Hoare2: Hoare Logic, Part II

Decorated Programs

The beauty of Hoare Logic is that it is structure-guided: the structure of proofs exactly follows the structure of programs. We can record the essential ideas of a Hoare-logic proof -- omitting low-level calculational details -- by "decorating" a program with appropriate assertions on each of its commands. Such a decorated program carries within itself 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 Here is one possible specification for this program, in the form of a Hoare triple: True X := m; Z := p; while X <> 0 do Z := Z - 1; X := X - 1 end Z = p - m (Note the parameters m and p, which stand for fixed-but-arbitrary numbers. Formally, they are simply Coq variables of type nat.) Here is a decorated version of this 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's text interleaved with assertions (sometimes multiple assertions separated by implications). A decorated program can be viewed as a compact representation of a proof in Hoare Logic: the assertions surrounding each command specify the Hoare triple to be proved for that part of the program using one of the Hoare Logic rules, and the structure of the program itself shows how to assemble all these individual steps into a proof for the whole program. Our goal is to verify such decorated programs "mostly automatically." But, before we can verify anything, we need to be able to find a proof for a given specification, and for this 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'll return to finding more interesting loop invariants later in the chapter.

Example: Swapping

Consider the following program, which 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 give a proof, in the form of decorations, that this program is correct -- i.e., it really swaps X and Y -- as follows. (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 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 -- by doing a bit of high-school algebra.

Example: Simple Conditionals

Here is a simple decorated program using conditionals: (1) True if 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 end (8) Z + X = Y \/ Z + Y = X These decorations can be constructed as follows:

• We start with the outer precondition (1) and postcondition (8).
• Following the format dictated by the hoare_if rule, we 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, optional (if_minus_plus_reloaded)

N.b.: Although this exercise is marked optional, it is an excellent warm-up for the (non-optional) if_minus_plus_correct exercise below! Fill in valid decorations for the following program:

Briefly justify each use of ->>. ☐

Example: Reduce to Zero

Here is a while loop that is so simple that True suffices as a loop invariant. (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). We also conjoin (1) with the negation of the guard to obtain (5).
• Because the final postcondition (6) does not syntactically match (5), we add an implication between them.
• Using the assignment rule with assertion (4), we trivially substitute and obtain assertion (3).
• We add the implication between (2) and (3).

Finally we check that the implications do hold; both are trivial.

Example: Division

Let's do one more example of simple reasoning about a loop. The following Imp program calculates the integer quotient and remainder of parameters m and n. X := m; Y := 0; while n <= X do X := X - n; Y := Y + 1 end; 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 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 loop 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 loop 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:

• (1) ->> (2): trivial, by algebra.
• (5) ->> (6): because n ≤ X, we are guaranteed that the subtraction in (6) does not get zero-truncated. We can therefore rewrite (6) as n × Y + n + X - n and cancel the ns, which results in the left conjunct of (5).
• (9) ->> (10): if ¬ (n ≤ X) then X < n. That's straightforward from high-school algebra.

So, we have a valid decorated program.

From Decorated Programs to Formal Proofs

From an informal proof in the form of a decorated program, it is "easy in principle" to read off a formal proof using the Coq theorems corresponding to the Hoare Logic rules, but these proofs can be a bit long and fiddly. Note that we do not unfold the definition of valid_hoare_triple anywhere in this proof: the point of the game we're playing now is to use the Hoare rules as a self-contained logic for reasoning about programs. For example...

In Hoare we introduced a series of tactics named assertion_auto to automate proofs involving assertions. The following declaration introduces a more sophisticated tactic that will help with proving assertions throughout the rest of this chapter. You don't need to understand the details, but briefly: it uses split repeatedly to turn all the conjunctions into separate subgoals, tries to use several theorems about booleans and (in)equalities, then uses eauto and lia to finish off as many subgoals as possible. What's left after verify_assertion does its thing should be just the "interesting parts" of the proof -- which, if we're lucky, might be nothing at all!

This makes it pretty easy to verify reduce_to_zero:

This example shows that it is conceptually straightforward to read off the main elements of a formal proof from a decorated program. Indeed, the process is so straightforward that it can be automated, as we will see next.

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 presentation style can be made completely formal -- and indeed that checking the validity of decorated programs can be largely automated.

Syntax

The first thing we need to do is to formalize a variant of the syntax of Imp commands that includes embedded assertions, which we'll call "decorations." 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 -- something like this...

But this would result in very verbose decorated programs with a lot of repeated annotations: even a simple program like skip;skip would be decorated like this, P (P skip P) ; (P skip P) P with pre- and post-conditions around each skip, plus identical pre- and post-conditions on the semicolon! In other words, 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, the formal syntax of decorated commands omits preconditions whenever possible, trying to embed just the postcondition.

• The skip command, for example, is decorated only with its postcondition
  skip Q
  on the assumption that the precondition will be provided by the context.
  We carry the same assumption through the other syntactic forms: each decorated command is assumed to carry its own postcondition within itself but take its precondition from its context in which it is used.

• Sequences d1 ; d2 need no additional decorations.
  Why?
  Because inside d2 there will be a postcondition; this serves as the postcondition of d1;d2.
  Similarly, inside d1 there will also be a postcondition, which additionally serves as the precondition for d2.

• An assignment X := a is decorated only with its postcondition:
  X := a Q

• A conditional if b then d1 else d2 is decorated with a postcondition for the entire statement, as well as preconditions for each branch:
  if b then P1 d1 else P2 d2 end Q

• A loop while b do d end is decorated with its postcondition and a precondition for the body:
  while b do P d end Q
  The postcondition embedded in d serves as the loop invariant.

• Implications ->> can be added as decorations either for a precondition
  • >> P d
  or for a postcondition
  d ->> Q
  The former is waiting for another precondition to be supplied by the context (e.g., {{ P'}} ->> {{ P }} d); the latter relies on the postcondition already embedded in d.

Putting this all together gives us the formal syntax of decorated commands:

To provide the initial precondition that goes at the very top of a decorated program, we introduce a new type decorated:

To avoid clashing with the existing Notations for ordinary commands, we introduce these notations in a new grammar scope called dcom.

Recall that you can Set Printing All to see how all that notation is desugared.

An example decorated program that decrements X to 0:

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

It is also straightforward to extract the precondition and postcondition from a decorated program.

We can then express what it means for a decorated program to be correct as follows:

For example:

The outer Hoare triple of a decorated program is just a Prop; thus, to show that it is valid, we need to produce a proof of this proposition. We will do this by extracting "proof obligations" from the decorations sprinkled through the program. These obligations are often called verification conditions, because they are the facts that must be verified to see that the decorations are locally consistent and thus constitute a proof of validity of the outer triple.

Extracting Verification Conditions

The function verification_conditions takes a decorated command d together with a precondition P and returns a proposition that, if it can be proved, implies that the triple P erase d post d is valid. It does this by walking over d and generating a big conjunction that includes

• local consistency checks for each form of command, plus
• uses of ->> to bridge the gap between the assertions found inside a decorated command and the assertions imposed by the precondition from its context; these uses correspond to applications of the consequence rule.

Local consistency is defined as follows...

• The decorated command
  skip Q
  is locally consistent with respect to a precondition P if P ->> Q.

• The sequential composition of d1 and d2 is locally consistent with respect to P if d1 is locally consistent with respect to P and d2 is locally consistent with respect to the postcondition of d1.

• An assignment
  X := a Q
  is locally consistent with respect to a precondition P if:
  P ->> Q X |-> a

• A conditional
  if b then P1 d1 else P2 d2 end Q
  is locally consistent with respect to precondition P if
  (1) P ∧ b ->> P1
  (2) P ∧ ¬b ->> P2
  (3) d1 is locally consistent with respect to P1
  (4) d2 is locally consistent with respect to P2
  (5) post d1 ->> Q
  (6) post d2 ->> Q

• A loop
  while b do Q d end R
  is locally consistent with respect to precondition P if:
  (1) P ->> post d
  (2) post d ∧ b ->> Q
  (3) post d ∧ ¬b ->> R
  (4) d is locally consistent with respect to Q

• A command with an extra assertion at the beginning
  • -> Q d
  is locally consistent with respect to a precondition P if:
  (1) P ->> Q
  (2) d is locally consistent with respect to Q

• A command with an extra assertion at the end
  d ->> Q
  is locally consistent with respect to a precondition P if:
  (1) d is locally consistent with respect to P
  (2) post d ->> Q

With all this in mind, we can write is a verification condition generator that takes a decorated command and reads off a proposition saying that all its decorations are locally consistent. Formally, since a decorated command is "waiting for its precondition" the main VC generator takes a dcom plus a given predondition as arguments.

The following key theorem states that verification_conditions does its job correctly. Not surprisingly, each of the Hoare Logic rules gets used at some point in the proof.

Now that all the pieces are in place, we can define what it means to verify an entire program.

This brings us to the main theorem of this section:

More Automation

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

Fortunately, our verify_assertion tactic can generally take care of most or all of them.

To automate the overall process of verification, we can use verification_correct to extract the verification conditions, use verify_assertion to verify them as much as it can, and finally tidy up any remaining bits by hand.

Here's the final, formal proof that dec_while is correct.

Similarly, here is the formal decorated program for the "swapping by adding and subtracting" example that we saw earlier.

And here is the formal decorated version of the "positive difference" program from earlier:

Exercise: 2 stars, standard, especially useful (if_minus_plus_correct)

Here is a skeleton of the formal decorated version of the if_minus_plus program that we saw earlier. Replace all occurrences of FILL_IN_HERE with appropriate assertions and fill in the proof (which should be just as straightforward as in the examples above).

☐

Exercise: 2 stars, standard, optional (div_mod_outer_triple_valid)

Fill in appropriate assertions for the division program from above.

☐

Finding Loop Invariants

Once the outermost precondition and postcondition are chosen, the only creative part of a verifying program 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 a loop invariant means that you have a stronger assumption to work with when trying to establish the postcondition of the loop body, but it also means that the loop body's postcondition is stronger and thus harder to prove.
• Similarly, strengthening an induction hypothesis means that you have a stronger assumption to work with when trying to complete the induction step of the proof, but it also means that 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 a single iteration of the loop, assuming that the loop guard also evaluates to true, i.e., (3) must imply (4).

These conditions are actually independent of the particular program and specification we are considering: every loop invariant has to satisfy them. One way to find a loop 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 a loop invariant did the job. Maybe it will work here too. To find out, 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, (b) is wrong: 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 loop invariant so that it implies the postcondition (8). One simple way to do this is to let the loop invariant be the postcondition. So let's return to our skeleton, instantiate Inv with Y = n - m, and try checking 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 loop 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 a loop invariant! To do better, we need to generalize (7) 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.) Here is the final version of the decorated program:

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. Fill in decorations and prove the decorated program correct. (The proof should be very simple.)

☐

Example: Parity

Here is a cute way of computing the parity of a value initially stored in X, due to Daniel Cristofani. X = m while 2 <= X do X := X - 2 end X = parity m The 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 hope to use that as a loop invariant. To find a loop 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 loop 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, optional (parity)

Translate the above informal decorated program into a formal one and prove it correct. Hint: There are actually several possible loop invariants that all lead to good proofs; one that leads to a particularly simple proof is parity X = parity m -- or more formally, using the ap operator to lift the application of the parity function into the syntax of assertions, {{ ap parity X = parity m }}.

If you use the suggested loop invariant, you may find the following lemmas helpful (as well as leb_complete and leb_correct).

☐

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*Z<=m /\ m<(Z+1)*(Z+1) As we did before, we can try to use the postcondition as a candidate loop 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*Z<=m /\ m<(Z+1)*(Z+1) /\ (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*Z<=m /\ m<(Z+1)*(Z+1) end (7) Z*Z<=m /\ m<(Z+1)*(Z+1) /\ ~((Z+1)*(Z+1)<=X) ->> (b - OK) (8) Z*Z<=m /\ 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*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 /\ ~((Z+1)*(Z+1)<=X) ->> (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, when 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 record the fact that it doesn't change in the loop invariant.

Exercise: 3 stars, standard, optional (sqrt)

Translate the above informal decorated program into a formal one and prove it correct. Hint: The loop invariant here must ensure that Z*Z is consistently less than or equal to X.

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 can be a good idea in such cases to add X = m to the loop invariant. The other thing that we know is often useful in the loop invariant is the postcondition, so let's add that too, leading to the candidate loop invariant 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 = 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 loop 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 loop 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 loop 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 loop invariants. ☐

Exercise: Factorial

Exercise: 4 stars, advanced (factorial_correct)

Recall that n! denotes the factorial of n (i.e., n! = 1*2*...*n). Formally, 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.

First, write the Imp program factorial that calculates the factorial of the number initially stored in the variable X and puts it in the variable Y. Using your definition factorial and slow_assignment_dec as a guide, write a formal decorated program factorial_dec that implements the factorial function. Hint: recall the use of ap in assertions to apply a function to an Imp variable. Fill in the blanks and finish the proof of correctness. Bear in mind that we are working with natural numbers, for which both division and subtraction can behave differently than with real numbers. Excluding both operations from your loop invariant is advisable! Then state a theorem named factorial_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: if those two subgoals become tedious to prove, give some thought to how you could restate your assertions such that the mathematical operations are more amenable to manipulation in Coq. For example, recall that 1 + ... is easier to work with than ... + 1.

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Exercise: Minimum

Exercise: 3 stars, standard (minimum_correct)

Fill in decorations for the following program and prove them correct. As with factorial, be careful about mathematical reasoning involving natural numbers, especially subtraction. Also, remember that applications of Coq functions in assertions need an ap or ap2 to be parsed correctly. E.g., min a b needs to be written ap2 min a b in an assertion. You may find andb_true_eq useful (perhaps after using symmetry to get an equality the right way around).

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Exercise: Two Loops

Exercise: 3 stars, standard (two_loops)

Here is a pretty 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 completing the following decorated program.

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Exercise: Power Series

Exercise: 4 stars, standard, optional (dpow2)

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 Turn this into a decorated program and prove it correct.