(** * Basics: Functional Programming in Coq *) (* IMPORTANT: | | | | | | | | | | | | | | | V V V V V V V V V V V V V V V ===> PLEASE DO NOT DISTRIBUTE SOLUTIONS PUBLICLY <=== ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ | | | | | | | | | | | | | | | (See the [Preface] for why.) *) (* [Admitted] is Coq's "escape hatch" that says accept this definition without proof. We use it to mark the 'holes' in the development that should be completed as part of your homework exercises. In practice, [Admitted] is useful when you're incrementally developing large proofs. *) Definition admit {T: Type} : T. Admitted. (* ###################################################################### *) (** * Introduction *) (** The functional programming style brings programming closer to simple, everyday mathematics: If a procedure or method has no side effects, then (ignoring efficiency) all we need to understand about it is how it maps inputs to outputs -- that is, we can think of it as just a concrete method for computing a mathematical function. This is one sense of the word "functional" in "functional programming." The direct connection between programs and simple mathematical objects supports both formal correctness proofs and sound informal reasoning about program behavior. The other sense in which functional programming is "functional" is that it emphasizes the use of functions (or methods) as _first-class_ values -- i.e., values that can be passed as arguments to other functions, returned as results, included in data structures, etc. The recognition that functions can be treated as data in this way enables a host of useful and powerful idioms. Other common features of functional languages include _algebraic data types_ and _pattern matching_, which make it easy to construct and manipulate rich data structures, and sophisticated _polymorphic type systems_ supporting abstraction and code reuse. Coq shares all of these features. The first half of this chapter introduces the most essential elements of Coq's functional programming language. The second half introduces some basic _tactics_ that can be used to prove simple properties of Coq programs. *) (* ###################################################################### *) (** * Enumerated Types *) (** One unusual aspect of Coq is that its set of built-in features is _extremely_ small. For example, instead of providing the usual palette of atomic data types (booleans, integers, strings, etc.), Coq offers a powerful mechanism for defining new data types from scratch, from which all these familiar types arise as instances. Naturally, the Coq distribution comes with an extensive standard library providing definitions of booleans, numbers, and many common data structures like lists and hash tables. But there is nothing magic or primitive about these library definitions. To illustrate this, we will explicitly recapitulate all the definitions we need in this course, rather than just getting them implicitly from the library. To see how this definition mechanism works, let's start with a very simple example. *) (* ###################################################################### *) (** ** Days of the Week *) (** The following declaration tells Coq that we are defining a new set of data values -- a _type_. *) Inductive day : Type := | monday : day | tuesday : day | wednesday : day | thursday : day | friday : day | saturday : day | sunday : day. (** The type is called [day], and its members are [monday], [tuesday], etc. The second and following lines of the definition can be read "[monday] is a [day], [tuesday] is a [day], etc." Having defined [day], we can write functions that operate on days. *) Definition next_weekday (d:day) : day := match d with | monday => tuesday | tuesday => wednesday | wednesday => thursday | thursday => friday | friday => monday | saturday => monday | sunday => monday end. (** One thing to note is that the argument and return types of this function are explicitly declared. Like most functional programming languages, Coq can often figure out these types for itself when they are not given explicitly -- i.e., it performs _type inference_ -- but we'll include them to make reading easier. *) (** Having defined a function, we should check that it works on some examples. There are actually three different ways to do this in Coq. First, we can use the command [Compute] to evaluate a compound expression involving [next_weekday]. *) Compute (next_weekday friday). (* ==> monday : day *) Compute (next_weekday (next_weekday saturday)). (* ==> tuesday : day *) (** (We show Coq's responses in comments, but, if you have a computer handy, this would be an excellent moment to fire up the Coq interpreter under your favorite IDE -- either CoqIde or Proof General -- and try this for yourself. Load this file, [Basics.v], from the book's accompanying Coq sources, find the above example, submit it to Coq, and observe the result.) Second, we can record what we _expect_ the result to be in the form of a Coq example: *) Example test_next_weekday: (next_weekday (next_weekday saturday)) = tuesday. (** This declaration does two things: it makes an assertion (that the second weekday after [saturday] is [tuesday]), and it gives the assertion a name that can be used to refer to it later. Having made the assertion, we can also ask Coq to verify it, like this: *) Proof. simpl. reflexivity. Qed. (** The details are not important for now (we'll come back to them in a bit), but essentially this can be read as "The assertion we've just made can be proved by observing that both sides of the equality evaluate to the same thing, after some simplification." Third, we can ask Coq to _extract_, from our [Definition], a program in some other, more conventional, programming language (OCaml, Scheme, or Haskell) with a high-performance compiler. This facility is very interesting, since it gives us a way to construct _fully certified_ programs in mainstream languages. Indeed, this is one of the main uses for which Coq was developed. We'll come back to this topic in later chapters. *) (* ###################################################################### *) (** ** Booleans *) (** In a similar way, we can define the standard type [bool] of booleans, with members [true] and [false]. *) Inductive bool : Type := | true : bool | false : bool. (** Although we are rolling our own booleans here for the sake of building up everything from scratch, Coq does, of course, provide a default implementation of the booleans in its standard library, together with a multitude of useful functions and lemmas. (Take a look at [Coq.Init.Datatypes] in the Coq library documentation if you're interested.) Whenever possible, we'll name our own definitions and theorems so that they exactly coincide with the ones in the standard library. Functions over booleans can be defined in the same way as above: *) Definition negb (b:bool) : bool := match b with | true => false | false => true end. Definition andb (b1:bool) (b2:bool) : bool := match b1 with | true => b2 | false => false end. Definition orb (b1:bool) (b2:bool) : bool := match b1 with | true => true | false => b2 end. (** The last two illustrate Coq's syntax for multi-argument function definitions. The corresponding multi-argument application syntax is illustrated by the following four "unit tests," which constitute a complete specification -- a truth table -- for the [orb] function: *) Example test_orb1: (orb true false) = true. Proof. simpl. reflexivity. Qed. Example test_orb2: (orb false false) = false. Proof. simpl. reflexivity. Qed. Example test_orb3: (orb false true) = true. Proof. simpl. reflexivity. Qed. Example test_orb4: (orb true true) = true. Proof. simpl. reflexivity. Qed. (** We can also introduce some familiar syntax for the boolean operations we have just defined. The [Infix] command defines new, infix notation for an existing definition. *) Infix "&&" := andb. Infix "||" := orb. Example test_orb5: false || false || true = true. Proof. simpl. reflexivity. Qed. (** _A note on notation_: In [.v] files, we use square brackets to delimit fragments of Coq code within comments; this convention, also used by the [coqdoc] documentation tool, keeps them visually separate from the surrounding text. In the html version of the files, these pieces of text appear in a [different font]. The special phrases [Admitted] and [admit] can be used as a placeholder for an incomplete definition or proof. We'll use them in exercises, to indicate the parts that we're leaving for you -- i.e., your job is to replace [admit] or [Admitted] with real definitions or proofs. *) (** **** Exercise: 1 star (nandb) *) (** Remove [admit] and complete the definition of the following function; then make sure that the [Example] assertions below can each be verified by Coq. (Remove "[Admitted.]" and fill in each proof, following the model of the [orb] tests above.) The function should return [true] if either or both of its inputs are [false]. *) Definition nandb (b1:bool) (b2:bool) : bool := (* FILL IN HERE *) admit. Example test_nandb1: (nandb true false) = true. (* FILL IN HERE *) Admitted. Example test_nandb2: (nandb false false) = true. (* FILL IN HERE *) Admitted. Example test_nandb3: (nandb false true) = true. (* FILL IN HERE *) Admitted. Example test_nandb4: (nandb true true) = false. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 1 star (andb3) *) (** Do the same for the [andb3] function below. This function should return [true] when all of its inputs are [true], and [false] otherwise. *) Definition andb3 (b1:bool) (b2:bool) (b3:bool) : bool := (* FILL IN HERE *) admit. Example test_andb31: (andb3 true true true) = true. (* FILL IN HERE *) Admitted. Example test_andb32: (andb3 false true true) = false. (* FILL IN HERE *) Admitted. Example test_andb33: (andb3 true false true) = false. (* FILL IN HERE *) Admitted. Example test_andb34: (andb3 true true false) = false. (* FILL IN HERE *) Admitted. (** [] *) (* ###################################################################### *) (** ** Function Types *) (** Every expression in Coq has a type, describing what sort of thing it computes. The [Check] command asks Coq to print the type of an expression. *) (** For example, the type of [negb true] is [bool]. *) Check true. (* ===> true : bool *) Check (negb true). (* ===> negb true : bool *) (** Functions like [negb] itself are also data values, just like [true] and [false]. Their types are called _function types_, and they are written with arrows. *) Check negb. (* ===> negb : bool -> bool *) (** The type of [negb], written [bool -> bool] and pronounced "[bool] arrow [bool]," can be read, "Given an input of type [bool], this function produces an output of type [bool]." Similarly, the type of [andb], written [bool -> bool -> bool], can be read, "Given two inputs, both of type [bool], this function produces an output of type [bool]." *) (* ###################################################################### *) (** ** Modules *) (** Coq provides a _module system_, to aid in organizing large developments. In this course we won't need most of its features, but one is useful: If we enclose a collection of declarations between [Module X] and [End X] markers, then, in the remainder of the file after the [End], these definitions are referred to by names like [X.foo] instead of just [foo]. Here, we use this feature to introduce the definition of the type [nat] in an inner module so that it does not interfere with the one from the standard library, which comes with a bit of special notational magic. *) Module Playground1. (* ###################################################################### *) (** ** Numbers *) (** The types we have defined so far are examples of "enumerated types": their definitions explicitly enumerate a finite set of elements. A more interesting way of defining a type is to give a collection of _inductive rules_ describing its elements. For example, we can define the natural numbers as follows: *) Inductive nat : Type := | O : nat | S : nat -> nat. (** The clauses of this definition can be read: - [O] is a natural number (note that this is the letter "[O]," not the numeral "[0]"). - [S] is a "constructor" that takes a natural number and yields another one -- that is, if [n] is a natural number, then [S n] is too. Let's look at this in a little more detail. Every inductively defined set ([day], [nat], [bool], etc.) is actually a set of _expressions_. The definition of [nat] says how expressions in the set [nat] can be constructed: - the expression [O] belongs to the set [nat]; - if [n] is an expression belonging to the set [nat], then [S n] is also an expression belonging to the set [nat]; and - expressions formed in these two ways are the only ones belonging to the set [nat]. The same rules apply for our definitions of [day] and [bool]. The annotations we used for their constructors are analogous to the one for the [O] constructor, indicating that they don't take any arguments. These three conditions are the precise force of the [Inductive] declaration. They imply that the expression [O], the expression [S O], the expression [S (S O)], the expression [S (S (S O))], and so on all belong to the set [nat], while other expressions like [true], [andb true false], and [S (S false)] do not. We can write simple functions that pattern match on natural numbers just as we did above -- for example, the predecessor function: *) Definition pred (n : nat) : nat := match n with | O => O | S n' => n' end. (** The second branch can be read: "if [n] has the form [S n'] for some [n'], then return [n']." *) End Playground1. Definition minustwo (n : nat) : nat := match n with | O => O | S O => O | S (S n') => n' end. (** Because natural numbers are such a pervasive form of data, Coq provides a tiny bit of built-in magic for parsing and printing them: ordinary arabic numerals can be used as an alternative to the "unary" notation defined by the constructors [S] and [O]. Coq prints numbers in arabic form by default: *) Check (S (S (S (S O)))). (* ===> 4 : nat *) Compute (minustwo 4). (* ===> 2 : nat *) (** The constructor [S] has the type [nat -> nat], just like the functions [minustwo] and [pred]: *) Check S. Check pred. Check minustwo. (** These are all things that can be applied to a number to yield a number. However, there is a fundamental difference between the first one and the other two: functions like [pred] and [minustwo] come with _computation rules_ -- e.g., the definition of [pred] says that [pred 2] can be simplified to [1] -- while the definition of [S] has no such behavior attached. Although it is like a function in the sense that it can be applied to an argument, it does not _do_ anything at all! For most function definitions over numbers, just pattern matching is not enough: we also need recursion. For example, to check that a number [n] is even, we may need to recursively check whether [n-2] is even. To write such functions, we use the keyword [Fixpoint]. *) Fixpoint evenb (n:nat) : bool := match n with | O => true | S O => false | S (S n') => evenb n' end. (** We can define [oddb] by a similar [Fixpoint] declaration, but here is a simpler definition that is a bit easier to work with: *) Definition oddb (n:nat) : bool := negb (evenb n). Example test_oddb1: oddb 1 = true. Proof. simpl. reflexivity. Qed. Example test_oddb2: oddb 4 = false. Proof. simpl. reflexivity. Qed. (** (You will notice if you step through these proofs that [simpl] actually has no effect on the goal -- all of the work is done by [reflexivity]. We'll see more about why that is shortly.) Naturally, we can also define multi-argument functions by recursion. *) Module Playground2. Fixpoint plus (n : nat) (m : nat) : nat := match n with | O => m | S n' => S (plus n' m) end. (** Adding three to two now gives us five, as we'd expect. *) Compute (plus 3 2). (** The simplification that Coq performs to reach this conclusion can be visualized as follows: *) (* [plus (S (S (S O))) (S (S O))] ==> [S (plus (S (S O)) (S (S O)))] by the second clause of the [match] ==> [S (S (plus (S O) (S (S O))))] by the second clause of the [match] ==> [S (S (S (plus O (S (S O)))))] by the second clause of the [match] ==> [S (S (S (S (S O))))] by the first clause of the [match] *) (** As a notational convenience, if two or more arguments have the same type, they can be written together. In the following definition, [(n m : nat)] means just the same as if we had written [(n : nat) (m : nat)]. *) Fixpoint mult (n m : nat) : nat := match n with | O => O | S n' => plus m (mult n' m) end. Example test_mult1: (mult 3 3) = 9. Proof. simpl. reflexivity. Qed. (** You can match two expressions at once by putting a comma between them: *) Fixpoint minus (n m:nat) : nat := match n, m with | O , _ => O | S _ , O => n | S n', S m' => minus n' m' end. (** The _ in the first line is a _wildcard pattern_. Writing _ in a pattern is the same as writing some variable that doesn't get used on the right-hand side. This avoids the need to invent a bogus variable name. *) End Playground2. Fixpoint exp (base power : nat) : nat := match power with | O => S O | S p => mult base (exp base p) end. (** **** Exercise: 1 star (factorial) *) (** Recall the standard mathematical factorial function: << factorial(0) = 1 factorial(n) = n * factorial(n-1) (if n>0) >> Translate this into Coq. *) Fixpoint factorial (n:nat) : nat := (* FILL IN HERE *) admit. Example test_factorial1: (factorial 3) = 6. (* FILL IN HERE *) Admitted. Example test_factorial2: (factorial 5) = (mult 10 12). (* FILL IN HERE *) Admitted. (** [] *) (** We can make numerical expressions a little easier to read and write by introducing _notations_ for addition, multiplication, and subtraction. *) Notation "x + y" := (plus x y) (at level 50, left associativity) : nat_scope. Notation "x - y" := (minus x y) (at level 50, left associativity) : nat_scope. Notation "x * y" := (mult x y) (at level 40, left associativity) : nat_scope. Check ((0 + 1) + 1). (** (The [level], [associativity], and [nat_scope] annotations control how these notations are treated by Coq's parser. The details are not important, but interested readers can refer to the optional "More on Notation" section at the end of this chapter.) Note that these do not change the definitions we've already made: they are simply instructions to the Coq parser to accept [x + y] in place of [plus x y] and, conversely, to the Coq pretty-printer to display [plus x y] as [x + y]. When we say that Coq comes with nothing built-in, we really mean it: even equality testing for numbers is a user-defined operation! *) (** The [beq_nat] function tests [nat]ural numbers for [eq]uality, yielding a [b]oolean. Note the use of nested [match]es (we could also have used a simultaneous match, as we did in [minus].) *) Fixpoint beq_nat (n m : nat) : bool := match n with | O => match m with | O => true | S m' => false end | S n' => match m with | O => false | S m' => beq_nat n' m' end end. (** The [leb] function tests whether its first argument is less than or equal to its second argument, yielding a boolean. *) Fixpoint leb (n m : nat) : bool := match n with | O => true | S n' => match m with | O => false | S m' => leb n' m' end end. Example test_leb1: (leb 2 2) = true. Proof. simpl. reflexivity. Qed. Example test_leb2: (leb 2 4) = true. Proof. simpl. reflexivity. Qed. Example test_leb3: (leb 4 2) = false. Proof. simpl. reflexivity. Qed. (** **** Exercise: 1 star (blt_nat) *) (** The [blt_nat] function tests [nat]ural numbers for [l]ess-[t]han, yielding a [b]oolean. Instead of making up a new [Fixpoint] for this one, define it in terms of a previously defined function. *) Definition blt_nat (n m : nat) : bool := (* FILL IN HERE *) admit. Example test_blt_nat1: (blt_nat 2 2) = false. (* FILL IN HERE *) Admitted. Example test_blt_nat2: (blt_nat 2 4) = true. (* FILL IN HERE *) Admitted. Example test_blt_nat3: (blt_nat 4 2) = false. (* FILL IN HERE *) Admitted. (** [] *) (* ###################################################################### *) (** * Proof by Simplification *) (** Now that we've defined a few datatypes and functions, let's turn to stating and proving properties of their behavior. Actually, we've already started doing this: each [Example] in the previous sections makes a precise claim about the behavior of some function on some particular inputs. The proofs of these claims were always the same: use [simpl] to simplify both sides of the equation, then use [reflexivity] to check that both sides contain identical values. The same sort of "proof by simplification" can be used to prove more interesting properties as well. For example, the fact that [0] is a "neutral element" for [+] on the left can be proved just by observing that [0 + n] reduces to [n] no matter what [n] is, a fact that can be read directly off the definition of [plus].*) Theorem plus_O_n : forall n : nat, 0 + n = n. Proof. intros n. simpl. reflexivity. Qed. (** (You may notice that the above statement looks different in the [.v] file in your IDE than it does in the HTML rendition in your browser, if you are viewing both. In [.v] files, we write the [forall] universal quantifier using the reserved identifier "forall." When the [.v] files are converted to HTML, this gets transformed into an upside-down-A symbol.) *) (** This is a good place to mention that [reflexivity] is a bit more powerful than we have admitted. In the examples we have seen, the calls to [simpl] were actually not needed, because [reflexivity] can perform some simplification automatically when checking that two sides are equal; [simpl] was just added so that we could see the intermediate state -- after simplification but before finishing the proof. Here is a shorter proof of the theorem: *) Theorem plus_O_n' : forall n : nat, 0 + n = n. Proof. intros n. reflexivity. Qed. (** Moreover, it will be useful later to know that [reflexivity] does somewhat _more_ simplification than [simpl] does -- for example, it tries "unfolding" defined terms, replacing them with their right-hand sides. The reason for this difference is that, if reflexivity succeeds, the whole goal is finished and we don't need to look at whatever expanded expressions [reflexivity] has created by all this simplification and unfolding; by contrast, [simpl] is used in situations where we may have to read and understand the new goal that it creates, so we would not want it blindly expanding definitions and leaving the goal in a messy state. *) (** The form of the theorem we just stated and its proof are almost exactly the same as the simpler examples we saw earlier; there are just a few differences. First, we've used the keyword [Theorem] instead of [Example]. This difference is purely a matter of style; the keywords [Example] and [Theorem] (and a few others, including [Lemma], [Fact], and [Remark]) mean exactly the same thing to Coq. Second, we've added the quantifier [forall n:nat], so that our theorem talks about _all_ natural numbers [n]. In order to prove theorems of this form, we need to to be able to reason by _assuming_ the existence of an arbitrary natural number [n]. This is achieved in the proof by [intros n], which moves the quantifier from the goal to a _context_ of current assumptions. In effect, we start the proof by saying "Suppose [n] is some arbitrary number..." The keywords [intros], [simpl], and [reflexivity] are examples of _tactics_. A tactic is a command that is used between [Proof] and [Qed] to guide the process of checking some claim we are making. We will see several more tactics in the rest of this chapter and yet more in future chapters. Other similar theorems can be proved with the same pattern. *) Theorem plus_1_l : forall n:nat, 1 + n = S n. Proof. intros n. reflexivity. Qed. Theorem mult_0_l : forall n:nat, 0 * n = 0. Proof. intros n. reflexivity. Qed. (** The [_l] suffix in the names of these theorems is pronounced "on the left." *) (** It is worth stepping through these proofs to observe how the context and the goal change. *) (** You may want to add calls to [simpl] before [reflexivity] to see the simplifications that Coq performs on the terms before checking that they are equal. Although simplification is powerful enough to prove some fairly general facts, there are many statements that cannot be handled by simplification alone. For instance, we cannot use it to prove that [0] is also a neutral element for [+] _on the right_. *) Theorem plus_n_O : forall n, n = n + 0. Proof. intros n. simpl. (* Doesn't do anything! *) (** (Can you explain why this happens? Step through both proofs with Coq and notice how the goal and context change.) When stuck in the middle of a proof, we can use the [Abort] command to give up on it for the moment. *) Abort. (** The next chapter will introduce _induction_, a powerful technique that can be used for proving this goal. For the moment, though, let's look at a few more simple tactics. *) (* ###################################################################### *) (** * Proof by Rewriting *) (** This theorem is a bit more interesting than the others we've seen: *) Theorem plus_id_example : forall n m:nat, n = m -> n + n = m + m. (** Instead of making a universal claim about all numbers [n] and [m], it talks about a more specialized property that only holds when [n = m]. The arrow symbol is pronounced "implies." As before, we need to be able to reason by assuming the existence of some numbers [n] and [m]. We also need to assume the hypothesis [n = m]. The [intros] tactic will serve to move all three of these from the goal into assumptions in the current context. Since [n] and [m] are arbitrary numbers, we can't just use simplification to prove this theorem. Instead, we prove it by observing that, if we are assuming [n = m], then we can replace [n] with [m] in the goal statement and obtain an equality with the same expression on both sides. The tactic that tells Coq to perform this replacement is called [rewrite]. *) Proof. (* move both quantifiers into the context: *) intros n m. (* move the hypothesis into the context: *) intros H. (* rewrite the goal using the hypothesis: *) rewrite -> H. reflexivity. Qed. (** The first line of the proof moves the universally quantified variables [n] and [m] into the context. The second moves the hypothesis [n = m] into the context and gives it the name [H]. The third tells Coq to rewrite the current goal ([n + n = m + m]) by replacing the left side of the equality hypothesis [H] with the right side. (The arrow symbol in the [rewrite] has nothing to do with implication: it tells Coq to apply the rewrite from left to right. To rewrite from right to left, you can use [rewrite <-]. Try making this change in the above proof and see what difference it makes.) *) (** **** Exercise: 1 star (plus_id_exercise) *) (** Remove "[Admitted.]" and fill in the proof. *) Theorem plus_id_exercise : forall n m o : nat, n = m -> m = o -> n + m = m + o. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** The [Admitted] command tells Coq that we want to skip trying to prove this theorem and just accept it as a given. This can be useful for developing longer proofs, since we can state subsidiary lemmas that we believe will be useful for making some larger argument, use [Admitted] to accept them on faith for the moment, and continue working on the main argument until we are sure it makes sense; then we can go back and fill in the proofs we skipped. Be careful, though: every time you say [Admitted] (or [admit]) you are leaving a door open for total nonsense to enter Coq's nice, rigorous, formally checked world! *) (** We can also use the [rewrite] tactic with a previously proved theorem instead of a hypothesis from the context. *) Theorem mult_0_plus : forall n m : nat, (0 + n) * m = n * m. Proof. intros n m. rewrite -> plus_O_n. reflexivity. Qed. (** **** Exercise: 2 stars (mult_S_1) *) Theorem mult_S_1 : forall n m : nat, m = S n -> m * (1 + n) = m * m. Proof. (* FILL IN HERE *) Admitted. (** [] *) (* ###################################################################### *) (** * Proof by Case Analysis *) (** Of course, not everything can be proved by simple calculation and rewriting: In general, unknown, hypothetical values (arbitrary numbers, booleans, lists, etc.) can block simplification. For example, if we try to prove the following fact using the [simpl] tactic as above, we get stuck. *) Theorem plus_1_neq_0_firsttry : forall n : nat, beq_nat (n + 1) 0 = false. Proof. intros n. simpl. (* does nothing! *) Abort. (** The reason for this is that the definitions of both [beq_nat] and [+] begin by performing a [match] on their first argument. But here, the first argument to [+] is the unknown number [n] and the argument to [beq_nat] is the compound expression [n + 1]; neither can be simplified. To make progress, we need to consider the possible forms of [n] separately. If [n] is [O], then we can calculate the final result of [beq_nat (n + 1) 0] and check that it is, indeed, [false]. And if [n = S n'] for some [n'], then, although we don't know exactly what number [n + 1] yields, we can calculate that, at least, it will begin with one [S], and this is enough to calculate that, again, [beq_nat (n + 1) 0] will yield [false]. The tactic that tells Coq to consider, separately, the cases where [n = O] and where [n = S n'] is called [destruct]. *) Theorem plus_1_neq_0 : forall n : nat, beq_nat (n + 1) 0 = false. Proof. intros n. destruct n as [| n']. - reflexivity. - reflexivity. Qed. (** The [destruct] generates _two_ subgoals, which we must then prove, separately, in order to get Coq to accept the theorem. The annotation "[as [| n']]" is called an _intro pattern_. It tells Coq what variable names to introduce in each subgoal. In general, what goes between the square brackets is a _list of lists_ of names, separated by [|]. In this case, the first component is empty, since the [O] constructor is nullary (it doesn't have any arguments). The second component gives a single name, [n'], since [S] is a unary constructor. The [-] signs on the second and third lines are called _bullets_, and they mark the parts of the proof that correspond to each generated subgoal. The proof script that comes after a bullet is the entire proof for a subgoal. In this example, each of the subgoals is easily proved by a single use of [reflexivity], which itself performs some simplification -- e.g., the first one simplifies [beq_nat (S n' + 1) 0] to [false] by first rewriting [(S n' + 1)] to [S (n' + 1)], then unfolding [beq_nat], and then simplifying the [match]. Marking cases with bullets is entirely optional: if bullets are not present, Coq simply asks you to prove each subgoal in sequence, one at a time. But it is a good idea to use bullets. For one thing, they make the structure of a proof apparent, making it more readable. Also, bullets instruct Coq to ensure that a subgoal is complete before trying to verify the next one, preventing proofs for different subgoals from getting mixed up. These issues become especially important in large developments, where fragile proofs lead to long debugging sessions. There are no hard and fast rules for how proofs should be formatted in Coq -- in particular, where lines should be broken and how sections of the proof should be indented to indicate their nested structure. However, if the places where multiple subgoals are generated are marked with explicit bullets at the beginning of lines, then the proof will be readable almost no matter what choices are made about other aspects of layout. This is also a good place to mention one other piece of somewhat obvious advice about line lengths. Beginning Coq users sometimes tend to the extremes, either writing each tactic on its own line or writing entire proofs on one line. Good style lies somewhere in the middle. One reasonable convention is to limit yourself to 80-character lines. The [destruct] tactic can be used with any inductively defined datatype. For example, we use it next to prove that boolean negation is involutive -- i.e., that negation is its own inverse. *) Theorem negb_involutive : forall b : bool, negb (negb b) = b. Proof. intros b. destruct b. - reflexivity. - reflexivity. Qed. (** Note that the [destruct] here has no [as] clause because none of the subcases of the [destruct] need to bind any variables, so there is no need to specify any names. (We could also have written [as [|]], or [as []].) In fact, we can omit the [as] clause from _any_ [destruct] and Coq will fill in variable names automatically. This is generally considered bad style, since Coq often makes confusing choices of names when left to its own devices. It is sometimes useful to invoke [destruct] inside a subgoal, generating yet more proof obligations. In this case, we use different kinds of bullets to mark goals on different "levels." For example: *) Theorem andb_commutative : forall b c, andb b c = andb c b. Proof. intros b c. destruct b. - destruct c. + reflexivity. + reflexivity. - destruct c. + reflexivity. + reflexivity. Qed. (** Each pair of calls to [reflexivity] corresponds to the subgoals that were generated after the execution of the [destruct c] line right above it. Besides [-] and [+], Coq proofs can also use [*] (asterisk) as a third kind of bullet. If we ever encounter a proof that generates more than three levels of subgoals, we can also enclose individual subgoals in curly braces ([{ ... }]): *) Theorem andb_commutative' : forall b c, andb b c = andb c b. Proof. intros b c. destruct b. { destruct c. { reflexivity. } { reflexivity. } } { destruct c. { reflexivity. } { reflexivity. } } Qed. (** Since curly braces mark both the beginning and the end of a proof, they can be used for multiple subgoal levels, as this example shows. Furthermore, curly braces allow us to reuse the same bullet shapes at multiple levels in a proof: *) Theorem andb3_exchange : forall b c d, andb (andb b c) d = andb (andb b d) c. Proof. intros b c d. destruct b. - destruct c. { destruct d. - reflexivity. - reflexivity. } { destruct d. - reflexivity. - reflexivity. } - destruct c. { destruct d. - reflexivity. - reflexivity. } { destruct d. - reflexivity. - reflexivity. } Qed. (** Before closing the chapter, let's mention one final convenience. As you may have noticed, many proofs perform case analysis on a variable right after introducing it: intros x y. destruct y as [|y]. This pattern is so common that Coq provides a shorthand for it: we can perform case analysis on a variable when introducing it by using an intro pattern instead of a variable name. For instance, here is a shorter proof of the [plus_1_neq_0] theorem above. *) Theorem plus_1_neq_0' : forall n : nat, beq_nat (n + 1) 0 = false. Proof. intros [|n]. - reflexivity. - reflexivity. Qed. (** If there are no arguments to name, we can just write [[]]. *) Theorem andb_commutative'' : forall b c, andb b c = andb c b. Proof. intros [] []. - reflexivity. - reflexivity. - reflexivity. - reflexivity. Qed. (** **** Exercise: 2 stars (andb_true_elim2) *) (** Prove the following claim, marking cases (and subcases) with bullets when you use [destruct]. *) Theorem andb_true_elim2 : forall b c : bool, andb b c = true -> c = true. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 1 star (zero_nbeq_plus_1) *) Theorem zero_nbeq_plus_1 : forall n : nat, beq_nat 0 (n + 1) = false. Proof. (* FILL IN HERE *) Admitted. (** [] *) (* ###################################################################### *) (** ** More on Notation (Optional) *) (** (In general, sections marked Optional are not needed to follow the rest of the book, except possibly other Optional sections. On a first reading, you might want to skim these sections so that you know what's there for future reference.) Recall the notation definitions for infix plus and times: *) Notation "x + y" := (plus x y) (at level 50, left associativity) : nat_scope. Notation "x * y" := (mult x y) (at level 40, left associativity) : nat_scope. (** For each notation symbol in Coq, we can specify its _precedence level_ and its _associativity_. The precedence level [n] is specified by writing [at level n]; this helps Coq parse compound expressions. The associativity setting helps to disambiguate expressions containing multiple occurrences of the same symbol. For example, the parameters specified above for [+] and [*] say that the expression [1+2*3*4] is shorthand for [(1+((2*3)*4))]. Coq uses precedence levels from 0 to 100, and _left_, _right_, or _no_ associativity. We will see more examples of this later, e.g., in the [Lists] chapter. Each notation symbol is also associated with a _notation scope_. Coq tries to guess what scope you mean from context, so when you write [S(O*O)] it guesses [nat_scope], but when you write the cartesian product (tuple) type [bool*bool] it guesses [type_scope]. Occasionally, you may have to help it out with percent-notation by writing [(x*y)%nat], and sometimes in Coq's feedback to you it will use [%nat] to indicate what scope a notation is in. Notation scopes also apply to numeral notation ([3], [4], [5], etc.), so you may sometimes see [0%nat], which means [O] (the natural number [0] that we're using in this chapter), or [0%Z], which means the Integer zero (which comes from a different part of the standard library). *) (* ###################################################################### *) (** ** Fixpoints and Structural Recursion (Optional) *) (** Here is a copy of the definition of addition: *) Fixpoint plus' (n : nat) (m : nat) : nat := match n with | O => m | S n' => S (plus' n' m) end. (** When Coq checks this definition, it notes that [plus'] is "decreasing on 1st argument." What this means is that we are performing a _structural recursion_ over the argument [n] -- i.e., that we make recursive calls only on strictly smaller values of [n]. This implies that all calls to [plus'] will eventually terminate. Coq demands that some argument of _every_ [Fixpoint] definition is "decreasing." This requirement is a fundamental feature of Coq's design: In particular, it guarantees that every function that can be defined in Coq will terminate on all inputs. However, because Coq's "decreasing analysis" is not very sophisticated, it is sometimes necessary to write functions in slightly unnatural ways. *) (** **** Exercise: 2 stars, optional (decreasing) *) (** To get a concrete sense of this, find a way to write a sensible [Fixpoint] definition (of a simple function on numbers, say) that _does_ terminate on all inputs, but that Coq will reject because of this restriction. *) (* FILL IN HERE *) (** [] *) (* ###################################################################### *) (** * More Exercises *) (** **** Exercise: 2 stars (boolean_functions) *) (** Use the tactics you have learned so far to prove the following theorem about boolean functions. *) Theorem identity_fn_applied_twice : forall (f : bool -> bool), (forall (x : bool), f x = x) -> forall (b : bool), f (f b) = b. Proof. (* FILL IN HERE *) Admitted. (** Now state and prove a theorem [negation_fn_applied_twice] similar to the previous one but where the second hypothesis says that the function [f] has the property that [f x = negb x].*) (* FILL IN HERE *) (** [] *) (** **** Exercise: 2 stars (andb_eq_orb) *) (** Prove the following theorem. (You may want to first prove a subsidiary lemma or two. Alternatively, remember that you do not have to introduce all hypotheses at the same time.) *) Theorem andb_eq_orb : forall (b c : bool), (andb b c = orb b c) -> b = c. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars (binary) *) (** Consider a different, more efficient representation of natural numbers using a binary rather than unary system. That is, instead of saying that each natural number is either zero or the successor of a natural number, we can say that each binary number is either - zero, - twice a binary number, or - one more than twice a binary number. (a) First, write an inductive definition of the type [bin] corresponding to this description of binary numbers. (Hint: Recall that the definition of [nat] from class, Inductive nat : Type := | O : nat | S : nat -> nat. says nothing about what [O] and [S] "mean." It just says "[O] is in the set called [nat], and if [n] is in the set then so is [S n]." The interpretation of [O] as zero and [S] as successor/plus one comes from the way that we _use_ [nat] values, by writing functions to do things with them, proving things about them, and so on. Your definition of [bin] should be correspondingly simple; it is the functions you will write next that will give it mathematical meaning.) (b) Next, write an increment function [incr] for binary numbers, and a function [bin_to_nat] to convert binary numbers to unary numbers. (c) Write five unit tests [test_bin_incr1], [test_bin_incr2], etc. for your increment and binary-to-unary functions. Notice that incrementing a binary number and then converting it to unary should yield the same result as first converting it to unary and then incrementing. *) (* FILL IN HERE *) (** [] *) (** $Date: 2016-02-18 15:17:32 -0800 (Thu, 18 Feb 2016) $ *)