(** * Logic: Logic in Coq *) Set Warnings "-notation-overridden,-parsing". From LF Require Export Tactics. (* ################################################################# *) (** * Back to Logic *) (** In the first lecture, we saw conjunction [/\], disjunction [\/], and falsehood [False]. All of these are build into Coq, and are common enough that Coq provides built-in tactics for using them. *) (** For [/\] and [\/] we have the tactics [split], [left] and [right]. Note that these are just shorthand for [apply conj], [apply or_introl] and [apply or_intror] *) Lemma and_symm : forall P Q, P /\ Q -> Q /\ P. intros P Q H. destruct H as [Hp Hq]. split. - apply Hq. - apply Hp. Qed. Lemma or_symm : forall P Q, P \/ Q -> Q \/ P. intros P Q H. destruct H as [Hp | Hq]. - right. apply Hp. - left. apply Hq. Qed. (** Expressions of the form [P -> False] or [m = n -> False] are common enough to have specific notations for them: *) Locate "~ _". Print not. Locate "_ <> _". (** We also have a tactic for dealing with [~A]: [contradict]. *) Lemma zero_not_even : ~ Even 0 -> 1 = 2. Proof. intros NE. contradict NE. apply E_O. Qed. (** This is equivalent to simply applying [ex_falso_quodlibet] (from ch. 1) and then applying [~A]. *) Lemma seven_not_four : 7 <> 4. Proof. intros F. discriminate F. Qed. (** **** Exercise: 2 stars, recommended (contrapositive) *) Theorem contrapositive : forall (P Q : Prop), (P -> Q) -> (~Q -> ~P). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 1 star (not_both_true_and_false) *) Theorem not_both_true_and_false : forall P : Prop, ~ (P /\ ~P). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 1 star, optional (informal_not_PNP) *) (** Write an informal proof (in English) of the proposition [forall P : Prop, ~(P /\ ~P)]. *) (* FILL IN HERE *) (* Do not modify the following line: *) Definition manual_grade_for_informal_not_PNP : option (nat*string) := None. (** [] *) (* ================================================================= *) (** ** Logical Equivalence *) (** The handy "if and only if" connective, which asserts that two propositions have the same truth value, is just the conjunction of two implications. *) Module MyIff. Definition iff (P Q : Prop) := (P -> Q) /\ (Q -> P). Notation "P <-> Q" := (iff P Q) (at level 95, no associativity) : type_scope. End MyIff. Theorem iff_sym : forall P Q : Prop, (P <-> Q) -> (Q <-> P). Proof. (* WORKED IN CLASS *) intros P Q [HAB HBA]. split. - (* -> *) apply HBA. - (* <- *) apply HAB. Qed. Lemma not_true_iff_false : forall b, b <> true <-> b = false. Proof. (* WORKED IN CLASS *) intros b. split. - (* -> *) intros H. destruct b. + contradict H. reflexivity. + reflexivity. - (* <- *) intros H1 H2. rewrite H1 in H2. discriminate H2. Qed. (** **** Exercise: 1 star, optional (iff_properties) *) (** Using the above proof that [<->] is symmetric ([iff_sym]) as a guide, prove that it is also reflexive and transitive. *) Theorem iff_refl : forall P : Prop, P <-> P. Proof. (* FILL IN HERE *) Admitted. Theorem iff_trans : forall P Q R : Prop, (P <-> Q) -> (Q <-> R) -> (P <-> R). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars, optional (or_distributes_over_and) *) Theorem or_distributes_over_and : forall P Q R : Prop, P \/ (Q /\ R) <-> (P \/ Q) /\ (P \/ R). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** Some of Coq's tactics treat [iff] statements specially, avoiding the need for some low-level proof-state manipulation. In particular, [rewrite] and [reflexivity] can be used with [iff] statements, not just equalities. To enable this behavior, we need to import a Coq library that supports it: *) Require Import Coq.Setoids.Setoid. (** Here is a simple example demonstrating how these tactics work with [iff]. First, let's recall one lemma and admit another: *) About or_assoc. (** **** Exercise: 2 stars, recommended (mult_0) *) Lemma mult_0 : forall n m, n * m = 0 <-> n = 0 \/ m = 0. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** We can now use these facts with [rewrite] and [reflexivity] to give smooth proofs of statements involving equivalences. Here is a ternary version of the previous [mult_0] result: *) Lemma mult_0_3 : forall n m p, n * m * p = 0 <-> n = 0 \/ m = 0 \/ p = 0. Proof. intros n m p. rewrite mult_0. rewrite mult_0. rewrite or_assoc. reflexivity. Qed. (** The [apply] tactic can also be used with [<->]. When given an equivalence as its argument, [apply] tries to guess which side of the equivalence to use. *) Lemma apply_iff_example : forall n m : nat, n * m = 0 -> n = 0 \/ m = 0. Proof. intros n m H. apply mult_0. apply H. Qed. (* ================================================================= *) (** ** Existential Quantification *) (** Another important logical connective is _existential quantification_. To say that there is some [x] of type [T] such that some property [P] holds of [x], we write [exists x : T, P]. As with [forall], the type annotation [: T] can be omitted if Coq is able to infer from the context what the type of [x] should be. *) (** To prove a statement of the form [exists x, P], we must show that [P] holds for some specific choice of value for [x], known as the _witness_ of the existential. This is done in two steps: First, we explicitly tell Coq which witness [t] we have in mind by invoking the tactic [exists t]. Then we prove that [P] holds after all occurrences of [x] are replaced by [t]. *) Lemma four_is_even : exists n : nat, 4 = n + n. Proof. exists 2. reflexivity. Qed. (** Conversely, if we have an existential hypothesis [exists x, P] in the context, we can destruct it to obtain a witness [x] and a hypothesis stating that [P] holds of [x]. *) Theorem exists_example_2 : forall n, (exists m, n = 4 + m) -> (exists o, n = 2 + o). Proof. (* WORKED IN CLASS *) intros n [m Hm]. (* note implicit [destruct] here *) exists (2 + m). apply Hm. Qed. (** **** Exercise: 1 star, recommended (dist_not_exists) *) (** Prove that "[P] holds for all [x]" implies "there is no [x] for which [P] does not hold." (Hint: [destruct H as [x E]] works on existential assumptions!) *) Theorem dist_not_exists : forall (X:Type) (P : X -> Prop), (forall x, P x) -> ~ (exists x, ~ P x). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 2 stars (dist_exists_or) *) (** Prove that existential quantification distributes over disjunction. *) Theorem dist_exists_or : forall (X:Type) (P Q : X -> Prop), (exists x, P x \/ Q x) <-> (exists x, P x) \/ (exists x, Q x). Proof. (* FILL IN HERE *) Admitted. (** [] *) (* ################################################################# *) (** * Programming with Propositions *) (** The logical connectives that we have seen provide a rich vocabulary for defining complex propositions from simpler ones. To illustrate, let's look at how to express the claim that an element [x] occurs in a list [l]. Notice that this property has a simple recursive structure: *) (** - If [l] is the empty list, then [x] cannot occur on it, so the property "[x] appears in [l]" is simply false. *) (** - Otherwise, [l] has the form [x' :: l']. In this case, [x] occurs in [l] if either it is equal to [x'] or it occurs in [l']. *) (** We can translate this directly into a straightforward recursive function taking an element and a list and returning a proposition: *) Fixpoint In {A : Type} (x : A) (l : list A) : Prop := match l with | [] => False | x' :: l' => x' = x \/ In x l' end. (** When [In] is applied to a concrete list, it expands into a concrete sequence of nested disjunctions. *) Example In_example_1 : In 4 [1; 2; 3; 4; 5]. Proof. (* WORKED IN CLASS *) simpl. right. right. right. left. reflexivity. Qed. Example In_example_2 : forall n, In n [2; 4] -> exists n', n = 2 * n'. Proof. (* WORKED IN CLASS *) simpl. intros n [H | [H | []]]. - exists 1. rewrite <- H. reflexivity. - exists 2. rewrite <- H. reflexivity. Qed. (** (Notice the use of the empty pattern to discharge the last case _en passant_.) *) (** We can also prove more generic, higher-level lemmas about [In]. Note, in the next, how [In] starts out applied to a variable and only gets expanded when we do case analysis on this variable: *) Lemma In_map : forall (A B : Type) (f : A -> B) (l : list A) (x : A), In x l -> In (f x) (map f l). Proof. intros A B f l x. induction l as [|x' l' IHl']. - (* l = nil, contradiction *) simpl. intros []. - (* l = x' :: l' *) simpl. intros [H | H]. + rewrite H. left. reflexivity. + right. apply IHl'. apply H. Qed. (** This way of defining propositions recursively, though convenient in some cases, also has some drawbacks. In particular, it is subject to Coq's usual restrictions regarding the definition of recursive functions, e.g., the requirement that they be "obviously terminating." In the next chapter, we will see how to define propositions _inductively_, a different technique with its own set of strengths and limitations. *) (** **** Exercise: 2 stars, recommended (In_map_iff) *) Lemma In_map_iff : forall (A B : Type) (f : A -> B) (l : list A) (y : B), In y (map f l) <-> exists x, f x = y /\ In x l. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 2 stars (In_app_iff) *) Lemma In_app_iff : forall A l l' (a:A), In a (l++l') <-> In a l \/ In a l'. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars, optional (All) *) (** Recall that functions returning propositions can be seen as _properties_ of their arguments. For instance, if [P] has type [nat -> Prop], then [P n] states that property [P] holds of [n]. Drawing inspiration from [In], write a recursive function [All] stating that some property [P] holds of all elements of a list [l]. To make sure your definition is correct, prove the [All_In] lemma below. (Of course, your definition should _not_ just restate the left-hand side of [All_In].) *) Fixpoint All {T : Type} (P : T -> Prop) (l : list T) : Prop (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted. Lemma All_In : forall T (P : T -> Prop) (l : list T), (forall x, In x l -> P x) <-> All P l. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars, recommended (combine_odd_even) *) (** Complete the definition of the [combine_odd_even] function below. It takes as arguments two properties of numbers, [Podd] and [Peven], and it should return a property [P] such that [P n] is equivalent to [Podd n] when [n] is odd and equivalent to [Peven n] otherwise. *) Definition combine_odd_even (Podd Peven : nat -> Prop) : nat -> Prop (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted. (** To test your definition, prove the following facts: *) Theorem combine_odd_even_intro : forall (Podd Peven : nat -> Prop) (n : nat), (oddb n = true -> Podd n) -> (oddb n = false -> Peven n) -> combine_odd_even Podd Peven n. Proof. (* FILL IN HERE *) Admitted. Theorem combine_odd_even_elim_odd : forall (Podd Peven : nat -> Prop) (n : nat), combine_odd_even Podd Peven n -> oddb n = true -> Podd n. Proof. (* FILL IN HERE *) Admitted. Theorem combine_odd_even_elim_even : forall (Podd Peven : nat -> Prop) (n : nat), combine_odd_even Podd Peven n -> oddb n = false -> Peven n. Proof. (* FILL IN HERE *) Admitted. (** [] *) (* ################################################################# *) (** * Coq vs. Set Theory *) (** Coq's logical core, the _Calculus of Inductive Constructions_, differs in some important ways from other formal systems that are used by mathematicians for writing down precise and rigorous proofs. For example, in the most popular foundation for mainstream paper-and-pencil mathematics, Zermelo-Fraenkel Set Theory (ZFC), a mathematical object can potentially be a member of many different sets; a term in Coq's logic, on the other hand, is a member of at most one type. This difference often leads to slightly different ways of capturing informal mathematical concepts, but these are, by and large, quite natural and easy to work with. For example, instead of saying that a natural number [n] belongs to the set of even numbers, we would say in Coq that [ev n] holds, where [ev : nat -> Prop] is a property describing even numbers. However, there are some cases where translating standard mathematical reasoning into Coq can be either cumbersome or sometimes even impossible, unless we enrich the core logic with additional axioms. We conclude this chapter with a brief discussion of some of the most significant differences between the two worlds. *) (* ================================================================= *) (** ** Functional Extensionality *) (** In common mathematical practice, two functions [f] and [g] are considered equal if they produce the same outputs: (forall x, f x = g x) -> f = g This is known as the principle of _functional extensionality_. *) (** Informally speaking, an "extensional property" is one that pertains to an object's observable behavior. Thus, functional extensionality simply means that a function's identity is completely determined by what we can observe from it -- i.e., in Coq terms, the results we obtain after applying it. *) (** Functional extensionality is not part of Coq's basic axioms. This means that some "reasonable" propositions are not provable. *) Example function_equality_ex2 : (fun x => plus x 1) = (fun x => plus 1 x). Proof. (* Stuck *) Abort. (** However, we can add functional extensionality to Coq's core logic using the [Axiom] command. *) Axiom functional_extensionality : forall {X Y: Type} {f g : X -> Y}, (forall (x:X), f x = g x) -> f = g. (** Using [Axiom] has the same effect as stating a theorem and skipping its proof using [Admitted], but it alerts the reader that this isn't just something we're going to come back and fill in later! *) (** We can now invoke functional extensionality in proofs: *) Example function_equality_ex2 : (fun x => plus x 1) = (fun x => plus 1 x). Proof. apply functional_extensionality. intros x. apply plus_comm. Qed. (** Naturally, we must be careful when adding new axioms into Coq's logic, as they may render it _inconsistent_ -- that is, they may make it possible to prove every proposition, including [False]! Unfortunately, there is no simple way of telling whether an axiom is safe to add: hard work is generally required to establish the consistency of any particular combination of axioms. Fortunately, it is known that adding functional extensionality, in particular, _is_ consistent. *) (** To check whether a particular proof relies on any additional axioms, use the [Print Assumptions] command. *) Print Assumptions function_equality_ex2. (* ===> Axioms: functional_extensionality : forall (X Y : Type) (f g : X -> Y), (forall x : X, f x = g x) -> f = g *) (** **** Exercise: 4 stars, optional (tr_rev_correct) *) (** One problem with the definition of the list-reversing function [rev] that we have is that it performs a call to [app] on each step; running [app] takes time asymptotically linear in the size of the list, which means that [rev] has quadratic running time. We can improve this with the following definition: *) Fixpoint rev_append {X} (l1 l2 : list X) : list X := match l1 with | [] => l2 | x :: l1' => rev_append l1' (x :: l2) end. Definition tr_rev {X} (l : list X) : list X := rev_append l []. (** This version is said to be _tail-recursive_, because the recursive call to the function is the last operation that needs to be performed (i.e., we don't have to execute [++] after the recursive call); a decent compiler will generate very efficient code in this case. Prove that the two definitions are indeed equivalent. *) Lemma tr_rev_correct : forall X, @tr_rev X = @rev X. (* FILL IN HERE *) Admitted. (** [] *) (* ================================================================= *) (** ** Propositions and Booleans *) (** We've seen two different ways of encoding logical facts in Coq: with _booleans_ (of type [bool]), and with _propositions_ (of type [Prop]). For instance, to claim that a number [n] is even, we can say either - (1) that [evenb n] returns [true], or - (2) that there exists some [k] such that [n = double k]. Indeed, these two notions of evenness are equivalent, as can easily be shown with a couple of auxiliary lemmas. Of course, it would be very strange if these two characterizations of evenness did not describe the same set of natural numbers! Fortunately, we can prove that they do... *) (** We first need two helper lemmas. *) Theorem evenb_double : forall k, evenb (double k) = true. Proof. intros k. induction k as [|k' IHk']. - reflexivity. - simpl. apply IHk'. Qed. (** **** Exercise: 3 stars (evenb_double_conv) *) Theorem evenb_double_conv : forall n, exists k, n = if evenb n then double k else S (double k). Proof. (* Hint: Use the [evenb_S] lemma from [Induction.v]. *) (* FILL IN HERE *) Admitted. (** [] *) Theorem even_bool_prop : forall n, evenb n = true <-> exists k, n = double k. Proof. intros n. split. - intros H. destruct (evenb_double_conv n) as [k Hk]. rewrite Hk. rewrite H. exists k. reflexivity. - intros [k Hk]. rewrite Hk. apply evenb_double. Qed. (** In view of this theorem, we say that the boolean computation [evenb n] is reflected in the truth of the proposition [exists k, n = double k]. *) (** Similarly, to state that two numbers [n] and [m] are equal, we can say either (1) that [n =? m] returns [true] or (2) that [n = m]. Again, these two notions are equivalent. *) Theorem eqb_eq : forall n1 n2 : nat, n1 =? n2 = true <-> n1 = n2. Proof. intros n1 n2. split. - apply eqb_true. - intros H. rewrite H. rewrite <- eqb_refl. reflexivity. Qed. (** However, even when the boolean and propositional formulations of a claim are equivalent from a purely logical perspective, they need not be equivalent _operationally_. Equality provides an extreme example: knowing that [n =? m = true] is generally of little direct help in the middle of a proof involving [n] and [m]; however, if we convert the statement to the equivalent form [n = m], we can rewrite with it. *) (** The case of even numbers is also interesting. Recall that, when proving the backwards direction of [even_bool_prop] (i.e., [evenb_double], going from the propositional to the boolean claim), we used a simple induction on [k]. On the other hand, the converse (the [evenb_double_conv] exercise) required a clever generalization, since we can't directly prove [(exists k, n = double k) -> evenb n = true]. *) (** For these examples, the propositional claims are more useful than their boolean counterparts, but this is not always the case. For instance, we cannot test whether a general proposition is true or not in a function definition; as a consequence, the following code fragment is rejected: *) Fail Definition is_even_prime n := if n = 2 then true else false. (** Coq complains that [n = 2] has type [Prop], while it expects an elements of [bool] (or some other inductive type with two elements). It would be nice if Coq provided a mechanism for testing the truth value of arbitrary propositions but unfortunately computably theory and Godel's incompleteness theorems say that this is impossible. (Complexity theory would add that if we limited that mechanism to computable propositions, it would still take forever.) Although general non-computable properties cannot be phrased as boolean computations, it is worth noting that even many _computable_ properties are easier to express using [Prop] than [bool], since recursive function definitions are subject to significant restrictions in Coq. For instance, the next chapter shows how to define the property that a regular expression matches a given string using [Prop]. Doing the same with [bool] would amount to writing a regular expression matcher, which would be more complicated, harder to understand, and harder to reason about. Conversely, an important side benefit of stating facts using booleans is enabling some proof automation through computation with Coq terms, a technique known as _proof by reflection_. Consider the following statement: *) Example even_1000 : exists k, 1000 = double k. (** The most direct proof of this fact is to give the value of [k] explicitly. *) Proof. exists 500. reflexivity. Qed. (** On the other hand, the proof of the corresponding boolean statement is even simpler: *) Example even_1000' : evenb 1000 = true. Proof. reflexivity. Qed. (** What is interesting is that, since the two notions are equivalent, we can use the boolean formulation to prove the other one without mentioning the value 500 explicitly: *) Example even_1000'' : exists k, 1000 = double k. Proof. apply even_bool_prop. reflexivity. Qed. (** Although we haven't gained much in terms of proof size in this case, larger proofs can often be made considerably simpler by the use of reflection. As an extreme example, the Coq proof of the famous _4-color theorem_ uses reflection to reduce the analysis of hundreds of different cases to a boolean computation. We won't cover reflection in great detail, but it serves as a good example showing the complementary strengths of booleans and general propositions. *) (** **** Exercise: 2 stars (logical_connectives) *) (** The following lemmas relate the propositional connectives to the corresponding boolean operations. *) Lemma andb_true_iff : forall b1 b2:bool, b1 && b2 = true <-> b1 = true /\ b2 = true. Proof. (* FILL IN HERE *) Admitted. Lemma orb_true_iff : forall b1 b2, b1 || b2 = true <-> b1 = true \/ b2 = true. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 1 star (eqb_neq) *) (** The following theorem is an alternate "negative" formulation of [eqb_eq] that is more convenient in certain situations (we'll see examples in later chapters). *) Theorem eqb_neq : forall x y : nat, x =? y = false <-> x <> y. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars, optional (eqb_list) *) (** Given a boolean operator [eqb] for testing equality of elements of some type [A], we can define a function [eqb_list] for testing equality of lists with elements in [A]. Complete the definition of the [eqb_list] function below. To make sure that your definition is correct, prove the lemma [eqb_list_true_iff]. *) Fixpoint eqb_list {A : Type} (eqb : A -> A -> bool) (l1 l2 : list A) : bool (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted. Lemma eqb_list_true_iff : forall A (eqb : A -> A -> bool), (forall a1 a2, eqb a1 a2 = true <-> a1 = a2) -> forall l1 l2, eqb_list eqb l1 l2 = true <-> l1 = l2. Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 2 stars, recommended (All_forallb) *) (** Recall the function [forallb], from the exercise [forall_exists_challenge] in chapter [Tactics]: *) Fixpoint forallb {X : Type} (test : X -> bool) (l : list X) : bool := match l with | [] => true | x :: l' => andb (test x) (forallb test l') end. (** Prove the theorem below, which relates [forallb] to the [All] property of the above exercise. *) Theorem forallb_true_iff : forall X test (l : list X), forallb test l = true <-> All (fun x => test x = true) l. Proof. (* FILL IN HERE *) Admitted. (** Are there any important properties of the function [forallb] which are not captured by this specification? *) (* FILL IN HERE *) (** [] *) (* ================================================================= *) (** ** Classical vs. Constructive Logic *) (** As we've seen earlier, the following proposition is not provable inside Coq: *) Definition excluded_middle := forall P : Prop, P \/ ~ P. (** However, if we happen to know that [P] is reflected in some boolean term [b], then knowing whether it holds or not is trivial: we just have to check the value of [b]. *) Theorem restricted_excluded_middle : forall P b, (P <-> b = true) -> P \/ ~ P. Proof. intros P [] H. - left. rewrite H. reflexivity. - right. rewrite H. intros contra. discriminate contra. Qed. (** In particular, the excluded middle is valid for equations [n = m], between natural numbers [n] and [m]. *) Theorem restricted_excluded_middle_eq : forall (n m : nat), n = m \/ n <> m. Proof. intros n m. apply (restricted_excluded_middle (n = m) (n =? m)). symmetry. apply eqb_eq. Qed. (** It may seem strange that the general excluded middle is not available by default in Coq; after all, any given claim must be either true or false. Nonetheless, there is an advantage in not assuming the excluded middle: statements in Coq can make stronger claims than the analogous statements in standard mathematics. Notably, if there is a Coq proof of [exists x, P x], it is possible to explicitly exhibit a value of [x] for which we can prove [P x] -- in other words, every proof of existence is necessarily _constructive_. *) (** Logics like Coq's, which do not assume the excluded middle, are referred to as _constructive logics_. More conventional logical systems such as ZFC, in which the excluded middle does hold for arbitrary propositions, are referred to as _classical_. *) (** The following example illustrates why assuming the excluded middle may lead to non-constructive proofs: _Claim_: There exist irrational numbers [a] and [b] such that [a ^ b] is rational. _Proof_: It is not difficult to show that [sqrt 2] is irrational. If [sqrt 2 ^ sqrt 2] is rational, it suffices to take [a = b = sqrt 2] and we are done. Otherwise, [sqrt 2 ^ sqrt 2] is irrational. In this case, we can take [a = sqrt 2 ^ sqrt 2] and [b = sqrt 2], since [a ^ b = sqrt 2 ^ (sqrt 2 * sqrt 2) = sqrt 2 ^ 2 = 2]. [] Do you see what happened here? We used the excluded middle to consider separately the cases where [sqrt 2 ^ sqrt 2] is rational and where it is not, without knowing which one actually holds! Because of that, we wind up knowing that such [a] and [b] exist but we cannot determine what their actual values are (at least, using this line of argument). As useful as constructive logic is, it does have its limitations: There are many statements that can easily be proven in classical logic but that have much more complicated constructive proofs, and there are some that are known to have no constructive proof at all! Fortunately, like functional extensionality, the excluded middle is known to be compatible with Coq's logic, allowing us to add it safely as an axiom. However, we will not need to do so in this book: the results that we cover can be developed entirely within constructive logic at negligible extra cost. It takes some practice to understand which proof techniques must be avoided in constructive reasoning, but arguments by contradiction, in particular, are infamous for leading to non-constructive proofs. Here's a typical example: suppose that we want to show that there exists [x] with some property [P], i.e., such that [P x]. We start by assuming that our conclusion is false; that is, [~ exists x, P x]. From this premise, it is not hard to derive [forall x, ~ P x]. If we manage to show that this intermediate fact results in a contradiction, we arrive at an existence proof without ever exhibiting a value of [x] for which [P x] holds! The technical flaw here, from a constructive standpoint, is that we claimed to prove [exists x, P x] using a proof of [~ ~ (exists x, P x)]. Allowing ourselves to remove double negations from arbitrary statements is equivalent to assuming the excluded middle, as shown in one of the exercises below. Thus, this line of reasoning cannot be encoded in Coq without assuming additional axioms. *) (** **** Exercise: 3 stars (excluded_middle_irrefutable) *) (** Proving the consistency of Coq with the general excluded middle axiom requires complicated reasoning that cannot be carried out within Coq itself. However, the following theorem implies that it is always safe to assume a decidability axiom (i.e., an instance of excluded middle) for any _particular_ Prop [P]. Why? Because we cannot prove the negation of such an axiom. If we could, we would have both [~ (P \/ ~P)] and [~ ~ (P \/ ~P)] (since [P] implies [~ ~ P], by the exercise below), which would be a contradiction. But since we can't, it is safe to add [P \/ ~P] as an axiom. *) Theorem excluded_middle_irrefutable: forall (P:Prop), ~ ~ (P \/ ~ P). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 3 stars, optional (not_exists_dist) *) (** It is a theorem of classical logic that the following two assertions are equivalent: ~ (exists x, ~ P x) forall x, P x The [dist_not_exists] theorem above proves one side of this equivalence. Interestingly, the other direction cannot be proved in constructive logic. Your job is to show that it is implied by the excluded middle. *) Theorem not_exists_dist : excluded_middle -> forall (X:Type) (P : X -> Prop), ~ (exists x, ~ P x) -> (forall x, P x). Proof. (* FILL IN HERE *) Admitted. (** [] *) (** **** Exercise: 5 stars, optional (classical_axioms) *) (** For those who like a challenge, here is an exercise taken from the Coq'Art book by Bertot and Casteran (p. 123). Each of the following four statements, together with [excluded_middle], can be considered as characterizing classical logic. We can't prove any of them in Coq, but we can consistently add any one of them as an axiom if we wish to work in classical logic. Prove that all five propositions (these four plus [excluded_middle]) are equivalent. *) Definition peirce := forall P Q: Prop, ((P->Q)->P)->P. Definition double_negation_elimination := forall P:Prop, ~~P -> P. Definition de_morgan_not_and_not := forall P Q:Prop, ~(~P /\ ~Q) -> P\/Q. Definition implies_to_or := forall P Q:Prop, (P->Q) -> (~P\/Q). (* FILL IN HERE *) (** [] *)