(** * Basics: Functional Programming *) (* ###################################################################### *) (** * Enumerated Types *) (** In Coq's programming language, almost nothing is built in -- not even booleans or numbers! Instead, it provides powerful tools for defining new types of data and functions that process and transform them. *) (* #########################################a############################# *) (** ** Days of the Week *) (** Let's start with a very simple example. The following declaration tells Coq that we are defining a new set of data values -- a "type." The type is called [day], and its members are [monday], [tuesday], etc. The lines of the definition can be read "[monday] is a [day], [tuesday] is a [day], etc." *) Inductive day : Type := | monday : day | tuesday : day | wednesday : day | thursday : day | friday : day | saturday : day | sunday : day. (** 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. Definition tomorrow (d:day) :day := match d with | monday => tuesday | tuesday => wednesday | wednesday => thursday | thursday => friday | friday => saturday | saturday => sunday | 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 work out these types even if they are not given explicitly -- i.e., it performs some _type inference_ -- but we'll always 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 [Eval simpl] to evaluate a compound expression involving [next_weekday]. Uncomment the following and see what they do. *) Eval simpl in (next_weekday friday). (* ==> monday : day *) Eval simpl in (next_weekday (next_weekday saturday)). (* ==> tuesday : day *) Eval simpl in (tomorrow friday). Eval simpl in (tomorrow sunday). (** If you have a computer handy, now 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. *) (** The keyword [simpl] ("simplify") tells Coq precisely how to evaluate the expression we give it. For the moment, [simpl] is the only one we'll need; later on we'll see some alternatives that are sometimes useful. *) (** 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. Example day_after_next: (tomorrow (tomorrow monday)) = wednesday. 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 are the same after simplification." *) (** Third, we can ask Coq to "extract," from a [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 won't have space to dig further into this topic, but more information can be found in the Coq'Art book by Bertot and Cast\u00c8ran, as well as the Coq reference manual. *) (* ###################################################################### *) (** ** Booleans *) (** In a similar way, we can define the 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 the syntax for multi-argument function definitions. *) (** The following four "unit tests" 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. (** _A note on notation_: We use square brackets to delimit fragments of Coq code in comments in .v files; 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 following bit of Coq hackery defines a magic value called [admit] that can fill a hole in an incomplete definition or proof. We'll use it in the definition of [nandb] in the following exercise. In general, your job in the exercises is to replace [admit] or [Admitted] with real definitions or proofs. *) Definition admit {T: Type} : T. Admitted. (** **** Exercise: 1 star (nandb) *) (** Complete the definition of the following function, then make sure that the [Example] assertions below each can be verified by Coq. *) (** This function should return [true] if either or both of its inputs are [false]. *) Definition nandb (b1:bool) (b2:bool) : bool := (negb (andb b1 b2)). (** Remove "[Admitted.]" and fill in each proof with "[Proof. simpl. reflexivity. Qed.]" *) Example test_nandb1: (nandb true false) = true. Proof. simpl. reflexivity. Qed. Example test_nandb2: (nandb false false) = true. Proof. simpl. reflexivity. Qed. Example test_nandb3: (nandb false true) = true. Proof. simpl. reflexivity. Qed. Example test_nandb4: (nandb true true) = false. Proof. simpl. reflexivity. Qed. (** [] *) (** **** Exercise: 1 star (andb3) *) Definition andb3 (b1:bool) (b2:bool) (b3:bool) : bool := (andb (andb b1 b2) b3). Example test_andb31: (andb3 true true true) = true. Proof. simpl. reflexivity. Qed. Example test_andb32: (andb3 false true true) = false. Proof. simpl. reflexivity. Qed. Example test_andb33: (andb3 true false true) = false. Proof. simpl. reflexivity. Qed. Example test_andb34: (andb3 true true false) = false. Proof. simpl. reflexivity. Qed. (** [] *) (* ###################################################################### *) (** ** Function Types *) (** The [Check] command causes 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 *) Check tomorrow. Check nandb. (** 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]." *) (* ###################################################################### *) (** ** Numbers *) (** _Technical digression_: Coq provides a fairly fancy module system, to aid in organizing large developments. In this course, we won't need most of its features, but one of them 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], all these definitions will be referred to by names like [X.foo] instead of just [foo]. This means that the new definition will not clash with the unqualified name [foo] later, which would otherwise be an error (a name can only be defined once in a given scope). Here, we use this feature to introduce the definition of the type [nat] in an inner module so that it does not shadow the one from the standard library. *) Module Playground1. (** 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 ([weekday], [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]. *) (** 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, predecessor: *) 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)))). Eval simpl in (minustwo 4). Eval simpl in (pred 4). Eval simpl in (Playground1.pred (Playground1.S (Playground1.S Playground1.O))). (** 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: functions like [pred] and [minustwo] come with _computation rules_ -- e.g., the definition of [pred] says that [pred n] can be simplified to [match n with | O => O | S m' => m' end] -- while the definition of [S] has no such behavior attached. Although it is 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, pure 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. (** When Coq checks this definition, it notes that [evenb] is "decreasing on 1st argument." What this means is that we are performing a _structural recursion_ (or _primitive 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 [evenb] will eventually terminate. Coq demands that some argument of _every_ [Fixpoint] definition is decreasing. *) (** We can define [oddb] by a similar [Fixpoint] declaration, but here is a simpler definition that will be a bit easier to work with: *) Definition oddb (n:nat) : bool := negb (evenb n). Example test_oddb1: (oddb (S O)) = true. Proof. simpl. reflexivity. Qed. Example test_oddb2: (oddb (S (S (S (S O))))) = false. Proof. simpl. reflexivity. Qed. (** Naturally, we can also define multi-argument functions by recursion. (Once again, we use a module to avoid polluting the namespace.) *) 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. *) Eval simpl in (plus (S (S (S O))) (S (S O))). (** 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. (** 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. Example test_mult1: (mult 3 3) = 9. Proof. simpl. reflexivity. Qed. Example test_exp: (exp 3 3) = 27. Proof. simpl. reflexivity. Qed. (** **** Exercise: 1 star (factorial) *) (** Recall the standard factorial function: << factorial(0) = 1 factorial(n) = n * factorial(n-1) (if n>0) >> Translate this into Coq. *) Fixpoint factorial (n:nat) : nat := match n with O => S O | S m => n * (factorial m) end. Example test_factorial1: (factorial 3) = 6. Proof. simpl. reflexivity. Qed. Example test_factorial2: (factorial 5) = (mult 10 12). Proof. simpl. reflexivity. Qed. (** [] *) (** 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). (** 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]. Each notation-symbol in Coq is active in a _notation scope_. Coq tries to guess what scope you mean, 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 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], or [0%Z] which means the Integer zero. *) Eval simpl in 1 + 5 * 4. (** 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. (** Similarly, the [ble_nat] function tests [nat]ural numbers for [l]ess-or-[e]qual, yielding a [b]oolean. *) Fixpoint ble_nat (n m : nat) : bool := match n with | O => true | S n' => match m with | O => false | S m' => ble_nat n' m' end end. Example test_ble_nat1: (ble_nat 2 2) = true. Proof. simpl. reflexivity. Qed. Example test_ble_nat2: (ble_nat 2 4) = true. Proof. simpl. reflexivity. Qed. Example test_ble_nat3: (ble_nat 4 2) = false. Proof. simpl. reflexivity. Qed. (** **** Exercise: 2 stars (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. Note: If you have trouble with the [simpl] tactic, try using [compute], which is like [simpl] on steroids. However, there is a simple, elegant solution for which [simpl] suffices. *) Definition blt_nat (n m : nat) : bool := (andb (ble_nat n m) (negb (beq_nat n m))). Example test_blt_nat1: (blt_nat 2 2) = false. Proof. simpl. reflexivity. Qed. Example test_blt_nat2: (blt_nat 2 4) = true. Proof. simpl. reflexivity. Qed. Example test_blt_nat3: (blt_nat 4 2) = false. Proof. simpl. reflexivity. Qed. (** [] *) (* ###################################################################### *) (** * Proof By Simplification *) (** Now that we've defined a few datatypes and functions, let's turn to the question of how to state and prove properties of their behavior. Actually, in a sense, 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 the function's definition to simplify the expressions on both sides of the [=] and notice that they become identical. 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, since the definition of [+] is recursive in its first argument. *) Theorem plus_O_n : forall n:nat, 0 + n = n. Proof. simpl. reflexivity. Qed. (** The [reflexivity] command implicitly simplifies both sides of the equality before testing to see if they are the same, so we can shorten the proof a little. *) (** (It will be useful later to know that [reflexivity] actually does somwhat more than [simpl] -- for example, it tries "unfolding" defined terms, replacing them with their right-hand sides. The reason for this difference is that, when reflexivity succeeds, the whole goal is finished and we don't need to look at whatever expanded expressions [reflexivity] has found; by contrast, [simpl] is used in situations where we may have to read and understand the new goal, so we would not want it blindly expanding definitions.) *) Theorem plus_O_n' : forall n:nat, 0 + n = n. Proof. reflexivity. Qed. (** The form of this theorem and proof are almost exactly the same as the examples above: the only differences are that we've added the quantifier [forall n:nat] and that we've used the keyword [Theorem] instead of [Example]. Indeed, the latter 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. The keywords [simpl] and [reflexivity] are examples of _tactics_. A tactic is a command that is used between [Proof] and [Qed] to tell Coq how it should check the correctness of some claim we are making. We will see several more tactics in the rest of this lecture, and yet more in future lectures. *) (** **** Exercise: 1 star, optional (simpl_plus) *) (** What will Coq print in response to this query? *) Eval compute in (forall n:nat, n + 0 = n). (** What about this one? *) Eval compute in (forall n:nat, 0 + n = n). (** Explain the difference. [] *) (* ###################################################################### *) (** * The [intros] Tactic *) (** Aside from unit tests, which apply functions to particular arguments, most of the properties we will be interested in proving about programs will begin with some quantifiers (e.g., "for all numbers [n], ...") and/or hypothesis ("assuming [m=n], ..."). In such situations, we will need to be able to reason by _assuming the hypothesis_ -- i.e., we start by saying "OK, suppose [n] is some arbitrary number," or "OK, suppose [m=n]." The [intros] tactic permits us to do this by moving one or more quantifiers or hypotheses from the goal to a "context" of current assumptions. For example, here is a slightly different proof of the same theorem. *) Theorem plus_O_n'' : forall n:nat, 0 + n = n. Proof. intros n. reflexivity. Qed. (** Step through this proof in Coq and notice how the goal and context change. *) 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." *) (* ###################################################################### *) (** * Proof by Rewriting *) (** Here is a slightly more interesting theorem: *) Theorem plus_id_example : forall n m:nat, n = m -> n + n = m + m. (** Instead of making a completely universal claim about all numbers [n] and [m], this theorem talks about a more specialized property that only holds when [n = m]. The arrow symbol is pronounced "implies." 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. intros n m. (* move both quantifiers into the context *) intros H. (* move the hypothesis into the context *) rewrite -> H. (* Rewrite the goal using the hypothesis *) 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 in Coq's behavior.) *) (** **** 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. intros n m o. intro. rewrite -> H. intro. rewrite -> H0. reflexivity. Qed. (** [] *) (** The [Admitted] command tells Coq that we want to give up 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 facts that we believe will be useful for making some larger argument, use [Admitted] to accept them on faith for the moment, and continue thinking about the larger 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 [admit] or [Admitted] 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, recommended (mult_1_plus) *) Theorem mult_1_plus : forall n m : nat, (1 + n) * m = m + (n * m). Proof. intros n m. simpl. reflexivity. Qed. (** [] *) (* ###################################################################### *) (** * Case Analysis *) (** Of course, not everything can be proved by simple calculation: In general, unknown, hypothetical values (arbitrary numbers, booleans, lists, etc.) can show up in the "head position" of functions that we want to reason about, blocking 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! *) Admitted. (** 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. What we need is to be able 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 as proved. (No special command is needed for moving from one subgoal to the other. When the first subgoal has been proved, it just disappears and we are left with the other "in focus.") In this proof, each of the subgoals is easily proved by a single use of [reflexivity]. 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 [|]. Here, the first component is empty, since the [O] constructor is nullary (it doesn't carry any data). The second component gives a single name, [n'], since [S] is a unary constructor. The [destruct] tactic can be used with any inductively defined datatype. For example, we use it here 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. Although this is convenient, it is arguably bad style, since Coq often makes confusing choices of names when left to its own devices. *) (** **** Exercise: 1 star (zero_nbeq_plus_1) *) Theorem zero_nbeq_plus_1 : forall n : nat, beq_nat 0 (n + 1) = false. Proof. intro. destruct n as [| n']. reflexivity. reflexivity. Qed. (** [] *) (* ###################################################################### *) (** * Naming Cases *) (** The fact that there is no explicit command for moving from one branch of a case analysis to the next can make proof scripts rather hard to read. In larger proofs, with nested case analyses, it can even become hard to stay oriented when you're sitting with Coq and stepping through the proof. (Imagine trying to remember that the first five subgoals belong to the inner case analysis and the remaining seven cases are what remains of the outer one...) Disciplined use of indentation and comments can help, but a better way is to use the [Case] tactic. [Case] is not built into Coq: we need to define it ourselves. There is no need to understand how it works -- just skip over the definition to the example that follows. It uses some facilities of Coq that we have not discussed -- the string library (just for the concrete syntax of quoted strings) and the [Ltac] command, which allows us to declare custom tactics. Kudos to Aaron Bohannon for this nice hack! *) Require String. Open Scope string_scope. Ltac move_to_top x := match reverse goal with | H : _ |- _ => try move x after H end. Tactic Notation "assert_eq" ident(x) constr(v) := let H := fresh in assert (x = v) as H by reflexivity; clear H. Tactic Notation "Case_aux" ident(x) constr(name) := first [ set (x := name); move_to_top x | assert_eq x name; move_to_top x | fail 1 "because we are working on a different case" ]. Tactic Notation "Case" constr(name) := Case_aux Case name. Tactic Notation "SCase" constr(name) := Case_aux SCase name. Tactic Notation "SSCase" constr(name) := Case_aux SSCase name. Tactic Notation "SSSCase" constr(name) := Case_aux SSSCase name. Tactic Notation "SSSSCase" constr(name) := Case_aux SSSSCase name. Tactic Notation "SSSSSCase" constr(name) := Case_aux SSSSSCase name. Tactic Notation "SSSSSSCase" constr(name) := Case_aux SSSSSSCase name. Tactic Notation "SSSSSSSCase" constr(name) := Case_aux SSSSSSSCase name. (** Here's an example of how [Case] is used. Step through the following proof and observe how the context changes. *) Theorem andb_true_elim1 : forall b c : bool, andb b c = true -> b = true. Proof. intros b c H. destruct b. Case "b = true". reflexivity. Case "b = false". rewrite <- H. reflexivity. Qed. (** [Case] does something very trivial: It simply adds a string that we choose (tagged with the identifier "Case") to the context for the current goal. When subgoals are generated, this string is carried over into their contexts. When the last of these subgoals is finally proved and the next top-level goal (a sibling of the current one) becomes active, this string will no longer appear in the context and we will be able to see that the case where we introduced it is complete. Also, as a sanity check, if we try to execute a new [Case] tactic while the string left by the previous one is still in the context, we get a nice clear error message. For nested case analyses (i.e., when we want to use a [destruct] to solve a goal that has itself been generated by a [destruct]), there is an [SCase] ("subcase") tactic. *) (** **** Exercise: 2 stars (andb_true_elim2) *) (** Prove [andb_true_elim2], marking cases (and subcases) when you use [destruct]. *) Theorem andb_true_elim2 : forall b c : bool, andb b c = true -> c = true. Proof. Admitted. (** [] *) (** 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 [Case] tactics placed 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 a good place to mention one other piece of (possibly obvious) advice about line lengths. Beginning Coq users sometimes tend to the extremes, either writing each tactic on its own line or entire proofs on one line. Good style lies somewhere in the middle. In particular, one reasonable convention is to limit yourself to 80-character lines. Lines longer than this are hard to read and can be inconvenient to display and print. Many editors have features that help enforce this. *) (* ###################################################################### *) (** * Induction *) (** We proved above that [0] is a neutral element for [+] on the left using a simple partial evaluation argument. The fact that it is also a neutral element on the _right_... *) Theorem plus_0_r_firsttry : forall n:nat, n + 0 = n. (** ... cannot be proved in the same simple way. Just applying [reflexivity] doesn't work: the [n] in [n + 0] is an arbitrary unknown number, so the [match] in the definition of [+] can't be simplified. And reasoning by cases using [destruct n] doesn't get us much further: the branch of the case analysis where we assume [n = 0] goes through, but in the branch where [n = S n'] for some [n'] we get stuck in exactly the same way. We could use [destruct n'] to get one step further, but since [n] can be arbitrarily large, if we try to keep on going this way we'll never be done. *) Proof. intros n. simpl. (* Does nothing! *) Admitted. (** Case analysis gets us a little further, but not all the way: *) Theorem plus_0_r_secondtry : forall n:nat, n + 0 = n. Proof. intros n. destruct n as [| n']. Case "n = 0". reflexivity. (* so far so good... *) Case "n = S n'". simpl. (* ...but here we are stuck again *) Admitted. (** To prove such facts -- indeed, to prove most interesting facts about numbers, lists, and other inductively defined sets -- we need a more powerful reasoning principle: _induction_. Recall (from high school) the principle of induction over natural numbers: If [P(n)] is some proposition involving a natural number [n] and we want to show that P holds for _all_ numbers [n], we can reason like this: - show that [P(O)] holds; - show that, for any [n'], if [P(n')] holds, then so does [P(S n')]; - conclude that [P(n)] holds for all [n]. In Coq, the steps are the same but the order is backwards: we begin with the goal of proving [P(n)] for all [n] and break it down (by applying the [induction] tactic) into two separate subgoals: first showing [P(O)] and then showing [P(n') -> P(S n')]. Here's how this works for the theorem we are trying to prove at the moment: *) Theorem plus_0_r : forall n:nat, n + 0 = n. Proof. intros n. induction n as [| n']. Case "n = 0". reflexivity. Case "n = S n'". simpl. rewrite -> IHn'. reflexivity. Qed. (** Like [destruct], the [induction] tactic takes an [as...] clause that specifies the names of the variables to be introduced in the subgoals. In the first branch, [n] is replaced by [0] and the goal becomes [0 + 0 = 0], which follows by simplification. In the second, [n] is replaced by [S n'] and the assumption [n' + 0 = n'] is added to the context (with the name [IHn'], i.e., the Induction Hypothesis for [n']). The goal in this case becomes [(S n') + 0 = S n'], which simplifies to [S (n' + 0) = S n'], which in turn follows from the induction hypothesis. *) Theorem minus_diag : forall n, minus n n = 0. Proof. (* WORKED IN CLASS *) intros n. induction n as [| n']. Case "n = 0". simpl. reflexivity. Case "n = S n'". simpl. rewrite -> IHn'. reflexivity. Qed. (** **** Exercise: 2 stars, recommended (basic_induction) *) Theorem mult_0_r : forall n:nat, n * 0 = 0. Proof. intro. induction n as [| n']. Case " n = 0". reflexivity. Case "n = S n'". simpl. rewrite -> IHn'. reflexivity. Qed. Theorem plus_n_Sm : forall n m : nat, S (n + m) = n + (S m). Proof. Admitted. Theorem plus_comm: forall n m : nat, n + m = m + n. Proof. Admitted. (** [] *) Fixpoint double (n:nat) := match n with | O => O | S n' => S (S (double n')) end. (** **** Exercise: 2 stars (double_plus) *) Lemma double_plus : forall n, double n = n + n . Proof. Admitted. (** [] *) (** **** Exercise: 1 star (destruct_induction) *) (** Briefly explain the difference between the tactics [destruct] and [induction]. (* FILL IN HERE *) *) (** [] *) (* ###################################################################### *) (** * Formal vs. Informal Proof *) (** "Informal proofs are algorithms; formal proofs are code." *) (** The question of what, exactly, constitutes a "proof" of a mathematical claim has challenged philosophers for millenia. A rough and ready definition, though, could be this: a proof of a mathematical proposition [P] is a written (or spoken) text that instills in the reader or hearer the certainty that [P] is true. That is, a proof is an act of communication. Now, acts of communication may involve different sorts of readers. On one hand, the "reader" can be a program like Coq, in which case the "belief" that is instilled is a simple mechanical check that [P] can be derived from a certain set of formal logical rules, and the proof is a recipe that guides the program in performing this check. Such recipes are _formal_ proofs. Alternatively, the reader can be a human being, in which case the proof will be written in English or some other natural language, thus necessarily _informal_. Here, the criteria for success are less clearly specified. A "good" proof is one that makes the reader believe [P]. But the same proof may be read by many different readers, some of whom may be convinced by a particular way of phrasing the argument, while others may not be. One reader may be particularly pedantic, inexperienced, or just plain thick-headed; the only way to convince them will be to make the argument in painstaking detail. But another reader, more familiar in the area, may find all this detail so overwhelming that they lose the overall thread. All they want is to be told the main ideas, because it is easier to fill in the details for themselves. Ultimately, there is no universal standard, because there is no single way of writing an informal proof that is guaranteed to convince every conceivable reader. In practice, however, mathematicians have developed a rich set of conventions and idioms for writing about complex mathematical objects that, within a certain community, make communication fairly reliable. The conventions of this stylized form of communication give a fairly clear standard for judging proofs good or bad. Because we are using Coq in this course, we will be working heavily with formal proofs. But this doesn't mean we can ignore the informal ones! Formal proofs are useful in many ways, but they are _not_ very efficient ways of communicating ideas between human beings. *) (** For example, here is a proof that addition is associative: *) Theorem plus_assoc : forall n m p : nat, n + (m + p) = (n + m) + p. Proof. Admitted. (** **** Exercise: 2 stars (plus_comm_informal) *) (** Translate your solution for [plus_comm] into an informal proof. *) (** Theorem: Addition is commutative. Proof: (* FILL IN HERE *) [] *) (** **** Exercise: 2 stars, optional (beq_nat_refl_informal) *) (** Write an informal proof of the following theorem, using the informal proof of [plus_assoc] as a model. Don't just paraphrase the Coq tactics into English! Theorem: [true = beq_nat n n] for any [n]. Proof: (* FILL IN HERE *) [] *) (** **** Exercise: 1 star, optional (beq_nat_refl) *) Theorem beq_nat_refl : forall n : nat, true = beq_nat n n. Proof. intro n. induction n as [| n']. Case "n = 0". simpl. reflexivity. Case "n = S n'". simpl. rewrite -> IHn'. reflexivity. Qed. (** [] *) (* ###################################################################### *) (** * Proofs Within Proofs *) (** In Coq, as in informal mathematics, large proofs are very often broken into a sequence of theorems, with later proofs referring to earlier theorems. Occasionally, however, a proof will need some miscellaneous fact that is too trivial (and of too little general interest) to bother giving it its own top-level name. In such cases, it is convenient to be able to simply state and prove the needed "sub-theorem" right at the point where it is used. The [assert] tactic allows us to do this. For example, our earlier proof of the [mult_0_plus] theorem referred to a previous theorem named [plus_O_n]. We can also use [assert] to state and prove [plus_O_n] in-line: *) Theorem mult_0_plus' : forall n m : nat, (0 + n) * m = n * m. Proof. intros n m. assert (H: 0 + n = n). Case "Proof of assertion". reflexivity. rewrite -> H. reflexivity. Qed. Theorem mult_0_lemma: forall n, 0 + n = n. Proof. intro n. reflexivity. Qed. (** The [assert] tactic introduces two sub-goals. The first is the assertion itself; by prefixing it with [H:] we name the assertion [H]. (Note that we could also name the assertion with [as] just as we did above with [destruct] and [induction], i.e., [assert (0 + n = n) as H]. Also note that we mark the proof of this assertion with a [Case], both for readability and so that, when using Coq interactively, we can see when we're finished proving the assertion by observing when the ["Proof of assertion"] string disappears from the context.) The second goal is the same as the one at the point where we invoke [assert], except that, in the context, we have the assumption [H] that [0 + n = n]. That is, [assert] generates one subgoal where we must prove the asserted fact and a second subgoal where we can use the asserted fact to make progress on whatever we were trying to prove in the first place. *) (** Actually, [assert] will turn out to be handy in many sorts of situations. For example, suppose we want to prove that [(n + m) + (p + q) = (m + n) + (p + q)]. The only difference between the two sides of the [=] is that the arguments [m] and [n] to the first inner [+] are swapped, so it seems we should be able to use the commutativity of addition ([plus_comm]) to rewrite one into the other. However, the [rewrite] tactic is a little stupid about _where_ it applies the rewrite. There are three uses of [+] here, and it turns out that doing [rewrite -> plus_comm] will affect only the _outer_ one. *) Theorem plus_rearrange_firsttry : forall n m p q : nat, (n + m) + (p + q) = (m + n) + (p + q). Proof. intros n m p q. (* We just need to swap (n + m) for (m + n)... it seems like plus_comm should do the trick! *) rewrite -> plus_comm. (* Doesn't work...Coq rewrote the wrong plus! *) Admitted. (** To get [plus_comm] to apply at the point where we want it, we can introduce a local lemma stating that [n + m = m + n] (for the particular [m] and [n] that we are talking about here), prove this lemma using [plus_comm], and then use this lemma to do the desired rewrite. *) Theorem plus_rearrange : forall n m p q : nat, (n + m) + (p + q) = (m + n) + (p + q). Proof. intros n m p q. assert (H: n + m = m + n). Case "Proof of assertion". rewrite -> plus_comm. reflexivity. rewrite -> H. reflexivity. Qed. (** **** Exercise: 4 stars (mult_comm) *) (** Use [assert] to help prove this theorem. You shouldn't need to use induction. *) Theorem plus_swap : forall n m p : nat, n + (m + p) = m + (n + p). Proof. Admitted. Theorem mult_comm : forall m n : nat, m * n = n * m. Proof. Admitted. Lemma evenb_1: forall n, evenb n = evenb (S (S n)). Proof. intro. simpl evenb. reflexivity. Qed. Lemma oddb_1: forall n, oddb n = oddb (S (S n)). Proof. intro. unfold oddb. simpl. reflexivity. Qed. Lemma foo: forall n, (evenb n = oddb (S n)). Proof. intro. unfold oddb. induction n. simpl. reflexivity. rewrite <- evenb_1. rewrite -> IHn. assert( R: forall p, p = (negb (negb p))). intro. unfold negb. case p. reflexivity. reflexivity. rewrite <- R. reflexivity. Qed. Theorem evenb_n__oddb_Sn : forall n : nat, evenb n = negb (evenb (S n)). Proof. intro. assert(W: forall n, negb (evenb n) = oddb n). intro. unfold oddb. reflexivity. rewrite -> W. rewrite -> foo. reflexivity. Qed. (** [] *) (* ###################################################################### *) (** * More Exercises *) (** **** Exercise: 3 stars, optional (more_exercises) *) (** Take a piece of paper. For each of the following theorems, first _think_ about whether (a) it can be proved using only simplification and rewriting, (b) it also requires case analysis ([destruct]), or (c) it also requires induction. Write down your prediction. Then fill in the proof. (There is no need to turn in your piece of paper; this is just to encourage you to reflect before hacking!) *) Theorem ble_nat_refl : forall n:nat, true = ble_nat n n. Proof. (* FILL IN HERE *) Admitted. Theorem zero_nbeq_S : forall n:nat, beq_nat 0 (S n) = false. Proof. (* FILL IN HERE *) Admitted. Theorem andb_false_r : forall b : bool, andb b false = false. Proof. (* FILL IN HERE *) Admitted. Theorem plus_ble_compat_l : forall n m p : nat, ble_nat n m = true -> ble_nat (p + n) (p + m) = true. Proof. (* FILL IN HERE *) Admitted. Theorem S_nbeq_0 : forall n:nat, beq_nat (S n) 0 = false. Proof. (* FILL IN HERE *) Admitted. Theorem mult_1_l : forall n:nat, 1 * n = n. Proof. (* FILL IN HERE *) Admitted. Theorem all3_spec : forall b c : bool, orb (andb b c) (orb (negb b) (negb c)) = true. Proof. (* FILL IN HERE *) Admitted. Theorem mult_plus_distr_r : forall n m p : nat, (n + m) * p = (n * p) + (m * p). Proof. (* FILL IN HERE *) Admitted. Theorem mult_assoc : forall n m p : nat, n * (m * p) = (n * m) * p. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 2 stars, optional (plus_swap') *) (** The [replace] tactic allows you to specify a particular subterm to rewrite and what you want it rewritten to. More precisely, [replace (t) with (u)] replaces (all copies of) expression [t] in the goal by expression [u], and generates [t = u] as an additional subgoal. This is often useful when a plain [rewrite] acts on the wrong part of the goal. Use the [replace] tactic to do a proof of [plus_swap'], just like [plus_swap] but without needing [assert (n + m = m + n)]. *) Theorem plus_swap' : forall n m p : nat, n + (m + p) = m + (n + p). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 4 stars, recommended (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 a nat (whatever that is), and if [n] is a nat 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 for binary numbers, and a function to convert binary numbers to unary numbers. (c) Finally, prove that your increment and binary-to-unary functions commute: that is, incrementing a binary number and then converting it to unary yields the same result as first converting it to unary and then incrementing. *) (* FILL IN HERE *) (** [] *) (** **** Exercise: 5 stars (binary_inverse) *) (** This exercise is a continuation of the previous exercise about binary numbers. You will need your definitions and theorems from the previous exercise to complete this one. (a) First, write a function to convert natural numbers to binary numbers. Then prove that starting with any natural number, converting to binary, then converting back yields the same natural number you started with. (b) You might naturally think that we should also prove the opposite direction: that starting with a binary number, converting to a natural, and then back to binary yields the same number we started with. However, it is not true! Explain what the problem is. (c) Define a function [normalize] from binary numbers to binary numbers such that for any binary number b, converting to a natural and then back to binary yields [(normalize b)]. Prove it. *) (* FILL IN HERE *) (** **** Exercise: 2 stars, optional (decreasing) *) (** The requirement that some argument to each function be "decreasing" 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. 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 _not_ accept because of this restriction. *) (* FILL IN HERE *) (** [] *)