Hoare2: Hoare Logic, Part II

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 (leaving out some low-level calculational details) by decorating programs with appropriate assertions around each statement. Such a decorated program carries with it an (informal) proof of its own correctness. For example, here is a complete decorated program: 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. To check that a decorated program represents a valid proof, we check that each individual command is locally consistent with its accompanying 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 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 and if the pre- and postconditions of its body are exactly P ∧ b and P: 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 ->> 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 involves logical and/or arithmetic reasoning.

We have seen above how verifying the correctness of a given proof 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 automatic way, with the exception of finding loop invariants, which is the subject of in the next section. In the reminder of this section we explain in detail how to construct decorations for several simple programs that don't involve non-trivial loop invariants.

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 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 The decorations were constructed as follows:

• 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) all the way 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

(Note that, since we are working with natural numbers, not fixed-size machine integers, we don't need to worry about the possibility of arithmetic overflow anywhere in this argument.)

Example: Simple Conditionals

Here is a simple decorated program using 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:

• 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

Example: Reduce to Zero (Trivial Loop)

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:

• 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.

From this informal proof, 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.

Example: Division

The following Imp program calculates the integer division 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 other words, if 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 reminder X and a quotient Y so 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 the just first conjunct n × Y + X = m, so we use that 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 /\ 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 two uses of the consequence rule are correct -- i.e., that (1) implies (2) and that (5) implies (6). 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 with Hoare Logic is finding the right loop invariants. The reason this is difficult is the same as the reason that doing inductive mathematical proofs requires creativity: 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 (complete the induction step of the proof), but it also means that the loop body postcondition itself is harder to prove! This section is dedicated to teaching you how to approach the challenge of finding loop invariants using 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 following specification: 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: (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:

• (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).

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 the process. 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 it again here! I.e., let's instantiate I with True in the skeleton above 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 (7) True ∧ X = 0 implies (8) Y = n - m. In fact, the two assertions are completely unrelated and it is easy to find a counterexample (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 very simple way to do this is to let the invariant be the postcondition. So let's return to our skeleton, instantiate I 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 does not hold (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, so this difference is always n - m. So let's try instantiating I 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 (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 is correct.

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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.

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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:

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 /\ X < 2 ->> (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, 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:

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Example: Finding Square Roots

The following program computes the square root of X by naive iteration: X=m Z ::= 0; WHILE (Z+1)*(Z+1) <= X DO Z ::= Z+1 END Z*Z<=m /\ 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<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) This didn't work very well: both conditions (a) and (c) 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 have X=m, but we didn't propagate this information from (1) into the loop invariant. Also, looking at the second conjunct of (8), it seems quite hopeless as an invariant -- and we don't even need it, since we can obtain it from the negation of the guard (third conjunct in (7)), again under the assumption that X=m. 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*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) 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, in such cases it is a good idea to add X = m to the invariant. The other thing we often use 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 <> 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 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 progesses 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, we derive the 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 <> 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 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 auxiliary variabes (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 (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: X = 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 : forall x y, (x=0 \/ y=0) -> min x y = 0. Lemma lemma2 : forall x y, min (x-1) (y-1) = (min x y) - 1. plus, as usual, standard high-school algebra. 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

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 Show that it does what it should by filling in the blanks in the following decorated program. 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

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.

Weakest Preconditions (Advanced)

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 a given 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}}, we have P' st implies P st for all states st.

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 Q 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

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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.

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Exercise: 2 stars, advanced (hoare_asgn_weakest)

Show that the precondition in the rule hoare_asgn is in fact the weakest precondition.

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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.

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Formal Decorated Programs (Advanced)

The 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 the 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.

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.

It is easy to go from a dcom to a com by erasing all annotations.

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! Instead, the rule we've followed is this:

• The post-condition expected by each dcom d is embedded in d
• The pre-condition is supplied by the context.

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:

Similarly, we can extract the "initial precondition" from a decorated program.

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:

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.

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.)

And now, the key theorem, which states 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. We have used in variants of several tactics before to apply them to values in the context rather than the goal. An extension of this idea is the syntax tactic in ×, which applies tactic in the goal and every hypothesis in the context. We most commonly use this facility in conjunction with the simpl tactic, as below.

Examples

The propositions generated by verification_conditions are fairly big, and they contain many conjuncts that are essentially trivial.

==> (((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) In principle, we could certainly work with them 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 applies splitting repeatedly to turn all the conjunctions into separate subgoals and then uses omega and eauto (a handy general-purpose automation tactic that we'll discuss in detail later) to deal with as many of them as possible.

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).

Another example (formalizing a decorated program we've seen before):

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.

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Exercise: 4 stars, advanced (factorial_dec)

Remember the factorial function we worked with before:

Following the pattern of subtract_slowly_dec, write a decorated program that implements the factorial function and prove it correct.

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