Existential Quantification

Another basic logical connective is existential quantification. To say that there is some x of type T such that some property P holds of x, we write ∃ x : T, P. As with ∀, the type annotation : T can be omitted if Coq is able to infer from the context what the type of x should be. To prove a statement of the form ∃ x, P, we must show that P holds for some specific choice of value for x, known as the witness of the existential. This is done in two steps: First, we explicitly tell Coq which witness t we have in mind by invoking the tactic ∃ t. Then we prove that P holds after all occurrences of x are replaced by t.

Conversely, if we have an existential hypothesis ∃ x, P in the context, we can destruct it to obtain a witness x and a hypothesis stating that P holds of x.

Exercise: 1 star, standard, especially useful (dist_not_exists)

Prove that "P holds for all x" implies "there is no x for which P does not hold." (Hint: destruct H as [x E] works on existential assumptions!)

Exercise: 2 stars, standard (dist_exists_or)

Prove that existential quantification distributes over disjunction.

Exercise: 3 stars, standard, optional (leb_plus_exists)

Programming with Propositions

The logical connectives that we have seen provide a rich vocabulary for defining complex propositions from simpler ones. To illustrate, let's look at how to express the claim that an element x occurs in a list l. Notice that this property has a simple recursive structure:

• If l is the empty list, then x cannot occur in it, so the property "x appears in l" is simply false.

• Otherwise, l has the form x' :: l'. In this case, x occurs in l if it is equal to x' or if it occurs in l'.

We can translate this directly into a straightforward recursive function taking an element and a list and returning a proposition (!):

When In is applied to a concrete list, it expands into a concrete sequence of nested disjunctions.

(Notice the use of the empty pattern to discharge the last case en passant.) We can also prove more generic, higher-level lemmas about In. (Note how In starts out applied to a variable and only gets expanded when we do case analysis on this variable.)

This way of defining propositions recursively, though convenient in some cases, also has some drawbacks. In particular, it is subject to Coq's usual restrictions regarding the definition of recursive functions, e.g., the requirement that they be "obviously terminating." In the next chapter, we will see how to define propositions inductively -- a different technique with its own set of strengths and limitations.

Exercise: 2 stars, standard (In_map_iff)

Exercise: 2 stars, standard (In_app_iff)

Exercise: 3 stars, standard, especially useful (All)

Recall that functions returning propositions can be seen as properties of their arguments. For instance, if P has type nat → Prop, then P n states that property P holds of n. Drawing inspiration from In, write a recursive function All stating that some property P holds of all elements of a list l. To make sure your definition is correct, prove the All_In lemma below. (Of course, your definition should not just restate the left-hand side of All_In.)

Exercise: 2 stars, standard, optional (combine_odd_even)

Complete the definition of the combine_odd_even function below. It takes as arguments two properties of numbers, Podd and Peven, and it should return a property P such that P n is equivalent to Podd n when n is odd and equivalent to Peven n otherwise.

To test your definition, prove the following facts:

Applying Theorems to Arguments

One feature that distinguishes Coq from some other popular proof assistants (e.g., ACL2 and Isabelle) is that it treats proofs as first-class objects. There is a great deal to be said about this, but it is not necessary to understand it all, in order to use Coq. This section gives just a taste, while a deeper exploration can be found in the optional chapters ProofTerms and IndPrinciples. We have seen that we can use Check to ask Coq to print the type of an expression. We can also use it to ask what theorem a particular identifier refers to.

Coq checks the type of the statements of the add_comm and plus_id_example theorems in the same way that it checks the type of any term (e.g., plus) that we ask it to Check. And if we leave off the colon and type, Coq will print these types for us. Why? The reason is that the identifier add_comm actually refers to a proof term, which represents a logical derivation establishing the truth of the statement ∀ n m : nat, n + m = m + n. The type of this term is the proposition that it is a proof of. The type of an ordinary function tells us what we can do with it.

• e.g., if we have a term of type nat → nat → nat, we can give it two nats as arguments and get a nat back.

Similarly, the statement of a theorem tells us what we can use that theorem for.

• if we have a term of type ∀ n m, n = m → n + n = m + m and we provide it two numbers n and m plus an "argument" of type n = m, we can derive n + n = m + m.

Operationally, this analogy goes even further: by applying a theorem as if it were a function, i.e., applying it to values and hypotheses with matching types, we can specialize its result without having to resort to intermediate assertions. For example, suppose we wanted to prove the following result:

It appears at first sight that we ought to be able to prove this by rewriting with add_comm twice to make the two sides match. The problem, however, is that the second rewrite will undo the effect of the first.

We saw similar problems back in Chapter Induction, and we saw one way to work around them by using assert to derive a specialized version of add_comm that can be used to rewrite exactly where we want.

A more elegant alternative is to apply add_comm directly to the arguments we want to instantiate it with, in much the same way as we apply a polymorphic function to a type argument.

We can in fact do it for both uses of rewrite

Let's see another example of using a theorem like a function. The following theorem says: any list l containing some element must be nonempty.

What makes this interesting is that one quantified variable (x) does not appear in the conclusion (l ≠ []). We should be able to use this theorem to prove the special case where x is 42. However, naively, the tactic apply in_not_nil will fail because it cannot infer the value of x.

There are several ways to work around this: Use apply ... with ...

Use apply ... in ...

Explicitly apply the lemma to the value for x.

Explicitly apply the lemma to a hypothesis.

You can "use a theorem as a function" in this way with almost any tactic that can take a theorem's name as an argument. Note, also, that theorem application uses the same inference mechanisms as function application; thus, it is possible, for example, to supply wildcards as arguments to be inferred, or to declare some arguments to a theorem as implicit by default. These features are illustrated in the proofs below.

We will see many more examples in later chapters.

Coq vs. Set Theory

Coq's logical core, the Calculus of Inductive Constructions, differs in some important ways from other formal systems that are used by mathematicians to write down precise and rigorous definitions and proofs -- in particular, from Zermelo-Fraenkel Set Theory (ZFC), the most popular foundation for paper-and-pencil mathematics. We conclude this chapter with a brief discussion of some of the most significant differences between the two worlds.

Functional Extensionality

Coq's logic is intentionally quite minimal. This means that there are occasionally some cases where translating standard mathematical reasoning into Coq can be cumbersome or even impossible, unless we enrich the core logic with additional axioms. The equality assertions that we have seen so far mostly have concerned elements of inductive types (nat, bool, etc.). But, since Coq's equality operator is polymorphic, we can use it at any type -- in particular, we can write propositions claiming that two functions are equal to each other: In certain cases Coq can successfully prove equality propositions stating that two functions are equal to each other:

This works when Coq can simplify the functions to the same expression, but this doesn't always work. In common mathematical practice, two functions f and g are considered equal if they produce the same output on every input:
    (∀ x, f x = g x) → f = g This is known as the principle of functional extensionality. Informally, an "extensional property" is one that pertains to an object's observable behavior. Thus, functional extensionality simply means that a function's identity is completely determined by what we can observe from it -- i.e., the results we obtain after applying it. However, functional extensionality is not part of Coq's built-in logic. This means that some apparently "obvious" propositions are not provable.

However, if we like, we can add functional extensionality to Coq using the Axiom command.

Defining something as an Axiom has the same effect as stating a theorem and skipping its proof using Admitted, but it alerts the reader that this isn't just something we're going to come back and fill in later! We can now invoke functional extensionality in proofs:

Naturally, we must be careful when adding new axioms into Coq's logic, as this can render it inconsistent -- that is, it may become possible to prove every proposition, including False, 2+2=5, etc.! Unfortunately, there is no simple way of telling whether an axiom is safe to add: hard work by highly trained mathematicians is often required to establish the consistency of any particular combination of axioms. Fortunately, it is known that adding functional extensionality, in particular, is consistent. To check whether a particular proof relies on any additional axioms, use the Print Assumptions command: Print Assumptions function_equality_ex₂.

(If you try this yourself, you may also see add_comm listed as an assumption, depending on whether the copy of Tactics.v in the local directory has the proof of add_comm filled in.)

Exercise: 4 stars, standard (tr_rev_correct)

One problem with the definition of the list-reversing function rev that we have is that it performs a call to app on each step; running app takes time asymptotically linear in the size of the list, which means that rev is asymptotically quadratic. We can improve this with the following definitions:

This version of rev is said to be tail-recursive, because the recursive call to the function is the last operation that needs to be performed (i.e., we don't have to execute ++ after the recursive call); a decent compiler will generate very efficient code in this case. Prove that the two definitions are indeed equivalent.

Propositions vs. Booleans

We've seen two different ways of expressing logical claims in Coq: with booleans (of type bool), and with propositions (of type Prop). Here are the key differences between bool and Prop:

                                           bool     Prop
                                           ====     ====
           decidable?                      yes       no
           useable with match?             yes       no
           equalities rewritable?          no        yes

The most essential difference between the two worlds is decidability. Every Coq expression of type bool can be simplified in a finite number of steps to either true or false -- i.e., there is a terminating mechanical procedure for deciding whether or not it is true. This means that, for example, the type nat → bool is inhabited only by functions that, given a nat, always return either true or false; and this, in turn, means that there is no function in nat → bool that checks whether a given number is the code of a terminating Turing machine. By contrast, the type Prop includes both decidable and undecidable mathematical propositions; in particular, the type nat → Prop does contain functions representing properties like "the nth Turing machine halts." The second row in the table above follow directly from this essential difference. To evaluate a pattern match (or conditional) on a boolean, we need to know whether the scrutinee evaluates to true or false; this only works for bool, not Prop. The third row highlights another important practical difference: equality functions like eqb_nat that return a boolean cannot be used directly to justify rewriting, whereas the propositional eq can be.

Working with Decidable Properties

Since Prop includes both decidable and undecidable properties, we have two options when when we want to formalize a property that happens to be decidable: we can express it as a boolean computation or as a function into Prop. For instance, to claim that a number n is even, we can say either... ... that even n evaluates to true...

... or that there exists some k such that n = double k.

Of course, it would be pretty strange if these two characterizations of evenness did not describe the same set of natural numbers! Fortunately, we can prove that they do... We first need two helper lemmas.

Exercise: 3 stars, standard (even_double_conv)

Now the main theorem:

In view of this theorem, we say that the boolean computation even n is reflected in the truth of the proposition ∃ k, n = double k. Similarly, to state that two numbers n and m are equal, we can say either

• (1) that n =? m returns true, or
• (2) that n = m.

Again, these two notions are equivalent.

Even when the boolean and propositional formulations of a claim are equivalent from a purely logical perspective, they are often not equivalent from the point of view of suitability for some specific purpose. In the case of even numbers above, when proving the backwards direction of even_bool_prop (i.e., even_double, going from the propositional to the boolean claim), we used a simple induction on k. On the other hand, the converse (the even_double_conv exercise) required a clever generalization, since we can't directly prove (even n = true) → Even n. Similarly, we cannot test whether or not a Prop is true in a function definition; as a consequence, the following code fragment is rejected:

Coq complains that n = 2 has type Prop, while it expects an element of bool (or some other inductive type with two constructors). The reason has to do with the computational nature of Coq's core language, which is designed so that every function it can express is computable and total. One reason for this is to allow the extraction of executable programs from Coq developments. As a consequence, Prop in Coq does not have a universal case analysis operation telling whether any given proposition is true or false, since such an operation would allow us to write non-computable functions. Beyond the fact that non-computable properties are impossible in general to phrase as boolean computations, even many computable properties are easier to express using Prop than bool, since recursive function definitions in Coq are subject to significant restrictions. For instance, the next chapter shows how to define the property that a regular expression matches a given string using Prop. Doing the same with bool would amount to writing a regular expression matching algorithm, which would be more complicated, harder to understand, and harder to reason about than a simple (non-algorithmic) definition of this property. Conversely, an important side benefit of stating facts using booleans is enabling some proof automation through computation with Coq terms, a technique known as proof by reflection. Consider the following statement:

The most direct way to prove this is to give the value of k explicitly.

The proof of the corresponding boolean statement is even simpler, because we don't have to invent the witness: Coq's computation mechanism does it for us!

What is interesting is that, since the two notions are equivalent, we can use the boolean formulation to prove the other one without mentioning the value 500 explicitly:

Although we haven't gained much in terms of proof-script size in this case, larger proofs can often be made considerably simpler by the use of reflection. As an extreme example, a famous Coq proof of the even more famous 4-color theorem uses reflection to reduce the analysis of hundreds of different cases to a boolean computation. Another notable difference is that the negation of a "boolean fact" is straightforward to state and prove: simply flip the expected boolean result.

In contrast, propositional negation can be more difficult to work with directly.

Equality provides a complementary example, where it is sometimes easier to work in the propositional world. Knowing that (n =? m) = true is generally of little direct help in the middle of a proof involving n and m; however, if we convert the statement to the equivalent form n = m, we can rewrite with it.

We won't discuss reflection any further for the moment, but it serves as a good example showing the different strengths of booleans and general propositions. Being able to cross back and forth between the boolean and propositional worlds will often be convenient in later chapters.

Exercise: 2 stars, standard (logical_connectives)

The following theorems relate the propositional connectives studied in this chapter to the corresponding boolean operations.

Exercise: 1 star, standard (eqb_neq)

The following theorem is an alternate "negative" formulation of eqb_eq that is more convenient in certain situations. (We'll see examples in later chapters.) Hint: not_true_iff_false.

Exercise: 3 stars, standard (eqb_list)

Given a boolean operator eqb for testing equality of elements of some type A, we can define a function eqb_list for testing equality of lists with elements in A. Complete the definition of the eqb_list function below. To make sure that your definition is correct, prove the lemma eqb_list_true_iff.

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

Prove the theorem below, which relates forallb, from the exercise forall_exists_challenge in chapter Tactics, to the All property defined above. Copy the definition of forallb from your Tactics here so that this file can be graded on its own.

(Ungraded thought question) Are there any important properties of the function forallb which are not captured by this specification?

Classical vs. Constructive Logic

We have seen that it is not possible to test whether or not a proposition P holds while defining a Coq function. You may be surprised to learn that a similar restriction applies in proofs! In other words, the following intuitive reasoning principle is not derivable in Coq:

To understand operationally why this is the case, recall that, to prove a statement of the form P ∨ Q, we use the left and right tactics, which effectively require knowing which side of the disjunction holds. But the universally quantified P in excluded_middle is an arbitrary proposition, which we know nothing about. We don't have enough information to choose which of left or right to apply, just as Coq doesn't have enough information to mechanically decide whether P holds or not inside a function. However, if we happen to know that P is reflected in some boolean term b, then knowing whether it holds or not is trivial: we just have to check the value of b.

In particular, the excluded middle is valid for equations n = m, between natural numbers n and m.

It may seem strange that the general excluded middle is not available by default in Coq, since it is a standard feature of familiar logics like ZFC. But there is a distinct advantage in not assuming the excluded middle: statements in Coq make stronger claims than the analogous statements in standard mathematics. Notably, when there is a Coq proof of ∃ x, P x, it is always possible to explicitly exhibit a value of x for which we can prove P x -- in other words, every proof of existence is constructive. Logics like Coq's, which do not assume the excluded middle, are referred to as constructive logics. More conventional logical systems such as ZFC, in which the excluded middle does hold for arbitrary propositions, are referred to as classical. The following example illustrates why assuming the excluded middle may lead to non-constructive proofs: Claim: There exist irrational numbers a and b such that a ^ b (a to the power b) is rational. Proof: It is not difficult to show that sqrt 2 is irrational. If sqrt 2 ^ sqrt 2 is rational, it suffices to take a = b = sqrt 2 and we are done. Otherwise, sqrt 2 ^ sqrt 2 is irrational. In this case, we can take a = sqrt 2 ^ sqrt 2 and b = sqrt 2, since a ^ b = sqrt 2 ^ (sqrt 2 × sqrt 2) = sqrt 2 ^ 2 = 2. ☐ Do you see what happened here? We used the excluded middle to consider separately the cases where sqrt 2 ^ sqrt 2 is rational and where it is not, without knowing which one actually holds! Because of that, we finish the proof knowing that such a and b exist but we cannot determine what their actual values are (at least, not from this line of argument). As useful as constructive logic is, it does have its limitations: There are many statements that can easily be proven in classical logic but that have only much more complicated constructive proofs, and there are some that are known to have no constructive proof at all! Fortunately, like functional extensionality, the excluded middle is known to be compatible with Coq's logic, allowing us to add it safely as an axiom. However, we will not need to do so here: the results that we cover can be developed entirely within constructive logic at negligible extra cost. It takes some practice to understand which proof techniques must be avoided in constructive reasoning, but arguments by contradiction, in particular, are infamous for leading to non-constructive proofs. Here's a typical example: suppose that we want to show that there exists x with some property P, i.e., such that P x. We start by assuming that our conclusion is false; that is, ¬ ∃ x, P x. From this premise, it is not hard to derive ∀ x, ¬ P x. If we manage to show that this intermediate fact results in a contradiction, we arrive at an existence proof without ever exhibiting a value of x for which P x holds! The technical flaw here, from a constructive standpoint, is that we claimed to prove ∃ x, P x using a proof of ¬ ¬ (∃ x, P x). Allowing ourselves to remove double negations from arbitrary statements is equivalent to assuming the excluded middle, as shown in one of the exercises below. Thus, this line of reasoning cannot be encoded in Coq without assuming additional axioms.

Exercise: 3 stars, standard (excluded_middle_irrefutable)

Proving the consistency of Coq with the general excluded middle axiom requires complicated reasoning that cannot be carried out within Coq itself. However, the following theorem implies that it is always safe to assume a decidability axiom (i.e., an instance of excluded middle) for any particular Prop P. Why? Because the negation of such an axiom leads to a contradiction. If ¬ (P ∨ ¬P) were provable, then by de_morgan_not_or as proved above, P ∧ ¬P would be provable, which would be a contradiction. So, it is safe to add P ∨ ¬P as an axiom for any particular P. Succinctly: for any proposition P, Coq is consistent ==> (Coq + P ∨ ¬P) is consistent.

Exercise: 3 stars, advanced (not_exists_dist)

It is a theorem of classical logic that the following two assertions are equivalent:
    ¬(∃ x, ¬P x)
    ∀ x, P x The dist_not_exists theorem above proves one side of this equivalence. Interestingly, the other direction cannot be proved in constructive logic. Your job is to show that it is implied by the excluded middle.

Exercise: 5 stars, standard, optional (classical_axioms)

For those who like a challenge, here is an exercise taken from the Coq'Art book by Bertot and Casteran (p. 123). Each of the following four statements, together with excluded_middle, can be considered as characterizing classical logic. We can't prove any of them in Coq, but we can consistently add any one of them as an axiom if we wish to work in classical logic. Prove that all five propositions (these four plus excluded_middle) are equivalent. Hint: Rather than considering all pairs of statements pairwise, prove a single circular chain of implications that connects them all.