Proving Theorems

Verifying the contract of a function is really proving a mathematical theorem. Stainless can be seen as a (mostly) automated theorem prover. It is automated in the sense that once the property stated, Stainless will proceed with searching for a proof without any user interaction. In practice, however, many theorems will be fairly difficult to prove, and it is possible for the user to provide hints to Stainless.

Hints typically take the form of simpler properties that combine in order to prove more complicated ones. In the remaining subsections, we provide code patterns and introduce simple domain-specific language operators that can help in constructing complex Stainless proofs.

A practical introduction to proofs

When writing logical propositions such as preconditions or postconditions in Stainless, one is basically writing Boolean predicates. They can be as simple as testing whether a list is empty or not, to more complex combinations of properties. Lemmas or theorems are simply logical tautologies, that is, propositions that always hold. They can be expressed using Boolean-valued methods that return true for all their inputs.

To make this more concrete, let’s take a simple lemma as an example [1]. Here we want to prove that the append operation (++) on lists preserves the content of the two lists being concatenated. This proposition is relatively straightforward and Stainless is able to verify that it always holds.

import stainless.collection._ // for List
import stainless.lang._       // for holds

object Example {
  def appendContent[A](l1: List[A], l2: List[A]): Boolean = {
    l1.content ++ l2.content == (l1 ++ l2).content
  }.holds
}

Here we wrote .holds which is a method implicitly available on Boolean that ensure the returned value is true. It is equivalent to writing ensuring { res => res }.

Now let’s look at another example that looks trivial but for which Stainless needs some help with the proof: we want to prove that adding Nil at the end of a list has no effect.

import stainless.collection._ // for List
import stainless.lang._       // for holds

object Example {
  def rightUnitAppend[T](l1: List[T]): Boolean = {
    l1 ++ Nil() == l1
  }.holds
}

If you try to verify this last example you’ll face a delicate situation: Stainless runs indeterminately until it is either killed or times out. But why does this happen? The proposition doesn’t seem more complicated than appendContent. Perhaps even more surprisingly, Stainless is able to verify the following:

def leftUnitAppend[T](l1: List[T]): Boolean = {
  Nil() ++ l1 == l1
}.holds

How is this possible? The two propositions are completely symmetric! The problem is that Stainless doesn’t know anything about lists, a priori. It can only reason about lists thanks to their definition in terms of the case classes Cons and Nil and associated methods such as ++. In particular, Stainless doesn’t know that Nil represents the empty list, and hence that appending it to some other list is a no-op. What then, is the difference between the two examples above? To answer this question, we need to have a look at the definition of the ++ method:

def ++(that: List[T]): List[T] = (this match {
  case Nil()       => that
  case Cons(x, xs) => Cons(x, xs ++ that)
}).ensuring { res => /* ... */ }

Note that the implementation of ++ is recursive in its first argument this but not in its second argument that. This is why Stainless was able to verify leftUnitAppend easily: it is true by definition, i.e. Nil() ++ l1 is actually defined to be l1. What about the symmetric case? How is l1 ++ Nil() defined? Well, it depends on whether l1 is the empty list or not. So in order to prove rightUnitAppend, we need to proceed with a case analysis. The resulting proof has a recursive (i.e. inductive) structure reminiscent of the definition of ++:

import stainless.collection._ // for List
import stainless.lang._       // for holds
import stainless.proof._      // for because

object Example {
  def rightUnitAppend[T](l1: List[T]): Boolean = {
    (l1 ++ Nil() == l1) because {
      l1 match {
        case Nil()       => true
        case Cons(x, xs) => rightUnitAppend(xs)
      }
    }
  }.holds
}

With this new implementation of the rightUnitAppend lemma, Stainless is capable of verifying that it holds. If you look closely at it, you can distinguish three parts:

1. the claim we want to prove l1 ++ Nil() == l1;

2. because, which is just some syntactic sugar for conjunction – remember, every proposition is a Boolean formula;

3. and a recursion on l1 that serves as a hint for Stainless to perform induction.

The recursion is based here on pattern matching, which Stainless will also check for exhaustiveness. It has essentially the same structure as the implementation of ++: the base case is when l1 is the empty list and the inductive case is performed on Cons objects.

Techniques for proving non-trivial propositions

In the previous section, we saw that “proof hints” can improve the odds of Stainless successfully verifying a given proposition. In this section, we will have a closer look at what constitutes such a proof and discuss a few techniques for writing them.

As mentioned earlier, propositions are represented by Boolean expressions in Stainless. But how are proofs represented? They are just Boolean expressions as well [2]. The difference between propositions and proofs is not their representation, but how they are used by Stainless. Intuitively, a proof p: Boolean for a proposition x: Boolean is an expression such that

1. Stainless is able to verify p, and

2. Stainless is able to verify that p implies x.

This is what we mean when we say that proofs are “hints”. Typically, a proof p of a proposition x is a more complex-looking but equivalent version of x, i.e. an expression such that p == x. This might seem a bit counter-intuitive: why should it be easier for Stainless to verify an equivalent but more complex expression? The answer is that the more complex version may consist of sub-expressions that more closely resemble the definitions of functions used in x. We have already seen an example of this principle in the previous section: let’s have another look at the proof of rightUnitAppend:

def rightUnitAppend[T](l1: List[T]): Boolean = {
  val x = l1 ++ Nil() == l1
  val p = l1 match {
    case Nil()       => true
    case Cons(x, xs) => rightUnitAppend(xs)
  }
  x because p
}.holds

Here, we have rewritten the example to make the distinction between the proposition x and its proof p more explicit. It’s easy to check that indeed x == p, and hence the overall result of rightUnitAppend is equivalent to x (recall that because is just an alias of &&, so (x because p) == (x && x) == x). However, the proof term p closely resembles the definition of ++ and its sub-expressions are easier to verify for Stainless than x itself. The only non-trivial expression is the recursive call to rightUnitAppend(xs), which serves as the inductive hypothesis. We will discuss induction in more detail in Section “Induction”.

Divide and Conquer

Before we delve into the details of particular proof techniques, it is worth revisiting a guiding principle for writing proofs – whether it be in Stainless, by hand, or using some other form of mechanized proof checker – namely to modularize proofs, i.e. to split the proofs of complex propositions into manageable sub-goals. This can be achieved in various ways.

• Use helper lemmas – these are propositions that are lemmas on their own, i.e. they state and prove simple but self-contained propositions that can be reused elsewhere. As such, they play a role akin to helper methods in normal programming, and indeed, are implemented in the same way, except that they carry a .holds suffix.

• Use case analysis to split complex propositions into simpler sub-cases. This is especially helpful in the presence of recursion, where it leads to inductive proofs (see Section “Induction”).

• Use relational reasoning to split complex relationships into conjunctions of elementary ones. This often requires one to make use of relational properties such as transitivity (e.g. to break a single equation a == b into a chain of equations a == x1 && x1 == x2 && ... && xN == b), symmetry (e.g. to use a previously proven inequality a <= b where b >= a is expected), anti-symmetry (to unify variables), and so on (see Section “Relational reasoning”).

• Separate specification form implementation. It is sometimes easier to prove the fact that a given function fulfills its specification as a separate lemma (although the proof techniques are roughly the same, see Section “Techniques for proving non-trivial postconditions”).

• Generalize (or specialize) propositions. Sometimes, propositions are more easily proved in a stronger (or weaker) form and subsequently instantiated (or combined with other propositions) to yield a proof of the original proposition.

While it is good practice to factor common propositions into helper lemmas, one sometimes wants to verify simple, specific sub-goals in a proof without going through the trouble of introducing an additional method. This is especially true while one is exploring the branches of a case analysis or wants to quickly check whether Stainless is able to prove a seemingly trivial statement automatically (we will see examples of such situations in the coming sections). For such cases, one can use the check function from stainless.proof. The check function behaves as the identity function on Booleans but additionally assumes its argument in its precondition. As a result, Stainless will check that x holds while verifying the call to check(x). For example, when verifying the following function:

import stainless.proof.check

def foo(x: BigInt): Boolean = {
  check(x >= 0 || x < 0) && check(x + 0 == 0)
}

Stainless will check (separately) that x >= 0 || x < 0 and x + 0 == 0 hold for all x, even though the function foo does not specify any pre or postconditions, and report a counter example for the second case:

[  Info  ]  - Now considering 'precond. (call check(x >= 0 || x < 0))' VC for foo @40:5...
[  Info  ]  => VALID
[  Info  ]  - Now considering 'precond. (call check(x + 0 == 0))' VC for foo @40:31...
[ Error  ]  => INVALID
[ Error  ] Found counter-example:
[ Error  ]   x -> 1

This is especially helpful when “debugging” proofs.

Proof control using assert and check

Both the assert and check keywords generate a verification condition for the corresponding formula in the current context. The difference in these keywords is in how they affect the context of other verification conditions. As a rule of thumb, assertions do not affect the context of verification conditions outside the block of the assertion, while check does. Thus you can use assertions to prove local properties, and use check to have the property (checked and) visible outside the block.

def foo(): Unit = {
  val x = {
    assert(b1) // verification condition: b1
    check(b2)  // verification condition: b1 ==> b2
  }
  assert(b3)   // verification condition: b2 ==> b3 (b1 not visible to the solver)
}

Similarly, assert’s are not guaranteed to be visible when generating verification conditions for postconditions, while check’s are.

 def foo(): Unit = {
   assert(b1) // verification condition: b1
   check(b2)  // verification condition: b1 ==> b2
}.ensuring(_ => b3) // verification condition b2 ==> b3 (b1 might not be visible to the solver)

Induction

The vast majority of functional programs are written as functions over ADTs, and consequently, Stainless comes with some special support for verifying properties of ADTs. Among other things, Stainless provides an annotation @induct, which can be used to automatically prove postconditions of recursive functions defined on ADTs by way of structural induction. We have already seen an example of such an inductive property, namely rightUnitAppend. In fact, using @induct, Stainless is able to prove rightUnitAppend directly:

import stainless.annotation._  // for @induct

@induct
def rightUnitAppend[T](l1: List[T]): Boolean = {
  l1 ++ Nil() == l1
}.holds

This is possible because the inductive step follows (more or less) directly from the inductive hypothesis and Stainless can verify the base case automatically. However, Stainless may fail to verify more complex functions with non-trivial base cases or inductive steps. In such cases, one may still try to provide proof hints by performing manual case analysis. Consider the following lemma about list reversal:

import stainless.collection._ // for List
import stainless.lang._       // for holds

object Example {
  def reverseReverse[T](l: List[T]): Boolean = {
    l.reverse.reverse == l
  }.holds
}

Stainless is unable to verify reverseReverse even using @induct. So let’s try and prove the lemma using manual case analysis. We start by adding an “unrolled” version of the proposition and inserting calls to check in each branch of the resulting pattern match:

def reverseReverse[T](l: List[T]): Boolean = {
  l.reverse.reverse == l because {
    l match {
      case Nil()       => check {  Nil[T]().reverse.reverse == Nil[T]()  }
      case Cons(x, xs) => check { (x :: xs).reverse.reverse == (x :: xs) }
    }
  }
}.holds

Clearly, the two versions of the lemma are equivalent: all we did was expand the proposition using a pattern match and add some calls to check (remember check acts as the identity function on its argument). Let’s see what output Stainless produces for the expanded version:

[  Info  ]  - Now considering 'postcondition' VC for reverseReverse @615:5...
[Warning ]  => UNKNOWN
[  Info  ]  - Now considering 'precond. (call check(List[T]().reverse().reverse() ...)' VC for reverseReverse @617:28...
[  Info  ]  => VALID
[  Info  ]  - Now considering 'precond. (call check({val x$27 = l.h; ...)' VC for reverseReverse @618:28...
[Warning ]  => UNKNOWN
[  Info  ]  - Now considering 'match exhaustiveness' VC for reverseReverse @616:7...
[  Info  ]  => VALID

As expected, Stainless failed to verify the expanded version. However, we get some additional information due to the extra pattern match and the calls to check. In particular, Stainless tells us that the match is exhaustive, which means we covered all the cases in our case analysis – that’s certainly a good start. Stainless was also able to automatically verify the base case, so we can either leave the call to check as is, or replace it by trivial. Unfortunately, Stainless wasn’t able to verify the inductive step, something is missing. Let’s try to manually reduce the inductive case and see where we get.

def reverseReverse[T](l: List[T]): Boolean = {
  l.reverse.reverse == l because {
    l match {
      case Nil()       => trivial
      case Cons(x, xs) => check { (xs.reverse :+ x).reverse == (x :: xs) }
    }
  }
}.holds

And now we’re stuck. We can’t apply the inductive hypothesis here, nor can we reduce the inductive case further, unless we perform case analysis on xs, which would grow the term further without changing its shape. To make any headway, we need to use an additional property of reverse, given by the following lemma (which Stainless is able to prove using @induct):

@induct
def snocReverse[T](l: List[T], t: T): Boolean = {
  (l :+ t).reverse == t :: l.reverse
}.holds

The lemma states that appending an element t to a list l and reversing it is equivalent to first reversing l and then prepending t. Using this lemma, we can write the proof of reverseReverse as

def reverseReverse[T](l: List[T]): Boolean = {
  l.reverse.reverse == l because {
    l match {
      case Nil()       => trivial
      case Cons(x, xs) => check {
        (xs.reverse :+ x).reverse == x :: xs.reverse.reverse &&
        x :: xs.reverse.reverse   == (x :: xs)               because
          snocReverse(xs.reverse, x) && reverseReverse(xs)
      }
    }
  }
}.holds

Stainless is able to verify this version of the lemma. Note that Stainless doesn’t actually require us to include the two equations as they are equivalent to the applications snocReverse(xs.reverse, x) and reverseReverse(xs). Similarly, the call to check is somewhat redundant now that Stainless is able to verify the entire proof. We could thus “simplify” the above to

def reverseReverse[T](l: List[T]): Boolean = {
  l.reverse.reverse == l because {
    l match {
      case Nil()       => trivial
      case Cons(x, xs) => snocReverse(xs.reverse, x) && reverseReverse(xs)
    }
  }
}.holds

However, the previous version is arguably more readable for a human being, and therefore preferable. In Section “Relational reasoning” we will see how readability can be improved even further through the use of a DSL for equational reasoning.

So far, we have only considered structurally inductive proofs. However, Stainless is also able to verify proofs using natural induction – the form of induction that is perhaps more familiar to most readers. Consider the following definition of the exponential function \(exp(x, y) = x^y\) over integers:

def exp(x: BigInt, y: BigInt): BigInt = {
  require(y >= 0)
  if      (x == 0) 0
  else if (y == 0) 1
  else             x * exp(x, y - 1)
}

The function exp is again defined recursively, but this time using if statements rather than pattern matching. Let’s try and prove some properties of this function using natural induction. One such property is that for any pair of positive numbers \(x, y \geq 0\), the exponential \(x^y\) is again a positive number.

def positive(x: BigInt, y: BigInt): Boolean = {
  require(y >= 0 && x >= 0)
  exp(x, y) >= 0
}

Since Stainless doesn’t know anything about exponentials, it isn’t able to verify the lemma without hints. As with the previous example, we start writing our inductive proof by expanding the top-level if statement in the definition of exp.

def positive(x: BigInt, y: BigInt): Boolean = {
  require(y >= 0 && x >= 0)
  exp(x, y) >= 0 because {
    if      (x == 0) check { exp(x, y) >= 0 }  // <-- valid
    else if (y == 0) check { exp(x, y) >= 0 }  // <-- valid
    else             check { exp(x, y) >= 0 }  // <-- unknown
  }
}.holds

Stainless was able to verify the first two (base) cases, but not the inductive step, so let’s continue unfolding exp for that case.

def positive(x: BigInt, y: BigInt): Boolean = {
  require(y >= 0 && x >= 0)
  exp(x, y) >= 0 because {
    if      (x == 0) trivial
    else if (y == 0) trivial
    else             check { x * exp(x, y - 1) >= 0 }
  }
}.holds

Although Stainless still isn’t able to verify the lemma, we now see a way to prove the inductive step: x is positive (by the second precondition) and so is exp(x, y - 1) (by the inductive hypothesis). Hence the product x * exp(x, y - 1) is again positive.

def positive(x: BigInt, y: BigInt): Boolean = {
  require(y >= 0 && x >= 0)
  exp(x, y) >= 0 because {
    if      (x == 0) trivial
    else if (y == 0) trivial
    else             check {
      x >= 0 && exp(x, y - 1) >= 0 because positive(x, y - 1)
    }
  }
}.holds

With these hints, Stainless is able to verify the proof. Again, we could shorten the proof by omitting inequalities that Stainless can infer directly, albeit at the expense of readability.

def positiveShort(x: BigInt, y: BigInt): Boolean = {
  require(y >= 0 && x > 0)
  exp(x, y) >= 0 because {
    if      (x == 0) trivial
    else if (y == 0) trivial
    else             positiveShort(x, y - 1)
  }
}.holds

We conclude the section with the inductive proof of another, somewhat more interesting property of the exponential function, namely that \((x y)^z = x^z y^z\).

def expMultCommute(x: BigInt, y: BigInt, z: BigInt): Boolean = {
  require(z >= 0)
  exp(x * y, z) == exp(x, z) * exp(y, z) because {
    if      (x == 0) trivial
    else if (y == 0) trivial
    else if (z == 0) trivial
    else             check {
      x * y * exp(x * y, z - 1) ==
        x * exp(x, z - 1) * y * exp(y, z - 1) because
        expMultCommute(x, y, z - 1)
    }
  }
}.holds

Relational reasoning

The majority of the example propositions we have seen so far related some expression (e.g. l.reverse ++ Nil() or exp(x, y)) to some other expression (e.g. ... == l1 or ... >= 0). This is certainly a common case among the sorts of propositions about functions and data structures one might wish to prove. The proofs of such propositions typically involve some form of relational reasoning, i.e. reasoning involving properties (such as transitivity, reflexivity or symmetry) of the relations in question. Stainless knows about these properties for built-in relations such as == or orders on numbers. For user-defined relations, they first need to be established as lemmas. In this section, we discuss how to make effective use of built-in relations, but the general principles extend to their user-defined counterparts.

When working with simple structural equality, we can rely on the default == operator and Stainless will happily understand when the reflexivity, symmetry and transitivity properties apply and use them to conclude bigger proofs. Similarly, when working on BigInt, Stainless knows about reflexivity, antisymmetry and transitivity over >= or <=, and also antireflexivity, antisymmetry and transitivity of > and <.

However, even for relatively simple proofs about ADTs, Stainless needs hints when combining, say equality, with user-defined operations, such as ++ or reverse on lists. For example, Stainless is not able to verify that the following holds for arbitrary pairs of lists l1 and l2:

(l1 ++ l2).reverse == l2.reverse ++ l1.reverse

The hard part of giving hints to Stainless is often to find them in the first place. Here we can apply a general principle on top of structural induction (as discussed in the previous section): we start from the left-hand side of an equation and build a chain of intermediate equations to the right-hand side. Using check statements we can identify where Stainless times out and hence potentially needs hints.

def reverseAppend[T](l1: List[T], l2: List[T]): Boolean = {
  ( (l1 ++ l2).reverse == l2.reverse ++ l1.reverse ) because {
    l1 match {
      case Nil() =>
        /* 1 */ check { (Nil() ++ l2).reverse == l2.reverse                  } &&
        /* 2 */ check { l2.reverse            == l2.reverse ++ Nil()         } &&
        /* 3 */ check { l2.reverse ++ Nil()   == l2.reverse ++ Nil().reverse }
      case Cons(x, xs) =>
        /* 4 */ check { ((x :: xs) ++ l2).reverse       == (x :: (xs ++ l2)).reverse       } &&
        /* 5 */ check { (x :: (xs ++ l2)).reverse       == (xs ++ l2).reverse :+ x         } &&
        /* 6 */ check { (xs ++ l2).reverse :+ x         == (l2.reverse ++ xs.reverse) :+ x } &&
        /* 7 */ check { (l2.reverse ++ xs.reverse) :+ x == l2.reverse ++ (xs.reverse :+ x) } &&
        /* 8 */ check { l2.reverse ++ (xs.reverse :+ x) == l2.reverse ++ (x :: xs).reverse }
    }
  }
}.holds

If we run the above code with a decent timeout, Stainless reports four UNKNOWN cases: the postcondition of the reverseAppend function itself and checks number 2, 6 and 7.

• Check #2 fails because, as we saw earlier, Stainless is not capable of guessing the rightUnitAppend lemma by itself. We fix this case by simply instantiating the lemma, i.e. by appending && rightUnitAppend(l2.reverse) to the base case.

• Check #6 fails because, at this point, we need to inject the induction hypothesis on xs and l2 by adding && reverseAppend(xs, l2).

• Finally, check #7 fails for a similar reason as check #2: we need an additional “associativity” lemma to prove that (l1 ++ l2) :+ t == l1 ++ (l2 :+ t) holds for any l1, l2 and t. We call this lemma snocAfterAppend and leave it as an exercise for the reader.

Once we have a valid proof, we can try to optimize it for readability. As it stands, the resulting code is rather verbose because both sides of most equations are duplicated. One option is to completely remove the equations (they are implied by the instantiations of the lemmas) and simply write

def reverseAppend[T](l1: List[T], l2: List[T]): Boolean = {
  ( (l1 ++ l2).reverse == l2.reverse ++ l1.reverse ) because {
    l1 match {
      case Nil() =>
        rightUnitAppend(l2.reverse)
      case Cons(x, xs) =>
        reverseAppend(xs, l2) && snocAfterAppend(l2.reverse, xs.reverse, x)
    }
  }
}.holds

Or we can employ the equational reasoning DSL provided by the stainless.proofs package to remove the duplicate expressions and interleave the equations with their associated proofs. This has the advantage of not losing information that is still useful for a human being reading the proof later on:

def reverseAppend[T](l1: List[T], l2: List[T]): Boolean = {
  ( (l1 ++ l2).reverse == l2.reverse ++ l1.reverse ) because {
    l1 match {
      case Nil() => {
        (Nil() ++ l2).reverse         ==| trivial                     |
        l2.reverse                    ==| rightUnitAppend(l2.reverse) |
        l2.reverse ++ Nil()           ==| trivial                     |
        l2.reverse ++ Nil().reverse
      }.qed
      case Cons(x, xs) => {
        ((x :: xs) ++ l2).reverse         ==| trivial               |
        (x :: (xs ++ l2)).reverse         ==| trivial               |
        (xs ++ l2).reverse :+ x           ==| reverseAppend(xs, l2) |
        (l2.reverse ++ xs.reverse) :+ x   ==|
          snocAfterAppend(l2.reverse, xs.reverse, x)                |
        l2.reverse ++ (xs.reverse :+ x)   ==| trivial               |
        l2.reverse ++ (x :: xs).reverse
      }.qed
    }
  }
}.holds

The idea is to group statements in a block ({ }) and call qed on it. Then, instead of writing a == b && b == c && hint1 && hint2 we write a ==| hint1 | b ==| hint2 | c. And when no additional hint is required, we can use trivial which simply stands for true.

Additionally, by using this DSL, we get the same feedback granularity from Stainless as if we had used check statements. This way we can construct proofs based on equality more easily and directly identify where hints are vital.

One shortcoming of the relational reasoning DSL is that it relies on Stainless’ knowledge of the relational properties of the built-in relations, and in particular those of equality. Consequently it works badly (if at all) with user-defined relations. However, since the DSL is defined as a library (in library/proof/package.scala), it can in principle be extended and modified to include specific user-defined relations on a case-by-case basis.

Limits of the approach: HOFs, quantifiers and termination

While the techniques discussed in this section are useful in general, their applicability has, of course, its limitations in practice. These limitations are mostly due to Stainless’ limited support for certain language constructs, such as higher-order functions (HOFs) or quantifiers (which in turn is due, mostly, to the limited support of the corresponding theories in the underlying SMT solvers).

Still, even using these “experimental” features, one manages to prove some interesting propositions. Here is another list example, which relates the foldLeft, foldRight and reverse operations defined on lists and makes crucial use of HOFs:

import stainless.collection._
import stainless.lang._
import stainless.proof._

def folds[A, B](xs: List[A], z: B, f: (B, A) => B): Boolean = {
  val f2 = (x: A, z: B) => f(z, x)
  xs.foldLeft(z)(f) == xs.reverse.foldRight(z)(f2) because {
    xs match {
      case Nil() => true
      case Cons(x, xs) => {
        (x :: xs).foldLeft(z)(f)              ==| trivial               |
        xs.foldLeft(f(z, x))(f)               ==| folds(xs, f(z, x), f) |
        xs.reverse.foldRight(f(z, x))(f2)     ==| trivial               |
        xs.reverse.foldRight(f2(x, z))(f2)    ==|
          snocFoldRight(xs.reverse, x, z, f2)                           |
        (xs.reverse :+ x).foldRight(z)(f2)    ==| trivial               |
        (x :: xs).reverse.foldRight(z)(f2)
      }.qed
    }
  }
}.holds

A rather different, more general issue that arises when proving propositions using Stainless is related to termination checking. When verifying inductive proofs (and more generally the postconditions of recursive methods), Stainless assumes that the corresponding proofs are well-founded, or equivalently, that the corresponding recursive methods terminate on all inputs. It is thus possible – and indeed rather easy – to write bogus proofs (intentionally or accidentally) which Stainless recognizes as valid, but which are not well-founded. Consider the following lemma, which apparently establishes that all lists are empty:

import stainless.collection._
import stainless.lang._
import stainless.proof._

object NotWellFounded {

  // This proof is not well-founded.  Since Stainless doesn't run the
  // termination checker by default, it will accept the proof as
  // valid.
  def allListsAreEmpty[T](xs: List[T]): Boolean = {
    xs.isEmpty because {
      xs match {
        case Nil()       => trivial
        case Cons(x, xs) => allListsAreEmpty(x :: xs)
      }
    }
  }.holds
}

Stainless hences performs termination checking by default to minimize the risk of accidentally writing bogus proofs such as the one above. It will thus emit a warning if it cannot prove that a function terminates, or if it can show that its measure (inferred or user-defined) does not decreases between recursive calls.

Techniques for proving non-trivial postconditions

When proving a mathematical lemma, the return type of the corresponding function is most of the time, if not always, Boolean. For those proofs it is rather easy to write a postcondition: using holds is generally enough.

But when it comes to writing postconditions for more general functions, such as the addition of rational numbers, we are no longer dealing with Boolean so we need a strategy to properly write ensuring statements.

Rationals: a simple example

Let’s take rational numbers as an example: we define them as a case class with two attributes, n for the numerator and d for the denominator. We also define three simple properties on them: isRational, isNonZero and isPositive.

case class Rational(n: BigInt, d: BigInt) {
  def isRational = d != 0
  def isPositive = isRational && (n * d >= 0)
  def isNonZero  = isRational && (n != 0)

  // ...
}

And on top of that, we want to support addition on Rational in a way that the rationality and positiveness properties are correctly preserved:

 def +(that: Rational): Rational = {
   require(isRational && that.isRational)
   Rational(n * that.d + that.n * d, d * that.d)
}.ensuring { res =>
   res.isRational &&
   (this.isPositive == that.isPositive ==> res.isPositive == this.isPositive)
 }

In this simple case, things work nicely and we can write the multiplication in a similar fashion:

 def *(that: Rational): Rational = {
   require(isRational && that.isRational)
   Rational(n * that.n, d * that.d)
}.ensuring { res =>
   res.isRational &&
   (res.isNonZero  == (this.isNonZero && that.isNonZero)) &&
   (res.isPositive == (!res.isNonZero || this.isPositive == that.isPositive))
 }

Measures: a slightly more complex example

Now let’s look at a slightly more complex example: measures on discrete probability spaces. We represent such measures using a List-like recursive data structure: a generic abstract class Meas[A] that has two subclasses, Empty[A] and Cons[A]. The constructor of the class Empty[A] takes no arguments; it represents an “empty” measure that evaluates to 0 when applied to any set of values of type A. The constructor of Cons[A], on the other hand, takes three parameters: a value x, its associated weight w expressed as a Rational (since Stainless doesn’t quite yet support real numbers out of the box), and another measure m on A. The value Cons(x, w, m) represents the measure obtained by adding to m the “single-point” measure that evaluates to w at x and to 0 everywhere else. We also define an isMeasure property – similar to the isRational property presented above – which recursively checks that all the weights in a measure are positive rationals (note that all our measures have finite support).

/** Measures on discrete probability spaces. */
sealed abstract class Meas[A] {

  /** All weights must be positive. */
  def isMeasure: Boolean = this match {
    case Empty()       => true
    case Cons(x, w, m) => w.isPositive && m.isMeasure
  }

  // ...
}

/** The empty measure maps every subset of the space A to 0. */
case class Empty[A]() extends Meas[A]

/**
 * The 'Cons' measure adjoins an additional element 'x' of type 'A'
 * to an existing measure 'm' over 'A'.  Note that 'x' might already
 * be present in 'm'.
 */
case class Cons[A](x: A, w: Rational, m: Meas[A]) extends Meas[A]

The defining operation on a measure m is its evaluation m(xs) (or equivalently m.apply(xs)) on some set xs: Set[A], i.e. on a subset of the “space” A. The value of m should be a positive rational for any such set xs, provided m.isMeasure holds. This suggests _.isPositive as the postcondition for apply, but simply claiming that the result is positive is not enough for Stainless to verify this postcondition.

We can provide the necessary hint to Stainless by performing structural induction on this inside the postcondition as follows:

 /** Compute the value of this measure on a subset of the space 'A'. */
 def apply(xs: Set[A]): Rational = {
   require (isMeasure)
   this match {
     case Empty()       => Rational(0, 1)
     case Cons(x, w, m) => if (xs contains x) w + m(xs) else m(xs)
   }
}.ensuring { res =>
   res.isPositive because {
     this match {
       case Empty()       => trivial
       case Cons(x, w, m) => m(xs).isPositive
     }
   }
 }

Notice the similarity between the pattern match in the body of the apply function and that in the postcondition. With this hint, Stainless is able to verify the postcondition.

A complex example: additivity of measures

Using the principles and techniques discussed so far, one can prove quite advanced propositions using Stainless. Returning to the measure-theoretic example from the previous section, we would like to prove that our implementation of measures is properly additive. Formally, a measure \(\mu \colon A \to \mathbb{R}\) on a countable set \(A\) must fulfill the following additivity property [3]:

\[\forall A_1, A_2 \subseteq A . A_1 \cap A_2 = \emptyset \Rightarrow \mu(A_1 \cup A_2) = \mu(A_1) + \mu(A_2)\]

which we can express in Stainless as

def additivity[A](m: Meas[A], xs: Set[A], ys: Set[A]): Boolean = {
  require(m.isMeasure && (xs & ys).isEmpty)
  m(xs ++ ys) == m(xs) + m(ys)
}.holds

We can prove this property using structural induction on the parameter m, case analysis on the parameters xs and ys, equational reasoning, and properties of rational numbers (in the form of user-defined lemmas) as well as sets (using Stainless’s built-in support).

def additivity[A](m: Meas[A], xs: Set[A], ys: Set[A]): Boolean = {
  require(m.isMeasure && (xs & ys).isEmpty)
  m(xs ++ ys) == m(xs) + m(ys) because {
    m match {
      case Empty()       => trivial
      case Cons(x, w, n) => if (xs contains x) {
        w + n(xs ++ ys)     ==| additivity(n, xs, ys)        |
        w + (n(xs) + n(ys)) ==| plusAssoc(w, n(xs), n(ys))   |
        (w + n(xs)) + n(ys) ==| !(ys contains x)             |
        m(xs)       + m(ys)
      }.qed else if (ys contains x) {
        w + n(xs ++ ys)     ==| additivity(n, xs, ys)        |
        w + (n(xs) + n(ys)) ==| plusComm(w, (n(xs) + n(ys))) |
        (n(xs) + n(ys)) + w ==| plusAssoc(n(xs), n(ys), w)   |
        n(xs) + (n(ys) + w) ==| plusComm(n(ys), w)           |
        n(xs) + (w + n(ys)) ==| !(xs contains x)             |
        m(xs) + m(ys)
      }.qed else {
        n(xs ++ ys)         ==| additivity(n, xs, ys)        |
        n(xs) + n(ys)
      }.qed
    }
  }
}.holds

The full proof (including the proofs of all helper lemmas) as well as its generalization to sub-additivity can be found in the testcases/verification/proof/measure/ directory of the Stainless distribution [1].

Quick Recap

Let’s summarize what we’ve learned here. To write proofs efficiently, it’s good to keep the following in mind:

1. Always use a proper timeout and ask Stainless for more information about what he tries to verify, e.g. --timeout=5 --debug=verification.

2. Use @induct when working on structurally inductive proofs to get a more precise feedback from Stainless: this will decompose the proof into a base case and an inductive case for the first argument of the function under consideration.

   If Stainless isn’t able to verify the proof using @induct, try performing manual case analysis.

3. Modularize your proofs and verify sub-goals!

   • use plenty of helper lemmas;

   • use check abundantly;

   • if possible use the relational reasoning DSL presented above.

   This is especially handy when you can connect the two sides of a relational claim with sub-statements.

Footnotes

[1] (1,2)

The source code of this example and all others in this chapter is included in the Stainless distribution. Examples about lists can be found in library/collection/List.scala, other examples are located in the testcases/verification/proof/ directory.

[2]

Propositions and proofs live in the same universe in Stainless. This is contrary to e.g. type-theoretic proof assistants where propositions are represented by types and proofs are terms inhabiting such types.

[3]

To be precise, we are assuming here the underlying measurable space \((A, \mathcal{P}(A))\), where \(A\) is countable and \(\mathcal{P}(A)\) denotes its discrete σ-algebra (i.e. the power set of \(A\)).