 (leb 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.
Here is the simplest invariant we've found for this program:
Definition parity_dec m : dcom :=
({{ fun st ⇒ st X = m}} ⇾
{{ fun st ⇒ parity (st X) = parity m }}
WHILE (BLe (ANum 2) (AId X)) DO
{{ fun st ⇒ parity (st X) = parity m ∧ 2 ≤ st X }} ⇾
{{ fun st ⇒ parity (st X - 2) = parity m }}
X ::= AMinus (AId X) (ANum 2)
{{ fun st ⇒ parity (st X) = parity m }}
END
{{ fun st ⇒ parity (st X) = parity m ∧ ~(2 ≤ st X) }} ⇾
{{ fun st ⇒ st X = parity m }})%dcom.
Theorem parity_dec_correct : ∀m,
dec_correct (parity_dec m).
Proof.
intros. verify;
(* simplification too aggressive ... reverting a bit *)
fold (leb 2 (st X)) in *;
try rewrite leb_iff in *;
try rewrite leb_iff_conv in *; eauto; try omega.
- (* invariant preserved *)
rewrite ← H. apply parity_ge_2. assumption.
- (* invariant strong enough *)
rewrite ← H. symmetry. apply parity_lt_2. assumption.
Qed.
Definition sqrt_dec m : dcom := (
{{ fun st ⇒ st X = m }} ⇾
{{ fun st ⇒ st X = m ∧ 0*0 ≤ m }}
Z ::= ANum 0
{{ fun st ⇒ st X = m ∧ st Z*st Z ≤ m }};;
WHILE BLe (AMult (APlus (AId Z) (ANum 1))
(APlus (AId Z) (ANum 1)))
(AId X) DO
{{ fun st ⇒ (st X = m ∧ st Z*st Z≤m)
∧ (st Z + 1)*(st Z + 1) ≤ st X }} ⇾
{{ fun st ⇒ st X = m ∧ (st Z+1)*(st Z+1)≤m }}
Z ::= APlus (AId Z) (ANum 1)
{{ fun st ⇒ st X = m ∧ st Z*st Z≤m }}
END
{{ fun st ⇒ (st X = m ∧ st Z*st Z≤m)
∧ ~((st Z + 1)*(st Z + 1) ≤ st X) }} ⇾
{{ fun st ⇒ st Z*st Z≤m ∧ m<(st Z+1)*(st Z+1) }})%dcom.
Theorem sqrt_correct : ∀m,
dec_correct (sqrt_dec m).
Proof. intro m. verify. Qed.
Squaring
Definition square_dec (m : nat) : dcom := (
{{ fun st ⇒ st X = m }}
Y ::= AId X
{{ fun st ⇒ st X = m ∧ st Y = m }};;
Z ::= ANum 0
{{ fun st ⇒ st X = m ∧ st Y = m ∧ st Z = 0}};;
{{ fun st ⇒ st Z + st X * st Y = m * m }}
WHILE BNot (BEq (AId Y) (ANum 0)) DO
{{ fun st ⇒ st Z + st X * st Y = m * m ∧ st Y ≠ 0 }} ⇾
{{ fun st ⇒ (st Z + st X) + st X * (st Y - 1) = m * m }}
Z ::= APlus (AId Z) (AId X)
{{ fun st ⇒ st Z + st X * (st Y - 1) = m * m }};;
Y ::= AMinus (AId Y) (ANum 1)
{{ fun st ⇒ st Z + st X * st Y = m * m }}
END
{{ fun st ⇒ st Z + st X * st Y = m * m ∧ st Y = 0 }} ⇾
{{ fun st ⇒ st Z = m * m }}
)%dcom.
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) : dcom := (
{{ fun st ⇒ True }}
X ::= ANum n
{{ fun st ⇒ st X = n }};;
Y ::= AId X
{{ fun st ⇒ st X = n ∧ st Y = n }};;
Z ::= ANum 0
{{ fun st ⇒ st X = n ∧ st Y = n ∧ st Z = 0 }};;
{{ fun st ⇒ st Z = st X * (st X - st Y)
∧ st X = n ∧ st Y ≤ st X }}
WHILE BNot (BEq (AId Y) (ANum 0)) DO
{{ fun st ⇒ (st Z = st X * (st X - st Y)
∧ st X = n ∧ st Y ≤ st X)
∧ st Y ≠ 0 }}
Z ::= APlus (AId Z) (AId X)
{{ fun st ⇒ st Z = st X * (st X - (st Y - 1))
∧ st X = n ∧ st Y ≤ st X }};;
Y ::= AMinus (AId Y) (ANum 1)
{{ fun st ⇒ st Z = st X * (st X - st Y)
∧ st X = n ∧ st Y ≤ st X }}
END
{{ fun st ⇒ (st Z = st X * (st X - st Y)
∧ st X = n ∧ st Y ≤ st X)
∧ st Y = 0 }} ⇾
{{ fun st ⇒ st Z = n * n }}
)%dcom.
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, m ≤ n → p ≤ m → n - (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.
- (* invarint + negation of guard imply desired postcondition *)
rewrite ← minus_n_O. reflexivity.
Qed.
Definition square_simpler_dec (m : nat) : dcom := (
{{ fun st ⇒ st X = m }} ⇾
{{ fun st ⇒ 0 = 0*m ∧ st X = m }}
Y ::= ANum 0
{{ fun st ⇒ 0 = (st Y)*m ∧ st X = m }};;
Z ::= ANum 0
{{ fun st ⇒ st Z = (st Y)*m ∧ st X = m }};;
{{ fun st ⇒ st Z = (st Y)*m ∧ st X = m }}
WHILE BNot (BEq (AId Y) (AId X)) DO
{{ fun st ⇒ (st Z = (st Y)*m ∧ st X = m)
∧ st Y ≠ st X }} ⇾
{{ fun st ⇒ st Z + st X = ((st Y) + 1)*m ∧ st X = m }}
Z ::= APlus (AId Z) (AId X)
{{ fun st ⇒ st Z = ((st Y) + 1)*m ∧ st X = m }};;
Y ::= APlus (AId Y) (ANum 1)
{{ fun st ⇒ st Z = (st Y)*m ∧ st X = m }}
END
{{ fun st ⇒ (st Z = (st Y)*m ∧ st X = m) ∧ st Y = st X }} ⇾
{{ fun st ⇒ st Z = m*m }}
)%dcom.
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.
Definition two_loops_dec (a b c : nat) :=
( {{ fun st ⇒ True }} ⇾
{{ fun st ⇒ c = 0 + c ∧ 0 = 0 }}
X ::= ANum 0
{{ fun st ⇒ c = st X + c ∧ 0 = 0 }};;
Y ::= ANum 0
{{ fun st ⇒ c = st X + c ∧ st Y = 0 }};;
Z ::= ANum c
{{ fun st ⇒ st Z = st X + c ∧ st Y = 0 }};;
WHILE BNot (BEq (AId X) (ANum a)) DO
{{ fun st ⇒ (st Z = st X + c ∧ st Y = 0) ∧ st X ≠ a }} ⇾
{{ fun st ⇒ st Z + 1 = st X + 1 + c ∧ st Y = 0 }}
X ::= APlus (AId X) (ANum 1)
{{ fun st ⇒ st Z + 1 = st X + c ∧ st Y = 0 }};;
Z ::= APlus (AId Z) (ANum 1)
{{ fun st ⇒ st Z = st X + c ∧ st Y = 0 }}
END
{{ fun st ⇒ (st Z = st X + c ∧ st Y = 0) ∧ st X = a }} ⇾
{{ fun st ⇒ st Z = a + st Y + c }};;
WHILE BNot (BEq (AId Y) (ANum b)) DO
{{ fun st ⇒ st Z = a + st Y + c ∧ st Y ≠ b }} ⇾
{{ fun st ⇒ st Z + 1 = a + st Y + 1 + c }}
Y ::= APlus (AId Y) (ANum 1)
{{ fun st ⇒ st Z + 1 = a + st Y + c }};;
Z ::= APlus (AId Z) (ANum 1)
{{ fun st ⇒ st Z = a + st Y + c }}
END
{{ fun st ⇒ (st Z = a + st Y + c) ∧ st Y = b }} ⇾
{{ fun st ⇒ st Z = a + b + c }}
)%dcom.
Theorem two_loops_correct : ∀a b c,
dec_correct (two_loops_dec a b c).
Proof. intros a b c. verify. Qed.
Fixpoint pow2 n :=
match n with
| 0 ⇒ 1
| S n' ⇒ 2 * (pow2 n')
end.
Definition dpow2_down n :=
( {{ fun st ⇒ True }} ⇾
{{ fun st ⇒ 1 = (pow2 (0 + 1))-1 ∧ 1 = pow2 0 }}
X ::= ANum 0
{{ fun st ⇒ 1 = (pow2 (0 + 1))-1 ∧ 1 = pow2 (st X) }};;
Y ::= ANum 1
{{ fun st ⇒ st Y = (pow2 (st X + 1))-1 ∧ 1 = pow2 (st X) }};;
Z ::= ANum 1
{{ fun st ⇒ st Y = (pow2 (st X + 1))-1 ∧ st Z = pow2 (st X) }};;
WHILE BNot (BEq (AId X) (ANum n)) DO
{{ fun st ⇒ (st Y = (pow2 (st X + 1))-1 ∧ st Z = pow2 (st X))
∧ st X ≠ n }} ⇾
{{ fun st ⇒ st Y + 2 * st Z = (pow2 (st X + 2))-1
∧ 2 * st Z = pow2 (st X + 1) }}
Z ::= AMult (ANum 2) (AId Z)
{{ fun st ⇒ st Y + st Z = (pow2 (st X + 2))-1
∧ st Z = pow2 (st X + 1) }};;
Y ::= APlus (AId Y) (AId Z)
{{ fun st ⇒ st Y = (pow2 (st X + 2))-1
∧ st Z = pow2 (st X + 1) }};;
X ::= APlus (AId X) (ANum 1)
{{ fun st ⇒ st 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 st ⇒ st Y = pow2 (n+1) - 1 }}
)%dcom.
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.