We have now seen many examples of factual claims (i.e., propositions) and ways of presenting evidence of their truth (proofs). In particular, we have worked extensively with equality propositions (e₁ = e₂), implications (P → Q), and quantified propositions (∀ x, P). In this chapter, we will see how Rocq can be used to carry out other familiar forms of logical reasoning. Before diving into details, we should talk a bit about the status of mathematical statements in Rocq. Rocq is a typed language, which means that every sensible expression has an associated type. Logical claims are no exception: any statement we might try to prove in Rocq has a type, namely Prop, the type of propositions. We can see this with the Check command:

Note that all syntactically well-formed propositions have type Prop in Rocq, regardless of whether they are true or not. Simply being a proposition is one thing; being provable is a different thing!

Indeed, propositions don't just have types -- they are first-class entities that can be manipulated in all the same ways as any of the other things in Rocq's world. So far, we've seen one primary place where propositions can appear: in Theorem (and Lemma and Example) declarations.

But propositions can be used in other ways. For example, we can give a name to a proposition using a Definition, just as we give names to other kinds of expressions.

We can later use this name in any situation where a proposition is expected -- for example, as the claim in a Theorem declaration.

We can also write parameterized propositions -- that is, functions that take arguments of some type and return a proposition. For instance, the following function takes a number and returns a proposition asserting that this number is equal to three:

In Rocq, functions that return propositions are said to define properties of their arguments. For instance, here's a (polymorphic) property defining the familiar notion of an injective function.

The familiar equality operator = is a (binary) function that returns a Prop. The expression n = m is syntactic sugar for eq n m (defined in Rocq's standard library using the Notation mechanism). Because eq can be used with elements of any type, it is also polymorphic:

(Notice that we wrote @eq instead of eq: The type argument A to eq is declared as implicit, and we need to turn off the inference of this implicit argument to see the full type of eq.)

Logical Connectives

Conjunction

The conjunction, or logical and, of propositions A and B is written A ∧ B; it represents the claim that both A and B are true.

To prove a conjunction, start with the split tactic. This will generate two subgoals, one for each part of the statement:

For any propositions A and B, if we assume that A and B are each true individually, we can conclude that A ∧ B is also true. The Rocq library provides a function conj that does this.

Since applying a theorem with hypotheses to some goal has the effect of generating as many subgoals as there are hypotheses for that theorem, we can apply conj to achieve the same effect as split.

Exercise: 2 stars, standard (plus_is_O)

So much for proving conjunctive statements. To go in the other direction -- i.e., to use a conjunctive hypothesis to help prove something else -- we can use our good old destruct tactic. When the current proof context contains a hypothesis H of the form A ∧ B, writing destruct H as [HA HB] will remove H from the context and replace it with two new hypotheses: HA, stating that A is true, and HB, stating that B is true.

As usual, we can also destruct H right at the point where we introduce it, instead of introducing and then destructing it:

You may wonder why we bothered packing the two hypotheses n = 0 and m = 0 into a single conjunction, since we could also have stated the theorem with two separate premises:

For this specific theorem, both formulations are fine. But it's important to understand how to work with conjunctive hypotheses because conjunctions often arise from intermediate steps in proofs, especially in larger developments. Here's a simple example:

Another common situation is that we know A ∧ B but in some context we need just A or just B. In such cases we can do a destruct (possibly implicitly, as part of an intros) and use an underscore pattern _ to indicate that the unneeded conjunct should just be thrown away.

Exercise: 1 star, standard, optional (proj2)

Finally, we sometimes need to rearrange the order of conjunctions and/or the grouping of multi-way conjunctions. We can see this at work in the proofs of the following commutativity and associativity theorems

Exercise: 1 star, standard (and_assoc)

In the following proof of associativity, notice how the nested intros pattern breaks the hypothesis H : P ∧ (Q ∧ R) down into HP : P, HQ : Q, and HR : R. Finish the proof.

The infix notation ∧ is actually just syntactic sugar for and A B. That is, and is a Rocq operator that takes two propositions as arguments and yields a proposition.

Disjunction

Another important connective is the disjunction, or logical or, of two propositions: A ∨ B is true when either A or B is. This infix notation stands for or A B, where or : Prop → Prop → Prop. To use a disjunctive hypothesis in a proof, we proceed by case analysis -- which, as with other data types like nat, can be done explicitly with destruct or implicitly with an intros pattern:

We can see in this example that, when we perform case analysis on a disjunction A ∨ B, we must separately discharge two proof obligations, each showing that the conclusion holds under a different assumption -- A in the first subgoal and B in the second. The case analysis pattern [Hn | Hm] allows us to name the hypotheses that are generated for the subgoals. Conversely, to show that a disjunction holds, it suffices to show that one of its sides holds. This can be done via the tactics left and right. As their names imply, the first one requires proving the left side of the disjunction, while the second requires proving the right side. Here is a trivial use...

... and here is a slightly more interesting example requiring both left and right:

Exercise: 2 stars, standard (mult_is_O)

Exercise: 1 star, standard (or_commut)

Falsehood and Negation

Up to this point, we have mostly been concerned with proving "positive" statements -- addition is commutative, appending lists is associative, etc. We are sometimes also interested in negative results, demonstrating that some proposition is not true. Such statements are expressed with the logical negation operator ¬. To see how negation works, recall the principle of explosion from the Tactics chapter, which asserts that, if we assume a contradiction, then any other proposition can be derived. Following this intuition, we could define ¬ P ("not P") as ∀ Q, P → Q. Rocq actually makes an equivalent but slightly different choice, defining ¬ P as P → False, where False is a specific un-provable proposition defined in the standard library.

Since False is a contradictory proposition, the principle of explosion also applies to it. If we can get False into the context, we can use destruct on it to complete any goal:

The Latin ex falso quodlibet means, literally, "from falsehood follows whatever you like"; this is another common name for the principle of explosion.

Exercise: 2 stars, standard, optional (not_implies_our_not)

Show that Rocq's definition of negation implies the intuitive one mentioned above. Hint: While getting accustomed to Rocq's definition of not, you might find it helpful to unfold not near the beginning of proofs.

Inequality is a very common form of negated statement, so there is a special notation for it:

For example:

The proposition 0 ≠ 1 is exactly the same as ~(0 = 1) -- that is, not (0 = 1) -- which unfolds to (0 = 1) → False. (We use unfold not explicitly here, to illustrate that point, but generally it can be omitted.)

To prove an inequality, we may assume the opposite equality...

... and deduce a contradiction from it. Here, the equality O = S O contradicts the disjointness of constructors O and S, so discriminate takes care of it.

It takes a little practice to get used to working with negation in Rocq. Even though you may see perfectly well why a claim involving negation holds, it can be a little tricky at first to see how to make Rocq understand it! Here are proofs of a few familiar facts to help get you warmed up.

Exercise: 2 stars, advanced, optional (double_neg_informal)

Write an informal proof of double_neg: Theorem: P implies ~~P, for any proposition P.

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

Exercise: 1 star, standard (not_both_true_and_false)

Exercise: 1 star, advanced (not_PNP_informal)

Write an informal proof (in English) of the proposition ∀ P : Prop, ~(P ∧ ¬P).

Exercise: 2 stars, standard (de_morgan_not_or)

De Morgan's Laws, named for Augustus De Morgan, describe how negation interacts with conjunction and disjunction. The following law says that "the negation of a disjunction is the conjunction of the negations." There is a dual law de_morgan_not_and_not to which we will return at the end of this chapter.

Exercise: 1 star, standard, optional (not_S_inverse_pred)

Since we are working with natural numbers, we can disprove that S and pred are inverses of each other:

Since inequality involves a negation, it also requires a little practice to be able to work with it fluently. Here is one useful trick. If you are trying to prove a goal that is nonsensical (e.g., the goal state is false = true), apply ex_falso_quodlibet to change the goal to False. This makes it easier to use assumptions of the form ¬P that may be available in the context -- in particular, assumptions of the form x≠y.

Since reasoning with ex_falso_quodlibet is quite common, Rocq provides a built-in tactic, exfalso, for applying it.

Truth

Besides False, Rocq's standard library also defines True, a proposition that is trivially true. To prove it, we use the constant I : True, which is also defined in the standard library:

Unlike False, which is used extensively, True is used relatively rarely: it is trivial (and therefore uninteresting) to prove as a goal, and it provides no useful information when it appears as a hypothesis. However, True can be quite useful when defining complex Props using conditionals or as a parameter to higher-order Props. We'll come back to this later. For now, let's take a look at how we can use True and False to achieve an effect similar to that of the discriminate tactic, without literally using discriminate. Pattern-matching lets us do different things for different constructors. If the result of applying two different constructors were hypothetically equal, then we could use match to convert an unprovable statement (like False) to one that is provable (like True).

To generalize this to other constructors, we simply have to provide an appropriate variant of disc_fn. To generalize it to other conclusions, we can use exfalso to replace them with False. The built-in discriminate tactic takes care of all this for us.

Exercise: 2 stars, advanced, optional (nil_is_not_cons)

Use the same technique as above to show that nil ≠ x :: xs. Do not use the discriminate tactic.

Logical Equivalence

The handy "if and only if" connective, which asserts that two propositions have the same truth value, is simply the conjunction of two implications.
      Print "<->".

We can also use apply with an ↔ in either direction, without explicitly thinking about the fact that it is really an and underneath.

Exercise: 1 star, standard, optional (iff_properties)

Using the above proof that ↔ is symmetric (iff_sym) as a guide, prove that it is also reflexive and transitive.

Exercise: 3 stars, standard (or_distributes_over_and)

Setoids and Logical Equivalence

Some of Rocq's tactics treat iff statements specially, avoiding some low-level proof-state manipulation. In particular, rewrite and reflexivity can be used with iff statements, not just equalities. To enable this behavior, we have to import the Rocq library that supports it:

A "setoid" is a set equipped with an equivalence relation -- that is, a relation that is reflexive, symmetric, and transitive. When two elements of a set are equivalent according to the relation, rewrite can be used to replace one by the other. We've seen this already with the equality relation = in Rocq: when x = y, we can use rewrite to replace x with y or vice-versa. Similarly, the logical equivalence relation ↔ is reflexive, symmetric, and transitive, so we can use it to replace one part of a proposition with another: if P ↔ Q, then we can use rewrite to replace P with Q, or vice-versa. Here is a simple example demonstrating how these tactics work with iff. First, let's prove a couple of basic iff equivalences. (For these proofs we are not using setoids yet.)

We can now use these facts with rewrite and reflexivity to prove a ternary version of the mult_eq_0 fact above without splitting the top-level iff:

Existential Quantification

Another fundamental 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 Rocq 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 for x, known as the witness of the existential. This is done in two steps: First, we explicitly tell Rocq 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 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 reason about more generic statements involving In.

(Note here 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 is very convenient in some cases, less so in others. In particular, it is subject to Rocq's usual restrictions regarding definitions 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 strengths and limitations.

Exercise: 2 stars, standard (In_map_iff)

Exercise: 2 stars, standard (In_app_iff)

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

We noted above that functions returning propositions can be seen as properties of their arguments. For instance, if P has type nat → Prop, then P n says 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 combine_odd_even 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 Rocq 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 Rocq. This section gives just a taste, leaving a deeper exploration for the optional chapters ProofObjects and IndPrinciples. We have seen that we can use Check to ask Rocq to check whether an expression has a given type:

We can also use it to check what theorem a particular identifier refers to:

Rocq checks 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). If we leave off the colon and type, Rocq will print these types for us. Why? The reason is that the identifier add_comm actually refers to a proof object -- a logical derivation establishing the truth of the statement ∀ n m : nat, n + m = m + n. The type of this object is the proposition that it is a proof of. The type of an ordinary function tells us what we can do with it.

• 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 and a third "argument" of type n = m, we get back a proof object of type 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 is that the second rewrite will undo the effect of the first.

We encountered similar issues back in 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.

If we really wanted, we could in fact do it for both rewrites.

Here's another example of using a theorem about lists like a function. Suppose we have proved the following simple fact about lists...

(I.e., if a list l contains some element x, then l must be nonempty.) Note that one quantified variable (x) does not appear in the conclusion (l ≠ []). Intuitively, we should be able to use this theorem to prove the special case where x is 42. However, simply invoking the tactic apply in_not_nil will fail because it cannot infer the value of x.

There are several ways to work around this: We can use apply ... with ...:

Or we can use apply ... in ...:

Or -- this is the new one -- we can explicitly apply the lemma to the value 42 for x:

We can also explicitly apply the lemma to a hypothesis, causing the values of the other parameters to be inferred:

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 hypotheses to a theorem as implicit by default. These features are illustrated in the proof below. (The details of how this proof works are not critical -- the goal here is just to illustrate applying theorems to arguments.)

We will see many more examples in later chapters.

Working with Decidable Properties

We've seen two different ways of expressing logical claims in Rocq: 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
           works with rewrite tactic?      no        yes

The crucial difference between the two worlds is decidability. Every (closed) Rocq 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 yield either true or false in finite time; and this, in turn, means (by a standard computability argument) 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 follows 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 an important practical difference: equality functions like eqb_nat that return a boolean cannot be used directly to justify rewriting with the rewrite tactic; propositional equality is required for this. Since Prop includes both decidable and undecidable properties, we have two options when we want to formalize a property that happens to be decidable: we can express it either as a boolean computation or as a function into Prop.

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

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

Exercise: 3 stars, standard (even_double_conv)

Now the main theorem:

In view of this theorem, we can 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:

So what should we do in situations where some claim could be formalized as either a proposition or a boolean computation? Which should we choose? In general, both can be useful. For example, booleans are more useful for defining functions. There is no effective way to test whether or not a Prop is true, so we cannot use Props in conditional expressions. The following definition is rejected:

Rocq complains that n = 2 has type Prop, while it expects an element of bool (or some other inductive type with two constructors). This has to do with the computational nature of Rocq'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 Rocq developments.) As a consequence, Prop in Rocq 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. Rather, we have to state this definition using a boolean equality test.

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 Rocq 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 Rocq 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 simpler, because we don't have to invent the witness 500: Rocq's computation mechanism does it for us!

Now, the useful observation 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 line count in this case, larger proofs can often be made considerably simpler by the use of reflection. As an extreme example, a famous Rocq 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 advantage of booleans is that the negation of a claim about booleans is straightforward to state and (when true) prove: simply flip the expected boolean result.

In contrast, propositional negation can be difficult to work with directly. For example, suppose we state the non-evenness of 1001 propositionally:

Proving this directly -- by assuming that there is some n such that 1001 = double n and then somehow reasoning to a contradiction -- would be rather complicated. But if we convert it to a claim about the boolean even function, we can let Rocq do the work for us.

Conversely, there are situations where it can be easier to work with propositions rather than booleans. In particular, knowing that (n =? m) = true is generally of little direct help in the middle of a proof involving n and m. But if we convert the statement to the equivalent form n = m, then we can easily 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?

The Logic of Rocq

Rocq'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 these two worlds.

Functional Extensionality

Rocq's logic is quite minimalistic. This means that one occasionally encounters cases where translating standard mathematical reasoning into Rocq is cumbersome -- or even impossible -- unless we enrich its core logic with additional axioms. For example, the equality assertions that we have seen so far mostly have concerned elements of inductive types (nat, bool, etc.). But, since Rocq'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 Rocq can successfully prove equality propositions stating that two functions are equal to each other:

This works when Rocq can simplify the functions to the same expression, but this doesn't always happen. These two functions are equal just by simplification, but in general functions can be equal for more interesting reasons. 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 Rocq's built-in logic. This means that some intuitively obvious propositions are not provable.

However, if we like, we can add functional extensionality to Rocq 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 need to be quite careful when adding new axioms into Rocq's logic, as this can render it inconsistent -- that is, it may become possible to prove every proposition, including False, 2+2=5, etc.! In general, 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 two-argument definition:

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.

Classical vs. Constructive Logic

We have seen that it is not possible to test whether or not a proposition P holds while defining a Rocq 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 Rocq:

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. However, in the special case where we happen to know that P is reflected in some boolean term b, 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.

Sadly, this trick only works for decidable propositions. It may seem strange that the general excluded middle is not available by default in Rocq, 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 Rocq make stronger claims than the analogous statements in standard mathematics. Notably, a Rocq proof of ∃ x, P x always includes a particular value of x for which we can prove P x -- in other words, every proof of existence is constructive. Logics like Rocq's, which do not assume the excluded middle, are referred to as constructive logics. 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. So 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 this, we finish the proof knowing that such a and b exist, but not being sure of their actual values. 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 Rocq'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 in Software Foundations 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 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 law, as shown in one of the exercises below. Thus, this line of reasoning cannot be encoded in Rocq without assuming additional axioms.

Exercise: 3 stars, standard (excluded_middle_irrefutable)

Proving the consistency of Rocq with the general excluded middle axiom requires complicated reasoning that cannot be carried out within Rocq 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, Rocq is consistent ==> Rocq + (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 adapted from the Coq'Art book by Bertot and Casteran (p. 123). Each of the following five statements, together with excluded_middle, can be considered as characterizing classical logic. We can't prove any of them in Rocq, but we can consistently add any one of them as an axiom if we wish to work in classical logic. To see this, prove that all six propositions (these five 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.