(** * Ind: Propositions and Evidence *) (* Version of 8/24/2010 *) Require Export Poly. (* ################################################################## *) (** * Review *) (** We've now seen a bunch of Coq's fundamental tactics -- enough, in fact, to do pretty much everything we'll want for a while. We'll introduce one or two more as we go along through the next few lectures, and later in the course we'll introduce some more powerful _automation_ tactics that make Coq do more of the low-level work in many cases. But basically we've got what we need to get work done. Here are the ones we've seen: - [intros]: move hypotheses/variables from goal to context - [reflexivity]: finish the proof (when the goal looks like [e = e]) - [apply]: prove goal using a hypothesis, lemma, or constructor - [apply... in H]: apply a hypothesis, lemma, or constructor to a hypothesis in the context (forward reasoning) - [apply... with...]: explicitly specify values for variables that cannot be determined by pattern matching - [simpl]: simplify computations in the goal - [simpl in H]: ... or a hypothesis - [rewrite]: use an equality hypothesis (or lemma) to rewrite the goal - [rewrite ... in H]: ... or a hypothesis - [unfold]: replace a defined constant by its right-hand side in the goal - [unfold... in H]: ... or a hypothesis - [destruct... as...]: case analysis on values of inductively defined types - [induction... with...]: induction on values of inductively defined types - [inversion]: reason by injectivity and distinctness of constructors - [remember (e) as x]: give a name ([x]) to an expression ([e]) so that we can destruct [x] without "losing" [e] - [assert (e) as H]: introduce a "local lemma" [e] and call it [H] *) (* ##################################################### *) (* ##################################################### *) (** * Programming with Propositions *) (** A _proposition_ is a statement expressing a factual claim, like "two plus two equals four." In Coq, propositions are written as expressions of type [Prop]. Although we haven't mentioned it explicitly, we have already seen numerous examples. *) Check (2 + 2 = 4). (* 2 + 2 = 4%nat : Prop *) Check (ble_nat 3 2 = false). (* ble_nat 3 2 = false : Prop *) (** Both provable and unprovable claims are perfectly good propositions. Simply _being_ a proposition is one thing; being _provable_ is something else! *) Check (2 + 2 = 4). Check (2 + 2 = 5). (* 2 + 2 = 5%nat : Prop *) (** Both [2 + 2 = 4] and [2 + 2 = 5] are legal expressions of type [Prop]. *) (** We've seen one way that propositions can be used in Coq: in [Theorem] (and [Lemma] and [Example]) declarations. *) Theorem plus_2_2_is_4 : 2 + 2 = 4. Proof. reflexivity. Qed. (** But they can be used in many other ways. For example, we can give a name to a proposition using a [Definition], just as we have given names to expressions of other sorts (numbers, functions, types, type functions, ...). *) Definition plus_fact : Prop := 2 + 2 = 4. Check plus_fact. (** Now we can use this name in any situation where a proposition is expected -- for example, as the subject of a theorem. *) Theorem plus_fact_is_true : plus_fact. Proof. reflexivity. Qed. (** So far, all the propositions we have seen are equality propositions. But can also form new propositions out of old ones. For example, given propositions [P] and [Q], we can form the proposition "[P] implies [Q]." *) Definition strange_prop1 := (2 + 2 = 5) -> (99 + 26 = 42). (** Also, given a proposition [P] with a free variable [n], we can form the proposition [forall n, P]. *) Definition strange_prop2 := forall n, (ble_nat n 17 = true) -> (ble_nat n 99 = true). (** We can also define _parameterized propositions_. For example, what does it mean to claim that "a number n is even"? We have written a function that tests evenness, so one possible definition of what it means to be even is "[n] is even iff [evenb n = true]." *) Definition even (n:nat) := evenb n = true. (** This defines [even] as a _function_ that, when applied to a number [n], yields a proposition asserting that [n] is even. *) Check even. Check (even 4). Check (even 3). (** The type of [even], [nat->Prop], can be pronounced in three ways: (1) "[even] is a _function_ from numbers to propositions," (2) "[even] is a _family_ of propositions, indexed by a number [n]," or (3) "[even] is a _property_ of numbers." *) (** Propositions -- including parameterized propositions -- are first-class citizens in Coq. We can use them in other definitions: *) Definition even_n__even_SSn (n:nat) := (even n) -> (even (S (S n))). (** We can define them to take multiple arguments... *) Definition between (n m o: nat) : Prop := andb (ble_nat n o) (ble_nat o m) = true. (** ... and then partially apply them: *) Definition teen : nat->Prop := between 13 19. (** We can even pass propositions -- including parameterized propositions -- as arguments to functions: *) Definition true_for_zero (P:nat->Prop) : Prop := P 0. Definition true_for_n__true_for_Sn (P:nat->Prop) (n:nat) : Prop := P n -> P (S n). (** (Names like [x__y] can be read "[x] implies [y].") *) Definition preserved_by_S (P:nat->Prop) : Prop := forall n', P n' -> P (S n'). Definition true_for_all_numbers (P:nat->Prop) : Prop := forall n, P n. Definition our_nat_induction (P:nat->Prop) : Prop := (true_for_zero P) -> (preserved_by_S P) -> (true_for_all_numbers P). (* ##################################################### *) (** * Evidence *) (** We've seen that well-formed expressions of type [Prop] can embody both provable and unprovable propositions. Naturally, though, we're particularly interested in the provable ones! When [P] is a proposition and [e] is a proof of [P] -- i.e., [e] is evidence that [P] is true -- we'll write "[e : P]." This overloading of the "has type" or "inhabits" notation is not accidental: indeed, we'll see that there is a fundamental and fruitful analogy between data values inhabiting types and evidence "inhabiting" propositions. *) (** The next question is "what are proofs?" -- i.e., what sorts of things we would be willing to accept as evidence that particular propositions are true. *) (* ##################################################### *) (** ** Inductively Defined Propositions *) (** One way to answer this question is to define a new proposition _together with_ a statement of what counts as evidence for its truth. For example, recalling the type [day] from Basics.v, we might introduce a parameterized proposition [good_day : day -> Prop] like this: *) Inductive good_day : day -> Prop := | gd_sat : good_day saturday | gd_sun : good_day sunday. (** That is, we're _defining_ what we mean by days being good by saying "Saturday is good, sunday is good, and that's all." Then someone can _prove_ that Sunday is good simply by remarking that this is what we said when we defined what [good_day] meant. *) Theorem gds : good_day sunday. Proof. apply gd_sun. Qed. (** In other words, the constructor [gd_sun] is "primitive evidence" -- an _axiom_ -- justifying the claim that Sunday is good. *) (** Similarly, we can define a proposition [day_before] parameterized by _two_ days, together with axioms stating that Monday comes before Tuesday, Tuesday before Wednesday, and so on. *) Inductive day_before : day -> day -> Prop := | db_tue : day_before tuesday monday | db_wed : day_before wednesday tuesday | db_thu : day_before thursday wednesday | db_fri : day_before friday thursday | db_sat : day_before saturday friday | db_sun : day_before sunday saturday | db_mon : day_before monday sunday. (** The axioms that we introduce along with an [Inductive]ly defined proposition can be richer than what we've seen above. For example, it is well known that _every_ day is a fine day for singing a song... *) Inductive fine_day_for_singing : day -> Prop := | fdfs_any : forall d:day, fine_day_for_singing d. (** In particular, Wednesday is a fine day for singing. *) Theorem fdfs_wed : fine_day_for_singing wednesday. Proof. apply fdfs_any. Qed. (** The first line here can be read "I'm about to show you some evidence for the proposition [fine_day_for_singing wednesday], and I want to introduce the name [fdfs_wed] to refer to that evidence later on." The second line then instructs Coq how to assemble the evidence. There's is another, more primitive way to accomplish the same thing: we can use a [Definition] whose left-hand side is the name we're introducing and whose right-hand side is the evidence _itself_, rather than a script for how to build it. *) Definition fdfs_wed' : fine_day_for_singing wednesday := fdfs_any wednesday. Check fdfs_wed. Check fdfs_wed'. (** The expression [fdfs_any wednesday] can be thought of as instantiating the parameterized axiom [fdfs_any] with the specific argument [wednesday]. Alternatively, we can think of [fdfs_any] as a primitive _function_ that takes a day and returns evidence that that day is a fine day for singing; its type, [forall d:day, fine_day_for_singing d], expresses this functionality, in the same way that the polymorphic type [forall X, list X] in the previous chapter expressed the fact that [nil] can be thought of as a function from types to empty lists with elements of that type. On this reading, [fdfs_any wednesday] is an ordinary application expression. *) (** Let's take another, slightly more interesting example. Let's say that a day of the week is "ok" if either (1) it is a good day or else (2) it falls the day before an ok day. *) Inductive ok_day : day -> Prop := | okd_gd : forall d, good_day d -> ok_day d | okd_before : forall d1 d2, ok_day d2 -> day_before d2 d1 -> ok_day d1. (** The first constructor can be read "To show that [d] is an OK day, it suffices to present evidence that [d] is good." The second can be read, "Another way to show that a day [d1] is ok is to present evidence that it is the day before some other day [d2] together with evidence that [d2] is ok." *) (** Now suppose that we want to prove that [wednesday] is an ok day. There are two ways to do it. First, we have the primitive way, where we simply write down an expression that has the right type -- a big nested application of constructors: *) Definition okdw : ok_day wednesday := okd_before wednesday thursday (okd_before thursday friday (okd_before friday saturday (okd_gd saturday gd_sat) db_sat) db_fri) db_thu. (** Second, we have the machine-assisted way, where we go into [Proof] mode and Coq guides us through a series of goals and subgoals until it is finally satisfied: *) Theorem okdw' : ok_day wednesday. Proof. (* wednesday is OK because it's the day before an OK day *) apply okd_before with (d2:=thursday). (* "subgoal: show that thursday is ok". *) (* thursday is OK because it's the day before an OK day *) apply okd_before with (d2:=friday). (* "subgoal: show that friday is ok". *) (* friday is OK because it's the day before an OK day *) apply okd_before with (d2:=saturday). (* "subgoal: show that saturday is ok". *) (* saturday is OK because it's good! *) apply okd_gd. apply gd_sat. (* "subgoal: show that the day before saturday is friday". *) apply db_sat. (* "subgoal: show that the day before friday is thursday". *) apply db_fri. (* "subgoal: show that the day before thursday is wednesday". *) apply db_thu. Qed. (** Fundamentally, though, these two ways of making proofs are the same, in the sense that what Coq is actually doing when it's following the commands in a [Proof] script is _literally_ attempting to construct an expression with the desired type. *) Print okdw'. (** These expressions are often called _proof objects_. *) (** Proof objects are the bedrock of reasoning in Coq. Tactic proofs are convenient shorthands that instruct Coq how to construct proof objects for us instead of our writing them out explicitly. Here, of course, the proof object is actually shorter than the tactic proof. But the proof objects for more interesting proofs can become quite large and complex -- building them by hand would be painful. Moreover, we'll see later on in the course that Coq has a number of automation tactics that can construct quite complex proof objects without our needing to specify every step. *) (* ##################################################### *) (** ** The Curry-Howard Correspondence *) (** The analogy << propositions ~ sets / types proofs ~ data values >> is called the _Curry-Howard correspondence_ (or _Curry-Howard isomorphism_). We will see that a great many things follow from it. *) (** Just as a set can be empty, a singleton, finite, or infinite -- it can have zero, one, or many inhabitants -- a proposition may be inhabited by zero, one, many, or infinitely many proofs. Each inhabitant of a proposition [P] is a different way of giving evidence for [P]. If there are none, then [P] is not provable. If there are many, then [P] has many different proofs. *) (* ##################################################### *) (** ** Implication *) (** We've seen that the [->] operator (implication) can be used to build bigger propositions from smaller ones. We need to think about what would constitute evidence for propositions built in this way. Consider this one: *) Definition okd_before2 := forall d1 d2 d3, ok_day d3 -> day_before d2 d1 -> day_before d3 d2 -> ok_day d1. (** In English: if we have three days, [d1] which is before [d2] which is before [d3], and if we know [d3] is ok, then so is [d1]. It should be easy to imagine how we'd construct a tactic proof of [okd_before2]... *) (** **** Exercise: 1 star, optional *) Theorem okd_before2_valid : okd_before2. Proof. (* FILL IN HERE *) Admitted. (** But what should the corresponding proof object look like? Answer: We've made a notational pun between [->] as implication and [->] as the type of functions. If we take this pun seriously, and if we're looking for an expression with _type_ [forall d1 d2 d3, ok_day d3 -> day_before d2 d1 -> day_before d3 d2 -> ok_day d1], then what we want is a _function_ that takes six arguments (three days and three pieces of evidence) and returns a piece of evidence! Here it is: *) Definition okd_before2_valid' : okd_before2 := fun (d1 d2 d3 : day) => fun (H : ok_day d3) => fun (H0 : day_before d2 d1) => fun (H1 : day_before d3 d2) => okd_before d1 d2 (okd_before d2 d3 H H1) H0. (* ##################################################### *) (** ** Induction Principles for Inductively Defined Types *) (** Every time we declare a new [Inductive] datatype, Coq automatically generates and proves a theorem that embodies the _induction principle_ for this type. The induction principle for a type [t] is called [t_ind]. Here is the one for natural numbers: *) Check nat_ind. (* ===> nat_ind : forall P : nat -> Prop, P 0 -> (forall n : nat, P n -> P (S n)) -> forall n : nat, P n *) (** Note that this is exactly the [our_nat_induction] property from above. *) (** The [induction] tactic is a straightforward wrapper that, at its core, simply performs [apply t_ind]. To see this more clearly, let's experiment a little with using [apply nat_ind] directly, instead of the [induction] tactic, to carry out some proofs. Here, for example, is an alternate proof of a theorem that we saw in the [Basics] chapter. *) Theorem mult_0_r' : forall n:nat, n * 0 = 0. Proof. apply nat_ind. Case "O". reflexivity. Case "S". simpl. intros n IHn. rewrite -> IHn. reflexivity. Qed. (** Several details in this proof are worth noting. First, in the induction step of the proof (the ["S"] case), we have to do a little bookkeeping manually (the [intros]) that [induction] does automatically. Second, we do not introduce [n] into the context before applying [nat_ind] -- the conclusion of [nat_ind] is a quantified formula, and [apply] needs this conclusion to exactly match the shape of the goal state, including the quantifier. The [induction] tactic works either with a variable in the context or a quantified variable in the goal. Third, the [apply] tactic automatically chooses variable names for us (in the second subgoal, here), whereas [induction] lets us specify (with the [as...] clause) what names should be used. The automatic choice is actually a little unfortunate, since it re-uses the name [n] for a variable that is different from the [n] in the original theorem. This is why the [Case] annotation is just [S] -- if we tried to write it out in the more explicit form that we've been using for most proofs, we'd have to write [n = S n], which doesn't make a lot of sense! All of these conveniences make [induction] nicer to use in practice than applying induction principles like [nat_ind] directly. But it is important to realize that, modulo this little bit of bookkeeping, applying [nat_ind] is what we are really doing. *) (** **** Exercise: 2 stars *) (** Complete this proof without using the [induction] tactic. *) Theorem plus_one_r' : forall n:nat, n + 1 = S n. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** The induction principles that Coq generates for other datatypes defined with [Inductive] follow a similar pattern. If we define a type [t] with constructors [c1] ... [cn], Coq generates and proves a theorem with this shape: [[ t_ind : forall P : t -> Prop, ... case for c1 ... -> ... case for c2 ... -> ... -> ... case for cn ... -> forall n : t, P n ]] The specific shape of each case depends on the arguments to the corresponding constructor. Before trying to write down a general rule, let's look at some more examples. First, an example where the constructors take no arguments: *) Inductive yesno : Type := | yes : yesno | no : yesno. Check yesno_ind. (* ===> yesno_ind : forall P : yesno -> Prop, P yes -> P no -> forall y : yesno, P y *) (** **** Exercise: 1 star *) (** Write out the induction principle that Coq will generate for the following datatype. Write down your answer on paper, and then compare it with what Coq prints. *) Inductive rgb : Type := | red : rgb | green : rgb | blue : rgb. (* Check rgb_ind. *) (** [] *) (** Here's another example, this time with one of the constructors taking some arguments. *) Inductive natlist : Type := | nnil : natlist | ncons : nat -> natlist -> natlist. Check natlist_ind. (* ===> (modulo a little tidying) natlist_ind : forall P : natlist -> Prop, P nnil -> (forall (n : nat) (l : natlist), P l -> P (ncons n l)) -> forall n : natlist, P n *) (** **** Exercise: 1 star *) (** Suppose we had written the above definition a little differently: *) Inductive natlist1 : Type := | nnil1 : natlist1 | nsnoc1 : natlist1 -> nat -> natlist1. (** Now what will the induction principle look like? *) (** [] *) (** From these examples, we can extract this general rule: - each constructor [c] takes argument types [a1]...[an] - each [ai] can be either [t] (the datatype we are defining) or some other type [s] - the corresponding case of the induction principle says (in English): - "for all values [x1]...[xn] of types [a1]...[an], if [P] holds for each of the [x]s of type [t], then [P] holds for [c x1 ... xn]". *) (** **** Exercise: 1 star (ExSet) *) (** Here is an induction principle for an inductively defined set. [[ ExSet_ind : forall P : ExSet -> Prop, (forall b : bool, P (con1 b)) -> (forall (n : nat) (e : ExSet), P e -> P (con2 n e)) -> forall e : ExSet, P e ]] Give an [Inductive] definition of [ExSet]: *) (* Inductive ExSet : Type := (* FILL IN HERE *) *) (** [] *) (** Next, what about polymorphic datatypes? The inductive definition of polymorphic lists [[ Inductive list (X:Type) : Type := | nil : list X | cons : X -> list X -> list X. ]] is very similar. The main difference is that, here, the whole definition is _parameterized_ on a set [X] -- that is, we are defining a _family_ of inductive types [list X], one for each [X]. Note that, wherever [list] appears in the body of the declaration, it is always applied to the parameter [X]. The induction principle is likewise parameterized on [X]: [[ list_ind : forall (X : Type) (P : list X -> Prop), P [] -> (forall (x : X) (l : list X), P l -> P (x :: l)) -> forall l : list X, P l ]] Note the wording here (and, accordingly, the form of [list_ind]): The _whole_ induction principle is parameterized on [X]. That is, [list_ind] can be thought of as a polymorphic function that, when applied to a type [X], gives us back an induction principle specialized to the type [list X]. *) (** **** Exercise: 1 star *) (** Write out the induction principle that Coq will generate for the following datatype. Compare your answer with what Coq prints. *) Inductive tree (X:Type) : Type := | leaf : X -> tree X | node : tree X -> tree X -> tree X. (* Check tree_ind. *) (** [] *) (** **** Exercise: 1 star (mytype) *) (** Find an inductive definition that gives rise to the following induction principle: [[ mytype_ind : forall (X : Type) (P : mytype X -> Prop), (forall x : X, P (constr1 X x)) -> (forall n : nat, P (constr2 X n)) -> (forall m : mytype X, P m -> forall n : nat, P (constr3 X m n)) -> forall m : mytype X, P m ]] *) (** [] *) (** **** Exercise: 1 star, optional (foo) *) (** Find an inductive definition that gives rise to the following induction principle: [[ foo_ind : forall (X Y : Type) (P : foo X Y -> Prop), (forall x : X, P (bar X Y x)) -> (forall y : Y, P (baz X Y y)) -> (forall f1 : nat -> foo X Y, (forall n : nat, P (f1 n)) -> P (quux X Y f1)) -> forall f2 : foo X Y, P f2 ]] *) (** [] *) (** **** Exercise: 1 star, optional *) (** Consider the following inductive definition: *) Inductive foo' (X:Type) : Type := | C1 : list X -> foo' X -> foo' X | C2 : foo' X. (** What induction principle will Coq generate for foo'? Fill in the blanks, then check your answer with Coq.) [[ foo'_ind : forall (X : Type) (P : foo' X -> Prop), (forall (l : list X) (f : foo' X), _______________________ -> _______________________ ) -> ___________________________________________ -> forall f : foo' X, ________________________ ]] *) (** [] *) (* ##################################################### *) (** ** Induction Hypotheses *) (** Where does the phrase "induction hypothesis" fit into this picture? The induction principle for numbers [[ forall P : nat -> Prop, P 0 -> (forall n : nat, P n -> P (S n)) -> forall n : nat, P n ]] is a generic statement that holds for all propositions [P] (strictly speaking, for all families of propositions [P] indexed by a number [n]). Each time we use this principle, we are choosing [P] to be a particular expression of type [nat->Prop]. We can make the proof more explicit by giving this expression a name. For example, instead of stating the theorem [mult_0_r] as "[forall n, n * 0 = 0]," we can write it as "[forall n, P_m0r n]", where [P_m0r] is defined as... *) Definition P_m0r (n:nat) : Prop := n * 0 = 0. (** ... or equivalently... *) Definition P_m0r' : nat->Prop := fun n => n * 0 = 0. (** Now when we do the proof it is easier to see where [P_m0r] appears. *) Theorem mult_0_r'' : forall n:nat, P_m0r n. Proof. apply nat_ind. Case "n = O". reflexivity. Case "n = S n'". (* Note the proof state at this point! *) unfold P_m0r. simpl. intros n' IHn'. apply IHn'. Qed. (** This extra naming step isn't something that we'll do in normal proofs, but it is useful to do it explicitly for an example or two, because it allows us to see exactly what the induction hypothesis is. If we prove [forall n, P_m0r n] by induction on [n] (using either [induction] or [apply nat_ind]), we see that the first subgoal requires us to prove [P_m0r 0] ("[P] holds for zero"), while the second subgoal requires us to prove [forall n', P_m0r n' -> P_m0r n' (S n')] (that is "[P] holds of [S n'] if it holds of [n']" or, more elegantly, "[P] is preserved by [S]"). The _induction hypothesis_ is the premise of this latter implication -- the assumption that [P] holds of [n'], which we are allowed to use in proving that [P] holds for [S n']. *) (* ##################################################### *) (** ** A Closer Look at the [induction] Tactic *) (** (This section can be skimmed.) The [induction] tactic actually does even more low-level bookkeeping for us than we discussed above. Recall the informal statement of the induction principle for 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, if [P n'] holds, then so does [P (S n')] - conclude that [P n] holds for all n. So, when we begin a proof with [intros n] and then [induction n], we are first telling Coq to consider a _particular_ [n] (by introducing it into the context) and then telling it to prove something about _all_ numbers (by using induction). What Coq actually does in this situation, internally, is to "re-generalize" the variable we perform induction on. For example, in the proof above that [plus] is associative... *) Theorem plus_assoc' : forall n m p : nat, n + (m + p) = (n + m) + p. Proof. (* ...we first introduce all 3 variables into the context, which amounts to saying "Consider an arbitrary [n], [m], and [p]..." *) intros n m p. (* ...We now use the [induction] tactic to prove [P n] (that is, [n + (m + p) = (n + m) + p]) for _all_ [n], and hence also for the particular [n] that is in the context at the moment. *) induction n as [| n']. Case "n = O". reflexivity. Case "n = S n'". (* In the second subgoal generated by [induction] -- the "inductive step" -- we must prove that [P n'] implies [P (S n')] for all [n']. The [induction] tactic automatically introduces [n'] and [P n'] into the context for us, leaving just [P (S n')] as the goal. *) simpl. rewrite -> IHn'. reflexivity. Qed. (** It also works to apply [induction] to a variable that is quantified in the goal. *) Theorem plus_comm' : forall n m : nat, n + m = m + n. Proof. induction n as [| n']. Case "n = O". intros m. rewrite -> plus_0_r. reflexivity. Case "n = S n'". intros m. simpl. rewrite -> IHn'. rewrite <- plus_n_Sm. reflexivity. Qed. (** Note that [induction n] leaves [m] still bound in the goal -- i.e., what we are proving inductively is a statement beginning with [forall m]. If we do [induction] on a variable that is quantified in the goal _after_ some other quantifiers, the [induction] tactic will automatically introduce the variables bound by these quantifiers into the context. *) Theorem plus_comm'' : forall n m : nat, n + m = m + n. Proof. (* Let's do induction on [m] this time, instead of [n]... *) induction m as [| m']. Case "m = O". simpl. rewrite -> plus_0_r. reflexivity. Case "m = S m'". simpl. rewrite <- IHm'. rewrite <- plus_n_Sm. reflexivity. Qed. (** **** Exercise: 1 star, optional (plus_explicit_prop) *) (** Rewrite both [plus_assoc'] and [plus_comm'] and their proofs in the same style as [mult_0_r''] above -- that is, for each theorem, give an explicit [Definition] of the proposition being proved by induction, and state the theorem and proof in terms of this defined proposition. *) (* FILL IN HERE *) (** [] *) (* ##################################################### *) (** One more quick digression, for adventurous souls: if we can define parameterized propositions using [Definition], then can we also define them using [Fixpoint]? Of course we can! However, this kind of "recursive parameterization" doesn't correspond to anything very familiar from everyday mathematics. The following exercise gives a slightly contrived example. *) (** **** Exercise: 4 stars, optional *) (** Define a recursive function [true_upto_n__true_everywhere] that makes [true_upto_n_example] work. *) (** << Fixpoint true_upto_n__true_everywhere (* FILL IN HERE *) Example true_upto_n_example : (true_upto_n__true_everywhere 3 (fun n => even n)) = (even 3 -> even 2 -> even 1 -> forall m : nat, even m). Proof. reflexivity. Qed. >> *) (** [] *) (* ####################################################### *) (** ** Reasoning by Induction Over Evidence *) (** Some of the examples in the opening discussion of propositions involved the concept of _evenness_. We began with a computation [evenb n] that _checks_ evenness, yielding a boolean. From this, we built a proposition [even n] (defined in terms of [evenb]) that _asserts_ that [n] is even. That is, we defined "[n] is even" to mean "[evenb] returns [true] when applied to [n]." An alternative is to define the concept of evenness directly. This allows us, instead of going "indirectly" via the [evenb] function, to say straight out what we would be willing to accept as evidence that a given number is even. *) Inductive ev : nat -> Prop := | ev_0 : ev O | ev_SS : forall n:nat, ev n -> ev (S (S n)). (** Informally, this definition says that there are two ways to give evidence that a number [m] is even: either observe that it is zero, or else observe that [n = S (S m)] for some [m] and give further evidence that [m] itself is even. The constructors [ev_0] and [ev_SS] are names for these different ways of giving evidence for evenness. *) (** The [Inductive] keyword means exactly the same thing whether we are using it to define sets of data values (in the [Type] world) or sets of evidence (in the [Prop] world). Consider the parts of the inductive definition of [ev] above: - The first line declares that [ev] is a proposition indexed by a natural number. - The second line declares that the constructor [ev_0] can be taken as evidence for the assertion [ev O]. - The third line declares that, if [n] is a number and [E] is evidence for the assertion [ev n]), then [ev_SS n E] can be taken as evidence for [ev (S (S n))]. That is, we can construct evidence that [S (S n)] is even by applying the constructor [ev_SS] to evidence that [n] is even. *) (** **** Exercise: 1 star, optional *) (** Give a tactic proof and a proof object showing that four is even. *) (* FILL IN HERE *) (** [] *) (** **** Exercise: 2 stars *) (** Give a tactic proof and a proof object showing that, if [n] is even, then so is [4+n]. *) (* FILL IN HERE *) (** [] *) (** **** Exercise: 2 stars *) (** Construct a tactic proof of the following proposition. *) Theorem double_even : forall n, ev (double n). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 4 stars, optional (double_even_pfobj) *) (** Try to predict what proof object is constructed by the above tactic proof. (Before checking your answer, you'll want to strip out any uses of [Case], as these will make the proof object a bit cluttered.) *) (** [] *) (** Besides _constructing_ evidence of evenness, we can also _reason about_ evidence of evenness. The fact that we introduced [ev] with an [Inductive] declaration tells us not only that the constructors [ev_O] and [ev_SS] are ways to build evidence of evenness, but also that these two constructors are the _only_ ways that evidence of evenness can be built. In other words, if someone gives us evidence [E] justifying the assertion [ev n], then we know that [E] can only have one of two forms: either [E] is [ev_0] (and [n] is [O]), or [E] is [ev_SS n' E'] (and [n] is [S (S n')]) and [E'] is evidence that [n'] is even. Thus, it makes sense to use the tactics that we have already seen for inductively defined _data_ to reason instead about inductively defined _evidence_. For example, here we use a [destruct] on evidence that [n] is even in order to show that [ev n] implies [ev (n-2)]. *) Theorem ev_minus2: forall n, ev n -> ev (pred (pred n)). Proof. intros n E. destruct E as [| n' E']. Case "E = ev_0". simpl. apply ev_0. Case "E = ev_SS n' E'". simpl. apply E'. Qed. (** **** Exercise: 1 star (ev_minus2_n) *) (** What happens if we try to [destruct] on [n] instead of [E]? *) (** [] *) (** We can also perform _induction_ on evidence that [n] is even. Here we use it show that the old [evenb] function returns [true] on [n] when [n] is even according to [ev]. *) Theorem ev_even : forall n, ev n -> even n. Proof. intros n E. induction E as [| n' E']. Case "E = ev_0". unfold even. reflexivity. Case "E = ev_SS n' E'". unfold even. simpl. unfold even in IHE'. apply IHE'. Qed. (** (Of course, we'd expect that [even n -> ev n] also holds. We'll see that it does in the next chapter.) *) (** **** Exercise: 1 star (ev_even_n) *) (** Could this proof be carried out by induction on [n] instead of [E]? *) (** [] *) (** The induction principle for inductively defined propositions does not follow quite the same form as that of inductively defined sets. For now, take the intuitive view that induction on evidence [ev n] is similar to induction on [n], but restricts our attention to only those numbers for which evidence [ev n] could be generated. We will look at the induction principle of [ev] in more depth below, to explain what's really going on. *) (** **** Exercise: 1 star (l_fails) *) (** The following proof attempt will not succeed. [[ Theorem l : forall n, ev n. Proof. intros n. induction n. Case "O". simpl. apply ev_0. Case "S". ... ]] Briefly explain why. (* FILL IN HERE *) *) (** [] *) (** **** Exercise: 2 stars *) (** Here's another exercise requiring induction. *) Theorem ev_sum : forall n m, ev n -> ev m -> ev (n+m). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** Here's another situation where we want to analyze evidence for evenness and build two sub-goals. *) Theorem SSev_ev_firsttry : forall n, ev (S (S n)) -> ev n. Proof. intros n E. destruct E as [| n' E']. (* Oops. [destruct] destroyed far too much! In the first sub-goal, we don't know that [n] is [0]. *) Admitted. (** The right thing to use here is [inversion] (!) *) Theorem SSev_even : forall n, ev (S (S n)) -> ev n. Proof. intros n E. inversion E as [| n' E']. apply E'. Qed. (* Print SSev_even. *) (** This use of [inversion] may seem a bit mysterious at first. Until now, we've only used [inversion] on equality propositions, to utilize injectivity of constructors or to discriminate between different constructors. But we see here that [inversion] can also be applied to analyzing evidence for inductively defined propositions. Here's how [inversion] works in general. Suppose the name [I] refers to an assumption [P] in the current context, where [P] has been defined by an [Inductive] declaration. Then, for each of the constructors of [P], [inversion I] generates a subgoal in which [I] has been replaced by the exact, specific conditions under which this constructor could have been used to prove [P]. Some of these subgoals will be self-contradictory; [inversion] throws these away. The ones that are left represent the cases that must be proved to establish the original goal. In this particular case, the [inversion] analyzed the construction [ev (S (S n))], determined that this could only have been constructed using [ev_SS], and generated a new subgoal with the arguments of that constructor as new hypotheses. (It also produced an auxiliary equality, which happens to be useless.) We'll begin exploring this more general behavior of inversion in what follows. *) (** **** Exercise: 1 star (inversion_practice) *) Theorem SSSSev_even : forall n, ev (S (S (S (S n)))) -> ev n. Proof. (* FILL IN HERE *) Admitted. (** [inversion] can also be used to derive goals by showing absurdity of a hypothesis. *) Theorem even5_nonsense : ev 5 -> 2 + 2 = 9. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** We can generally use [inversion] instead of [destruct] on inductive propositions. This illustrates that in general, we get one case for each possible constructor. Again, we also get some auxiliary equalities that are rewritten in the goal but not in the other hypotheses. *) Theorem ev_minus2': forall n, ev n -> ev (pred (pred n)). Proof. intros n E. inversion E as [| n' E']. Case "E = ev_0". simpl. apply ev_0. Case "E = ev_SS n' E'". simpl. apply E'. Qed. (** **** Exercise: 3 stars *) (** Finding the appropriate thing to do induction on is a bit tricky here. *) Theorem ev_ev_even : forall n m, ev (n+m) -> ev n -> ev m. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars, optional *) (** Here's an exercise that just requires applying existing lemmas. No induction or even case analysis is needed, but some of the rewriting may be tedious. You'll want the [replace] tactic used for [plus_swap'] in Basics.v *) Theorem ev_plus_plus : forall n m p, ev (n+m) -> ev (n+p) -> ev (m+p). Proof. (* FILL IN HERE *) Admitted. (** [] *) (* ##################################################### *) (** ** Why Define Propositions Inductively? *) (** We have seen that the proposition "some number is even" can be phrased in two different ways -- indirectly, via a testing function [evenb], or directly, by inductively describing what constitutes evidence for evenness. These two ways of defining evenness are about equally easy to state and work with. Which we choose is basically a question of taste. However, for many other properties of interest, the direct inductive definition is preferable, since writing a testing function may be awkward or even impossible. For example, consider the property [MyProp] defined as follows: - the number [4] has property [MyProp] - if [n] has property [MyProp], then so does [4+n] - if [2+n] has property [MyProp], then so does [n] - no other numbers have property [MyProp] This is a perfectly sensible definition of a set of numbers, but we cannot translate this definition directly as a Coq Fixpoint (or translate it directly into a recursive function in any other programming language). We might be able to find a clever way of testing this property using a [Fixpoint] (indeed, it is not even too hard to find one in this case), but in general this could require arbitrarily much thinking. In fact, if the property we are interested in is uncomputable, then we cannot define it as a [Fixpoint] no matter how hard we try, because Coq requires that all [Fixpoint]s correspond to terminating computations. On the other hand, writing an inductive definition of what it means to give evidence for the property [MyProp] is straightforward: *) Inductive MyProp : nat -> Prop := | MyProp1 : MyProp 4 | MyProp2 : forall n:nat, MyProp n -> MyProp (4 + n) | MyProp3 : forall n:nat, MyProp (2 + n) -> MyProp n. (** The first three clauses in the informal definition of MyProp above are reflected in the first three clauses of the inductive definition. The fourth clause is the precise force of the keyword [Inductive]. *) (** As we did with evenness, we can now construct evidence that certain numbers satisfy [MyProp]. *) Theorem MyProp_ten : MyProp 10. Proof. apply MyProp3. simpl. assert (12 = 4 + 8) as H12. Case "Proof of assertion". reflexivity. rewrite -> H12. apply MyProp2. assert (8 = 4 + 4) as H8. Case "Proof of assertion". reflexivity. rewrite -> H8. apply MyProp2. apply MyProp1. Qed. (** **** Exercise: 2 stars (MyProp) *) (** Here are two useful facts about MyProp. The proofs are left to you. *) Theorem MyProp_0 : MyProp 0. Proof. (* FILL IN HERE *) Admitted. Theorem MyProp_plustwo : forall n:nat, MyProp n -> MyProp (S (S n)). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** With these, we can show that MyProp holds of all even numbers, and vice versa. *) Theorem MyProp_ev : forall n:nat, ev n -> MyProp n. Proof. intros n E. induction E as [| n' E']. Case "E = ev_0". apply MyProp_0. Case "E = ev_SS n' E'". apply MyProp_plustwo. apply IHE'. Qed. (** Here's an informal proof of this theorem: Theorem : For any nat [n], if [ev n] then [MyProp n]. Proof: Let a nat [n] and a derivation [E] of [ev n] be given. We go by induction on [E]. There are two case: - If the last step in [E] is a use of [ev_0], then [n] is [0]. But we know by lemma [MyProp_0] that [MyProp 0], so the theorem is satisfied in this case. - If the last step in [E] is by [ev_SS], then [n = S (S n')] for some [n'] and there is a derivation of [ev n']. By lemma [MyProp_plustwo], it's enough to show [MyProp n']. This follows directly from the inductive hypothesis for the derivation of [ev n']. [] *) (** **** Exercise: 3 stars *) Theorem ev_MyProp : forall n:nat, MyProp n -> ev n. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars, optional (ev_MyProp_informal) *) (** Write an informal proof corresponding to your formal proof of [ev_MyProp]: Theorem: For any nat [n], if [MyProp n] then [ev n]. Proof: (* FILL IN HERE *) [] *) (** **** Exercise: 4 stars, optional (MyProp_pfobj) *) (** Prove [MyProp_ev] and [ev_MyProp] again by constructing explicit proof objects by hand (as we did above in [ev_plus4], for example). *) (* FILL IN HERE *) (** [] *) (** **** Exercise: 3 stars, optional *) (** Complete this proof using [apply MyProp_ind] instead of the [induction] tactic. *) Theorem ev_MyProp' : forall n:nat, MyProp n -> ev n. Proof. (* FILL IN HERE *) Admitted. (** [] *) (* ##################################################### *) (** ** Induction Principles in [Prop] *) (** Earlier, we looked in detail at the induction principles that Coq generates for inductively defined _sets_. Inductively defined _propositions_ are handled slightly differently. For example, from what we've said so far, you might expect the inductive definition of [ev] [[ Inductive ev : nat -> Prop := | ev_0 : ev O | ev_SS : forall n:nat, ev n -> ev (S (S n)). ]] to give rise to an induction principle that looks like this... [[ ev_ind_max : forall P : (forall n : nat, ev n -> Prop), P O ev_0 -> (forall (n : nat) (e : ev n), P n e -> P (S (S n)) (ev_SS n e)) -> forall (n : nat) (e : ev n), P n e ]] ... because: - Since [ev] is indexed by a number [n] (every [ev] object [e] is a piece of evidence that some particular number [n] is even), the proposition [P] is parameterized by both [n] and [e] -- that is, the induction principle can be used to prove assertions involving both an even number and the evidence that it is even. - Since there are two ways of giving evidence of evenness ([ev] has two constructors), applying the induction principle generates two subgoals: - We must prove that [P] holds for [O] and [ev_0]. - We must prove that, whenever [n] is an even number and [e] is evidence of its evenness, if [P] holds of [n] and [e], then it also holds of [S (S n)] and [ev_SS n e]. - If these subgoals can be proved, then the induction principle tells us that [P] is true for _all_ even numbers [n] and evidence [e] of their evenness. But this is a little more flexibility than we actually need or want: it is giving us a way to prove logical assertions where the assertion involves properties of some piece of _evidence_ of evenness, while all we really care about is proving properties of _numbers_ that are even -- we are interested in assertions about numbers, not about evidence. It would therefore be more convenient to have an induction principle for proving propositions [P] that are parameterized just by [n] and whose conclusion establishes [P] for all even numbers [n]: [[ forall P : nat -> Prop, ... -> forall n : nat, ev n -> P n ]] For this reason, Coq actually generates the following simplified induction principle for [ev]: [[ ev_ind : forall P : nat -> Prop, P O -> (forall n : nat, ev n -> P n -> P (S (S n))) -> forall n : nat, ev n -> P n ]] In particular, Coq has dropped the evidence term [e] as a parameter of the the proposition [P], and consequently has rewritten the assumption [forall (n:nat) (e:ev n), ...] to be [forall (n:nat), ev n -> ...]; i.e., we no longer require explicit evidence of the provability of [ev n]. *) Module P. (** **** Exercise: 3 stars (p_provability) *) (** Consider the following inductively defined proposition: *) Inductive p : (tree nat) -> nat -> Prop := | c1 : forall n, p (leaf _ n) 1 | c2 : forall t1 t2 n1 n2, p t1 n1 -> p t2 n2 -> p (node _ t1 t2) (n1 + n2) | c3 : forall t n, p t n -> p t (S n). (** Describe, in English, the conditions under which the proposition [p t n] is provable. (* FILL IN HERE *) *) (** [] *) End P. (** **** Exercise: 3 stars, optional (prop_ind) *) (** Write out the induction principles that Coq will generate for the inductive declarations [list] and [MyProp]. Compare your answers against the results Coq prints for the following queries. *) (* Check list_ind. *) (* Check MyProp_ind. *) (** [] *) (* ##################################################### *) (** * The Big Picture: Coq's Two Universes *) (** Expressions in Coq live in two distinct universes: - [Type] is the universe of _computations_ and _data_. - [Prop] is the universe of _logical assertions_ and _evidence_. The two universes have some deep similarities -- in each, we can talk about values, inductive definitions, quantification, etc. -- but they play quite different roles in defining and reasoning about mathematical structures. *) (** ** Values *) (** Both universes start with an infinite set of _constructors_. Constructors have no internal structure: they are just atomic symbols. Examples: - [true], [false], [O], [S], [nil], [cons], [ev_0], [ev_SS], ... The simplest values are expressions consisting entirely of constructor applications. Examples: - [true] - [O] - [S (S (S O))] - [ev_0] - [ev_SS (S (S O)) ev_0] - [ev_SS (S (S (S (S O)))) (ev_SS (S (S O)) ev_0)] Such expressions can be thought of as trees. Their leaves are nullary constructors (applied to no arguments), and their internal nodes are applications of constructors to one or more values. In the universe [Type], we think of values as _data_. In [Prop], we think of values as _evidence_. Values in [Prop] are sometimes called _derivation trees_. *) (** ** Inductive Definitions *) (** [Inductive] declarations give names to subsets of the set of all values. For example, the declaration of the inductive type [nat] defines a _set_ whose _elements_ are values representing natural numbers. That is, it picks out a subset [nat] of the set of all values that satisfies the following properties: - the value [O] is in this set; - the set is _closed_ under applications of [S] (i.e., if a value [n] is in the set, then [S n] is too); - it is the smallest set satisfying these conditions (i.e., the only values in [nat] are the ones that _must_ be, according to the previous two conditions... there is no other "junk"). Inductively defined sets can themselves appear as arguments to constructors in compound values. Examples: - [nat] - [nil nat] - [cons nat O (cons nat (S O) (nil nat))] Also, we can write functions that take sets as arguments and return sets as results. For example, [list] is a function of type [Type->Type]: it takes a set [X] as argument and returns as result the set [list X] (whose members are lists with elements drawn from [X]). Similarly, the declaration of the inductive type [ev] defines a _family of propositions_ whose _elements_ are values representing evidence that numbers are even. That is, for each [n], the definition picks out a subset [ev n] of the set of all values satisfying the following properties: - the value [ev_0] is in the set [ev O] - the sets are _closed_ under well-typed applications of [ev_SS] -- i.e., if [e] is in the set [ev n], then [ev_SS n e] is in the set [ev (S (S n))] - it is the smallest family of sets satisfying these conditions (i.e., the only values in any set [ev n] are the ones that _must_ be, according to the previous two conditions... there is no other junk). *) (** ** Functions and Quantifiers *) (** The types [A->B] and [forall x:A, B] both describe functions from [A] to [B]. The only difference is that, in the second case, the expression [B] -- the type of the result -- can mention the argument [x] by name. For example: - The function [fun x:nat => x + x] has type [nat->nat] -- that is, it maps each number [n] to a number. - The function [fun X:Type => nil (list X)] has type [forall X:Type, list (list X)] -- that is, it maps each set [X] to a particular list of lists of [X]s. (Of course, [nil] is usually written as [[]] instead of [nil X].) In fact, the two ways of writing function types are really the same: In Coq, [A->B] is actually just an abbreviation for [forall x:A, B], where [x] is some variable name not occurring in [B]. For example, the type of [fun x:nat => x + x] can be written, if we like, as [forall x:nat, nat]. *) (** ** Functions and Implications *) (** In both [Type] and [Prop], we can write functions that transform values into other values. Also, functions themselves are values; this means we can - write higher-order functions that take functions as arguments or return functions as results, and - apply constructors to functions to build complex values containing functions. A function of type [P->Q] in [Prop] is something that takes evidence for [P] as an argument and yields evidence for [Q] as its result. Such a function can be regarded as _evidence_ that [P] implies [Q], since, whenever we have evidence that [P] is true, we can apply the function and get back evidence that [Q] is true: evidence for an implication is a function on evidence. This is why we use the same notation for functions and logical implications in Coq: they are exactly the same thing! This coincidence between functions and implications is an instance of a deep connection between programming languages and mathematical logic known as the _curry-howard correspondence_. *) (** ** Propositions and Booleans *) (** Propositions and booleans are superficially similar, but they are really quite different things! - Booleans are _values_ in the _computational_ world. Every expression of type [bool] (with no free variables) can be simplified to either [true] or [false]. - Propositions are _types_ in the _logical_ world. They are either _provable_ (i.e., there is some expression that has this type) or not (there is no such expression). It doesn't make sense to say that a proposition is "equivalent to [true]," and it is not necessarily the case that, if a proposition [P] is not provable, then [~P] is provable. We do sometimes use the words "true" and "false" informally when referring to propositions. Strictly speaking, this is wrong: a proposition is either provable or it is not. *) (* ##################################################### *) (** * Generalizing Induction Hypotheses *) (** There is one last point that needs to be discussed in order to turn you loose on this chapter's homework problems. In the previous chapter, we saw a proof that the [double] function is injective. The way we _start_ this proof is a little bit delicate: if we begin it with [[ intros n. induction n. ]] all is well. But if we begin it with [[ intros n m. induction n. ]] we get stuck in the middle of the inductive case... *) Theorem double_injective_FAILED : forall n m, double n = double m -> n = m. Proof. intros n m. induction n as [| n']. Case "n = O". simpl. intros eq. destruct m as [| m']. SCase "m = O". reflexivity. SCase "m = S m'". inversion eq. Case "n = S n'". intros eq. destruct m as [| m']. SCase "m = O". inversion eq. SCase "m = S m'". assert (n' = m') as H. SSCase "Proof of assertion". (* Here we are stuck. We need the assertion in order to rewrite the final goal (subgoal 2 at this point) to an identity. But the induction hypothesis, [IHn'], does not give us [n' = m'] -- there is an extra [S] in the way -- so the assertion is not provable. *) Admitted. (** What went wrong here? The problem is that, at the point we invoke the induction hypothesis, we have already introduced [m] into the context -- intuitively, we have told Coq, "Let's consider some particular [n] and [m]..." and we now have to prove that, if [double n = double m] for _this particular_ [n] and [m], then [n = m]. The next tactic, [induction n] says to Coq: We are going to show the goal by induction on [n]. That is, we are going to prove that the proposition - [P n] = "if [double n = double m], then [n = m]" holds for all [n] by showing - [P O] (i.e., "if [double O = double m] then [O = m]") - [P n -> P (S n)] (i.e., "if [double n = double m] then [n = m]" implies "if [double (S n) = double m] then [S n = m]"). If we look closely at the second statement, it is saying something rather strange: it says that, for a _particular_ [m], if we know - "if [double n = double m] then [n = m]" then we can prove - "if [double (S n) = double m] then [S n = m]". To see why this is strange, let's think of a particular [m] -- say, [5]. The statement is then saying that, if we can prove - [Q] = "if [double n = 10] then [n = 5]" then we can prove - [R] = "if [double (S n) = 10] then [S n = 5]". But knowing [Q] doesn't give us any help with proving [R]! (If we tried to prove [R] from [Q], we would say something like "Suppose [double (S n) = 10]..." but then we'd be stuck: knowing that [double (S n)] is [10] tells us nothing about whether [double n] is [10], so [Q] is useless at this point.) To summarize: Trying to carry out this proof by induction on [n] when [m] is already in the context doesn't work because we are trying to prove a relation involving _every_ [n] but just a _single_ [m]. *) (** The good proof of [double_injective] leaves [m] in the goal statement at the point where the [induction] tactic is invoked on [n]: *) Theorem double_injective' : forall n m, double n = double m -> n = m. Proof. intros n. induction n as [| n']. Case "n = O". simpl. intros m eq. destruct m as [| m']. SCase "m = O". reflexivity. SCase "m = S m'". inversion eq. Case "n = S n'". (* Notice that both the goal and the induction hypothesis have changed: the goal asks us to prove something more general (i.e., to prove the statement for *every* [m]), but the IH is correspondingly more flexible, allowing us to choose any [m] we like when we apply the IH. *) intros m eq. (* Now we choose a particular [m] and introduce the assumption that [double n = double m]. Since we are doing a case analysis on [n], we need a case analysis on [m] to keep the two "in sync". *) destruct m as [| m']. SCase "m = O". inversion eq. (* The 0 case is trivial *) SCase "m = S m'". (* At this point, since we are in the second branch of the [destruct m], the [m'] mentioned in the context at this point is actually the predecessor of the one we started out talking about. Since we are also in the [S] branch of the induction, this is perfect: if we instantiate the generic [m] in the IH with the [m'] that we are talking about right now (this instantiation is performed automatically by [apply]), then [IHn'] gives us exactly what we need to finish the proof. *) assert (n' = m') as H. SSCase "Proof of assertion". apply IHn'. inversion eq. reflexivity. rewrite -> H. reflexivity. Qed. (** So what we've learned is that we need to be careful about using induction to try to prove something too specific: If we're proving a property of [n] and [m] by induction on [n], we may need to leave [m] generic. However, this strategy doesn't always apply directly; sometimes a little rearrangement is needed. Suppose, for example, that we had decided we wanted to prove [double_injective] by induction on [m] instead of [n]. *) Theorem double_injective_take2_FAILED : forall n m, double n = double m -> n = m. Proof. intros n m. induction m as [| m']. Case "m = O". simpl. intros eq. destruct n as [| n']. SCase "n = O". reflexivity. SCase "n = S n'". inversion eq. Case "m = S m'". intros eq. destruct n as [| n']. SCase "n = O". inversion eq. SCase "n = S n'". assert (n' = m') as H. SSCase "Proof of assertion". (* Here we are stuck again, just like before. *) Admitted. (** The problem is that, to do induction on [m], we must first introduce [n]. (If we simply say [induction m] without introducing anything first, Coq will automatically introduce [n] for us!) What can we do about this? One possibility is to rewrite the statement of the lemma so that [m] is quantified before [n]. This will work, but it's not nice: We don't want to have to mangle the statements of lemmas to fit the needs of a particular strategy for proving them -- we want to state them in the most clear and natural way. What we can do instead is to first introduce all the quantified variables and then _re-generalize_ one or more of them, taking them out of the context and putting them back at the beginning of the goal. The [generalize dependent] tactic does this. *) Theorem double_injective_take2 : forall n m, double n = double m -> n = m. Proof. intros n m. (* [n] and [m] are both in the context *) generalize dependent n. (* Now [n] is back in the goal and we can do induction on [m] and get a sufficiently general IH. *) induction m as [| m']. Case "m = O". simpl. intros n eq. destruct n as [| n']. SCase "n = O". reflexivity. SCase "n = S n'". inversion eq. Case "m = S m'". intros n eq. destruct n as [| n']. SCase "n = O". inversion eq. SCase "n = S n'". assert (n' = m') as H. SSCase "Proof of assertion". apply IHm'. inversion eq. reflexivity. rewrite -> H. reflexivity. Qed. (** Let's look at an informal proof of this theorem. Note that the proposition we prove by induction leaves n quantified, corresponding to the use of generalize dependent in our formal proof. Theorem: For any nats [n] and [m], if [double n = double m], then [n = m]. Proof: Let a nat [m] be given. We prove that for any [n], if [double n = double m] then [n = m] by induction on [m]. - In the base case, we have [m = 0]. Let a nat [n] be given such that [double n = double m]. Since [m = 0], we have [double n = 0]. If [n = S n'] for some [n'], then [double n = S (S (double n'))] by the definition of [double]. This would be a contradiction of the assumption that [double n = 0], so [n = 0], and thus [n = m]. - In the inductive case, we have [m = S m'] for some nat [m']. Let a nat [n] be given such that [double n = double m]. By the definition of [double], we therefore have: [[ double n = S (S (double m')) ]] If [n = 0], then [double n = 0] (by the definition of [double]), which we have just seen is not the case. Thus, [n = S n'] for some [n'], and we have: [[ S (S (double n')) = S (S (double m')) ]] which implies that [double n' = double m']. But observe that our inductive hypothesis here is: for any [n], if [double n = double m'] then [n = m']. Applying this for [n'] then yields [n' = m'], and it follows directly that [S n' = S m']. Since [S n' = n] and [S m' = m], this is just what we wanted to show. [] *) (** **** Exercise: 3 stars (gen_dep_practice) *) (** Carry out this proof by induction on [m]. *) Theorem plus_n_n_injective_take2 : forall n m, n + n = m + m -> n = m. Proof. (* FILL IN HERE *) Admitted. (** Now prove this by induction on [l]. *) Theorem index_after_last : forall (n : nat) (X : Type) (l : list X), length l = n -> index (S n) l = None. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars, optional (index_after_last_informal) *) (** Write an informal proof corresponding to your Coq proof of [index_after_last]: Theorem: For any nat [n] and list [l], if [length l = n] then [index (S n) l = None]. Proof: (* FILL IN HERE *) [] *) (** **** Exercise: 3 stars, optional (gen_dep_practice_opt) *) (** Prove this by induction on [l]. *) Theorem length_snoc''' : forall (n : nat) (X : Type) (v : X) (l : list X), length l = n -> length (snoc l v) = S n. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars, optional *) (** Prove this by induction on [m]. *) Theorem eqnat_false_S : forall n m, beq_nat n m = false -> beq_nat (S n) (S m) = false. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars, optional *) (** Prove this by induction on [l1], without using [app_length]. *) Theorem app_length_cons : forall (X : Type) (l1 l2 : list X) (x : X) (n : nat), length (l1 ++ (x :: l2)) = n -> S (length (l1 ++ l2)) = n. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 4 stars, optional *) (** Prove this by induction on [l], without using app_length. *) Theorem app_length_twice : forall (X:Type) (n:nat) (l:list X), length l = n -> length (l ++ l) = n + n. Proof. (* FILL IN HERE *) Admitted. (** [] *) (* ####################################################### *) (** * Templates for Informal Proofs by Induction *) (** In the real world of mathematical communication, written proofs range from extremely longwinded and pedantic to extremely brief and telegraphic. The ideal is somewhere in between, of course, but while you are getting used to the style it is better to start out at the pedantic end. Also, during the learning phase, it is probably helpful to have a clear standard to compare against. So, for the rest of the course, please make _all_ your inductive proofs match one of the two following templates. *) (** ** Proof by Induction Over an Inductively Defined Set *) (** _Template_: - _Theorem_: _Proof_: By induction on [n]. - Suppose [n = c a1 ... ak], where <...and here we state the IH for each of the [a]'s that has type [S], if any>. We must show <...and here we restate [P(c a1 ... ak)]>. - [] _Example_: - _Theorem_: For all sets [X], lists [l : list X], and numbers [n], if [length l = n] then [index (S n) l = None]. _Proof_: By induction on [l]. - Suppose [l = []]. We must show, for all numbers [n], that, if length [[] = n], then [index (S n) [] = None]. This follows immediately from the definition of index. - Suppose [l = x :: l'] for some [x] and [l'], where [length l' = n'] implies [index (S n') l' = None], for any number [n']. We must show, for all [n], that, if [length (x::l') = n] then [index (S n) (x::l') = None]. Let [n] be a number with [length l = n]. Since [[ length l = length (x::l') = S (length l'), ]] it suffices to show that [[ index (S (length l')) l' = None. ]] But this follows directly from the induction hypothesis, picking [n'] to be length [l']. [] *) (** ** Induction Over an Inductively Defined Proposition *) (** Since inductively defined proof objects are often called "derivation trees," this form of proof is also known as _induction on derivations_. _Template_: - _Theorem_: P]," where [Q] is some inductively defined proposition (more generally, "For all [x] [y] [z], [Q x y z -> P x y z]")> _Proof_: By induction on a derivation of [Q]. - Suppose the final rule used to show [Q] is [c]. Then <...and here we state the types of all of the [a]'s together with any equalities that follow from the definition of the constructor and the IH for each of the [a]'s that has type [Q], if there are any>. We must show <...and here we restate [P]>. - [] _Example_ - _Theorem_: The [<=] relation is transitive -- i.e., for all numbers [n], [m], and [o], if [n <= m] and [m <= o], then [n <= o]. _Proof_: By induction on a derivation of [m <= o]. - Suppose the final rule used to show [m <= o] is [le_n]. Then [m = o] and the result is immediate. - Suppose the final rule used to show [m <= o] is [le_S]. Then [o = S o'] for some [o'] with [m <= o']. By induction hypothesis, [n <= o']. But then, by [le_S], [n <= o]. [] *) (* ####################################################### *) (** * Additional Exercises *) (** **** Exercise: 4 stars *) (** A palindrome is a sequence that reads the same backwards as forwards. - Define an inductive proposition [pal] on [list X] that captures what it means to be a palindrome. (Hint: You'll need three cases. Your definition should be based on the structure of the list; just having a single constructor [[ c : forall l, l = rev l -> pal l ]] may seem obvious, but will not work very well.) - Prove that [[ forall l, pal (l ++ rev l). ]] - Prove that [[ forall l, pal l -> l = rev l. ]] - (Much harder, for those who like a challenge.) Prove that [[ forall l, l = rev l -> pal l. ]] *) (* FILL IN HERE *) (** [] *) (** **** Exercise: 4 stars, optional (subsequence) *) (** A list is a _subsequence_ of another list if all of the elements in the first list occur in the same order in the second list, possibly with some extra elements in between. For example, [[ [1,2,3] ]] is a subsequence of each of the lists [[ [1,2,3] [1,1,1,2,2,3] [1,2,7,3] [5,6,1,9,9,2,7,3,8] ]] but it is _not_ a subsequence of any of the lists [[ [1,2] [1,3] [5,6,2,1,7,3,8] ]] - Define an inductive proposition [subseq] on [list nat] that captures what it means to be a subsequence. (Hint: You'll need three cases.) - Prove that subsequence is reflexive, that is, any list is a subsequence of itself. - Prove that for any lists [l1], [l2], and [l3], if [l1] is a subsequence of [l2], then [l1] is also a subsequence of [l2 ++ l3]. - Optional, harder: prove that subsequence is transitive, that is, if [l1] is a subsequence of [l2] and [l2] is a subsequence of [l3], then [l1] is a subsequence of [l3]. Hint: choose your induction carefully! *) (* FILL IN HERE *) (** [] *) (** **** Exercise: 2 stars, optional *) (** Suppose we make the following inductive definition: [[ Inductive foo (X : Set) (Y : Set) : Set := | foo1 : X -> foo X Y | foo2 : Y -> foo X Y | foo3 : foo X Y -> foo X Y. ]] Fill in the blanks to complete the induction principle that will be generated by Coq. [[ foo_ind : forall (X Y : Set) (P : foo X Y -> Prop), (forall x : X, __________________________________) -> (forall y : Y, __________________________________) -> (________________________________________________) -> ________________________________________________ ]] *) (** [] *) (** **** Exercise: 2 stars, optional *) (** Consider the following induction principle: [[ bar_ind : forall P : bar -> Prop, (forall n : nat, P (bar1 n)) -> (forall b : bar, P b -> P (bar2 b)) -> (forall (b : bool) (b0 : bar), P b0 -> P (bar3 b b0)) -> forall b : bar, P b ]] Write out the corresponding inductive set definition. [[ Inductive bar : Set := | bar1 : ________________________________________ | bar2 : ________________________________________ | bar3 : ________________________________________. ]] *) (** [] *) (** **** Exercise: 2 stars, optional *) (** Given the following inductively defined proposition: [[ Inductive no_longer_than (X : Set) : (list X) -> nat -> Prop := | nlt_nil : forall n, no_longer_than X [] n | nlt_cons : forall x l n, no_longer_than X l n -> no_longer_than X (x::l) (S n) | nlt_succ : forall l n, no_longer_than X l n -> no_longer_than X l (S n). ]] write the induction principle generated by Coq. [[ no_longer_than_ind : forall (X : Set) (P : list X -> nat -> Prop), (forall n : nat, ____________________) -> (forall (x : X) (l : list X) (n : nat), no_longer_than X l n -> ____________________ -> _____________________________ -> (forall (l : list X) (n : nat), no_longer_than X l n -> ____________________ -> _____________________________ -> forall (l : list X) (n : nat), no_longer_than X l n -> ____________________ ]] *) (** [] *) (** **** Exercise: 1 star, optional *) (** Suppose we give Coq the following definition: [[ Inductive R : nat -> list nat -> Prop := | c1 : R 0 [] | c2 : forall n l, R n l -> R (S n) (n :: l) | c3 : forall n l, R (S n) l -> R n l. ]] Which of the following propositions are provable? - [R 2 [1,0]] - [R 1 [1,2,1,0]] - [R 6 [3,2,1,0]] *) (** [] *)