(** * Induction in Coq ----- ## Topics: - recursive functions - induction on lists - induction on natural numbers - rings and fields - induction principles ## ----- We'll need the list library for these notes. *) Require Import List. Import ListNotations. (** (**********************************************************************) ** Recursive functions The [List] library defines the list append operator, which in Coq is written as infix operator [++], or as prefix function [app]. In OCaml, the same operator is written [@]. In OCaml's standard library, you'll recall that append is defined as follows: << let rec append lst1 lst2 = match lst1 with | [] -> lst2 | h::t -> h :: (append t lst2) >> The Coq equivalent of that would be: *) Fixpoint append {A : Type} (lst1 : list A) (lst2 : list A) := match lst1 with | nil => lst2 | h::t => h :: (append t lst2) end. (** The [Fixpoint] keyword is similar to a [let rec] definition in OCaml. The braces around [A : Type] in that definition make the [A] argument implicit, hence we don't have to provide it at the recursive call to [append] in the second branch of the pattern match. Without the braces, we'd have to write the function as follows: *) Fixpoint append' (A : Type) (lst1 : list A) (lst2 : list A) := match lst1 with | nil => lst2 | h::t => h :: (append' A t lst2) end. (** The actual definition of [++] in the Coq standard library is a little more complicated, but it's essentially the same idea. Here's that definition: << Definition app (A : Type) : list A -> list A -> list A := fix app' lst1 lst2 := match lst1 with | nil => lst2 | h :: t => h :: app' t lst2 end. >> The Coq [fix] keyword is similar to a [let rec] expression in OCaml, but where the body of the expression is implicit: Coq [fix f x1 .. xn := e] is like OCaml [let rec f x1 .. xn = e in f]. So in OCaml we could rephrase the above definition as follows: << let app : 'a list -> 'a list -> 'a list = let rec app' lst1 lst2 = match lst1 with | [] -> lst2 | h :: t -> h :: app' t lst2 in app' >> Now that we know how Coq defines [++], let's prove a theorem about it. *) Theorem nil_app : forall (A:Type) (lst : list A), [] ++ lst = lst. (** Intuition: appending the empty list to [lst] immediately returns [lst], by the definition of [++]. *) Proof. intros A lst. simpl. trivial. Qed. (** The second step in that proof simplifies [[] ++ lst] to [lst]. That's because of how [++] is defined: it pattern matches against its first argument, which here is [[]], hence simply returns its second argument. Next, let's prove that appending nil on the right-hand side also results in [lst]: *) Theorem app_nil : forall (A:Type) (lst : list A), lst ++ [] = lst. (* Intuition (incomplete): by case analysis on [lst]. - if [lst] is [[]], then trivially [[] ++ []] is [[]]. - if [lst] is [h::t], then ...? *) Proof. intros A lst. destruct lst as [ | h t]. - trivial. - simpl. (* can't proceed *) Abort. (** When we get to the end of that proof, we are trying to show that [h :: (t ++ []) = h :: t]. There's no way to make progress on that, because we can't simplify [t ++ []] to just [t]. Of course as humans we know that holds. But to Coq, that's a fact that hasn't yet been proved. Indeed, it is an instance of the theorem we're currently trying to prove! What's going wrong here is that case analysis is not a sufficiently powerful proof technique for this theorem. We need to be able to _recursively_ apply the theorem we're trying to prove to smaller lists. That's where _induction_ comes into play. (**********************************************************************) ** Induction on lists _Induction_ is a proof technique that you will have encountered in math classes before---including CS 2800. It is useful when you want to prove that some property holds of all the members of an infinite set, such as the natural numbers, as well as lists, trees, and other data types. One classic metaphor for induction is _falling dominos_: if you arrange some dominos such that each domino, when it falls, will knock over the next domino, and if you knock over the first domino, then all the dominos will fall. Another classic metaphor for induction is a _ladder_: if you can reach the first rung, and if for any given rung the next rung can be reached, then you can reach any rung you wish. (As long as you're not afraid of heights.) What both of those metaphors have in common is - a _base case_, in which something is done first. For the dominos, it's knocking over the first domino; for the ladder, it's climbing the first rung. And, - an _inductive case_, in which a step is taken from one thing to the the next. For the dominos, it's one domino knocking over the next; for the ladder, it's literally taking the step from one rung to the next. In both cases, it must actually be possible for the action to occur: if the dominos or the rungs were spaced too far apart, then progress would stop. A proof by induction likewise has a base case and an inductive case. Here's the structure of a proof by induction on a list: << Theorem. For all lists lst, P(lst). Proof. By induction on lst. Case: lst = nil Show: P(nil) Case: lst = h::t IH: P(t) Show: P(h::t) QED. >> The _base case_ of a proof by induction on lists is for when the list is empty. The _inductive case_ is when the list is non-empty, hence is the cons of a head to a tail. In the inductive case, we get to assume an _inductive hypothesis_, which is that the property [P] holds of the tail. In the metaphors above, the inductive hypothesis is the assumption that we've already reached some particular domino or rung. From there, the metaphorical work we do in the inductive case of the proof is to show that from that domino or rung, we can reach the next one. An inductive hypothesis is exactly the kind of assumption we needed to get our proof about appending nil to go through. Here's how that proof could be written: << Theorem: for all lists lst, lst ++ nil = lst. Proof: by induction on lst. P(lst) = lst ++ nil = lst. Case: lst = nil Show: P(nil) = nil ++ nil = nil = nil = nil Case: lst = h::t IH: P(t) = (t ++ nil = t) Show P(h::t) = (h::t) ++ nil = h::t = h::(t ++ nil) = h::t // by definition of ++ = h::t = h::t // by IH QED >> In Coq, we could prove that theorem as follows: *) Theorem app_nil : forall (A:Type) (lst : list A), lst ++ nil = lst. Proof. intros A lst. induction lst as [ | h t IH]. - simpl. trivial. - simpl. rewrite -> IH. trivial. Qed. (** The tactics used in that proof correspond exactly to the non-Coq proof above. The [induction] tactic is new to us. It initiates a proof by induction on its argument, in this case [lst], and provides names for the variables to be used in the cases. There aren't any new variables in the base case, but the inductive case has variables for the head of the list, the tail, and the inductive hypothesis. You could leave out those variables and simply write [induction lst.], but that leads to a less human-readable proof. In the inductive case, we use the [rewrite ->] tactic to rewrite [t ++ nil] to [t]. The [IH] says those terms are equal. That tactic replaces the left-hand side of the equality with the right-hand side, wherever the left-hand side appears in the subgoal. It's also possible to rewrite from right to left with the [rewrite <-] tactic. If you leave out the arrow, Coq assumes that you mean [->]. Here's another theorem we can prove in exactly the same manner. This theorem shows that append is _associative_. << Theorem: forall lists l1 l2 l3, l1 ++ (l2 ++ l3) = (l1 ++ l2) ++ l3. Proof: by induction on l1. P(l1) = l1 ++ (l2 ++ l3) = (l1 ++ l2) ++ l3 Case: l1 = nil Show: P(nil) = nil ++ (l2 ++ l3) = (nil ++ l2) ++ l3 = l2 ++ l3 = l2 ++ l3 // simplifying ++, twice Case: l1 = h::t IH: P(t) = t ++ (l2 ++ l3) = (t ++ l2) ++ l3 Show: P(h::t) = h::t ++ (l2 ++ l3) = (h::t ++ l2) ++ l3 = h::(t ++ (l2 ++ l3)) = h::((t ++ l2) ++ l3) // simplifying ++, thrice = h::((t ++ l2) ++ l3) = h::((t ++ l2) ++ l3) // by IH QED >> In Coq, that proof looks more or less identical to our previous Coq proof about append and nil: *) Theorem app_assoc : forall (A:Type) (l1 l2 l3 : list A), l1 ++ (l2 ++ l3) = (l1 ++ l2) ++ l3. (* Intuition: above *) Proof. intros A l1 l2 l3. induction l1 as [ | h t IH]. - trivial. - simpl. rewrite -> IH. trivial. Qed. (** (**********************************************************************) ** Induction on natural numbers One of the classic theorems proved by induction is that [0 + 1 + ... + n] is equal to [n * (n+1) / 2]. It uses proof by induction on the natural numbers, which are the non-negative integers. The structure of a proof by induction on the naturals is as follows: << Theorem. For all natural numbers n, P(n). Proof. By induction on n. Case: n = 0 Show: P(0) Case: n = k+1 IH: P(k) Show: P(k+1) QED. >> The base case is for zero, the smallest natural number. The inductive case assumes that P holds of k, then shows that P holds of k+1. The [induction] tactic in Coq works for inductive types---that is, types defined with the [Inductive] keyword. You might, therefore, suspect that if we're going to do induction over [nat]s, the type [nat] must be inductively defined. Indeed it is. There is a famous inductive definition of the natural numbers that is credited to Giuseppe Peano (1858-1932). In OCaml, that definition would be: << type nat = O | S of nat >> The [O] constructor (that's the letter capital O) represents zero. The [S] constructor represents the successor function---that is, the function that adds one to its argument. So: - 0 is [O] - 1 is [S O] - 2 is [S (S O)] - 3 is [S (S (S O))] - etc. This is a kind of _unary_ representation of the naturals, in which we repeat the symbol [S] a total of [n] times to represent the natural number [n]. The Coq definition of [nat] is much the same: *) Print nat. (** Coq responds with output that is equivalent to the following: << Inductive nat : Set := | O : nat | S : nat -> nat >> That is, [nat] has two constructors, [O] and [S], which are just like the OCaml constructors we examined above. And [nat] has type [Set], meaning that [nat] is a specification for program computations. (Or, a little more loosely, that [nat] is a type representing program values.) Anywhere we write something that looks like an integer literal in Coq, Coq actually understand that as its expansion in the Peano representation defined above. For example,  is understood by Coq as just syntactic sugar for [S (S O)]. We can even write computations using those constructors: *) Compute S (S O). (** Coq responds, though, by reintroducing the syntactic sugar: << = 2 : nat >> The Coq standard library defines many functions over [nat] using those constructors and pattern matching, including addition, subtraction, and multiplication. For example, addition is defined like this: *) Fixpoint my_add (a b : nat) : nat := match a with | 0 => b | S c => S (my_add c b) end. (** Note that we're allowed to use either  or [O] as a pattern, because the former is just syntactic sugar for the latter. The second branch of the pattern match is effectively calling [my_add] recursively with [a-1] as its first argument, since [a = S c], meaning that [a] is the successor of [c]. Now that we know how [nat] is defined inductively, let's try to prove the classic theorem mentioned above about summation. Moreover, let's prove that a program that computes the sum [0 + 1 + ... + n] does in fact compute [n * (n+1) / 2]. First, we need to write that program. In OCaml, we could write the program quite easily: << let rec sum_to n = if n=0 then 0 else n + sum_to (n-1) >> In Coq, it will turn out to be surprisingly tricky... (**********************************************************************) ** Recursive functions, revisited Here's a first attempt at defining [sum_to], which is just a direct translation of the OCaml code into Coq. The [Fail] keyword before it tells Coq to expect the definition to fail to compile. *) Fail Fixpoint sum_to (n:nat) : nat := if n = 0 then 0 else n + sum_to (n-1). (** Coq responds: << The command has indeed failed with message: The term "n = 0" has type "Prop" which is not a (co-)inductive type. >> The problem is the the equality operator [=] returns a proposition (i.e., something we could try to prove), whereas the [if] expression expects a Boolean (i.e., [true] or [false]) as its guard. (Actually [if] is willing to accept any value of an inductive type with exactly two constructors as its guard, and [bool] is an example of such a type.) To fix this problem, we need to use an equality operator that returns a [bool], rather than a [Prop], when applied to two [nat]s. Such an operator is defined in the [Arith] library for us: *) Require Import Arith. Locate "=?". Check Nat.eqb. (** Coq responds: << Nat.eqb : nat -> nat -> bool >> We can now try to use that operator. Unfortunately, we discover a new problem: *) Fail Fixpoint sum_to (n:nat) : nat := if n =? 0 then 0 else n + sum_to (n-1). (** Coq responds with output that contains the following lines: << Recursive definition of sum_to is ill-formed. ... Recursive call to sum_to has principal argument equal to "n - 1" instead of a subterm of "n". ... >> Although the error message might be cryptic, you can see that Coq is complaining about the recursive call in the [else] branch. For some reason, Coq is unhappy about the argument [n-1] provided at that call. Coq wants that argument to be a "subterm" of [n]. The words _term_ and _expression_ are synonymous here, so Coq is saying that it wants the argument to be a subexpression of [n]. Of course [n] doesn't have any subexpressions. So why is Coq giving us this error? Before we can answer that question, let's look at a different recursive function---one that implements an infinite loop: *) Fail Fixpoint inf (x:nat) : nat := inf x. (** Coq responds very similarly to how it did with [sum_to]: << Recursive definition of inf is ill-formed. ... Recursive call to inf has principal argument equal to "x" instead of a subterm of "x". >> The reason Coq rejects [inf] is that #Coq does not permit any infinite loops#. That might seem strange, but there's an excellent reason for it. Consider how [inf] would be defined in OCaml: << # let rec inf x = inf x val inf : 'a -> 'b = >> Let's look at the type of that function, using what we learned about propositions-as-types in the previous lecture. The type ['a -> 'b] corresponds to the proposition [A -> B], where [A] and [B] could be any propositions. In particular, [A] could be [True] and [B] could be [False], leading to the proposition [True -> False]. That's a proposition that should never be provable: truth does not imply falsehood. And yet, since [inf] is a program of that type, [inf] corresponds to a proof of that proposition. So using [inf] we could actually prove [False]: << type void = {nope : 'a . 'a};; let rec inf x = inf x;; let ff : void = inf ();; >> The [void] type is how we represented [False] in the previous lecturev. The value [ff] above corresponds to a proof of [False]. So infinite loops are able to prove [False]. In OCaml we don't mind that phenomenon, because OCaml's purpose is not to be a proof assistant. But in Coq it would be deadly: we should never allow the proof assistant to prove false propositions. Coq therefore wants to prohibit all infinite loops. But that's easier said than done! Recall from CS 2800 that the _halting problem_ is undecidable: we can't write a program that precisely determines whether another program will terminate. Well, the Coq compiler is a program, and it wants to detect which programs terminate and which programs do not---which is exactly what the halting problem says is impossible. So instead of trying to do something impossible, Coq settles for doing something possible but imprecise, specifically, something that prohibits all non-terminating programs as well as prohibiting some terminating programs. Coq enforces a syntactic restriction on recursive functions: there must always be an argument that is _syntactically smaller_ at every recursive function application. An expression [e1] is syntactically smaller than [e2] if [e1] is a subexpression of [e2]. For example,  is syntactically smaller than [1-x], but [n-1] is not syntactically smaller than [n]. It turns out this restriction is sufficient to guarantee that programs must terminate: eventually, if every call results in something smaller, you must reach something that is small enough that you cannot make a recursive call on it, hence evaluation must terminate. A synonym for "syntactically smaller" is _structurally decreasing_. But that does rule out some programs that we as humans know will terminate yet do not meet the syntactic restriction. And [sum_to] is one of them. Here is the definition we previously tried: *) Fail Fixpoint sum_to (n:nat) : nat := if n =? 0 then 0 else n + sum_to (n-1). (** The recursive call to [sum_to] has as its argument [n-1], which syntactically is actually bigger than the original argument of [n]. So Coq rejects the program. To finally succeed in definining [sum_to], we can make use of what we know about how [nat] is defined: since it's an inductive type, we can pattern match on it: *) Fixpoint sum_to (n:nat) : nat := match n with | 0 => 0 | S k => n + sum_to k end. (** The second branch of the pattern match recurses on an argument that is both syntactically and arithmetically smaller, just as our definition of [my_add] did, above. So Coq's syntactic restriction on recursion is satisfied, and the definition is accepted as a program that definitely terminates. (**********************************************************************) ** Inductive proof of the summation formula Now that we've finally succeeded in defining [sum_to], we can prove the classic theorem about summation. Here's how we would write the proof mathematically: << Theorem: for all natural numbers n, sum_to n = n * (n+1) / 2. Proof: by induction on n. P(n) = sum_to n = n * (n+1) / 2 Case: n=0 Show: P(0) = sum_to 0 = 0 * (0+1) / 2 = 0 = 0 * (0+1) / 2 // simplifying sum_to 0 = 0 = 0 // 0 * x = 0 Case: n=k+1 IH: P(k) = sum_to k = k * (k+1) / 2 Show: P(k+1) = sum_to (k+1) = (k+1) * (k+1+1) / 2 = k + 1 + sum_to k = (k+1) * (k+1+1) / 2 // simplifying sum_to (k+1) = k + 1 + k * (k+1) / 2 = (k+1) * (k+1+1) / 2 // using IH = 2 + 3k + k*k = 2 + 3k + k*k // simplifying terms on each side QED >> Now let's do the proof in Coq. *) Theorem sum_sq : forall n : nat, sum_to n = n * (n+1) / 2. Proof. intros n. induction n as [ | k IH]. - trivial. - simpl. rewrite -> IH. Abort. (** The proof is working fine so far, but now we have a complicated algebraic equation we need to prove: << S (k + k * (k + 1) / 2) = fst (Nat.divmod (k + 1 + k * S (k + 1)) 1 0 0) >> ([divmod] is part of how [/] is implemented in Coq.) Although we could try to prove that manually using the definitions of all the operators, it would be much nicer to get Coq to find the proof for us. It turns out that Coq does have great support for finding proofs that involve _rings_: algebraic structures that support addition and multiplication operations. (We'll discuss rings in detail after we finish the current proof.) But we can't use that automation here, because the equation we want to prove also involves division, and rings do not support division operations. To avoid having to reason about division, we could rewrite the theorem we want to prove: by multiplying both sides by 2, the division goes away: *) Theorem sum_sq_no_div : forall n : nat, 2 * sum_to n = n * (n+1). Proof. intros n. induction n as [ | k IH]. - trivial. - simpl. Abort. (** Now, after the call to [simpl], we don't have any division, but we also don't have any expressions that look exactly like the left-hand side of the inductive hypothesis. The problem is that [simpl] was too agressive in simplifying all the expressions. All we really want is to transform the left-hand side of the subgoal, [2 * sum_to (S k)], into an expression that contains the left-hand side of the inductive hypothesis, [2 * sum_to k]. Thinking about the definition of [sum_to], we ought to be able to transform [2 * sum_to (S k)] into [2 * (S k + sum_to k)], which equals [2 * (S k) + 2 * sum_to k]. That final expression does have the left-hand side of the inductive hypothesis in it, as desired. Let's factor out that reasoning as a separate "helper" theorem. In math, helper theorems are usually called _lemmas_. The Coq keyword [Lemma] is synonymous with [Theorem]. *) Lemma sum_helper : forall n : nat, 2 * sum_to (S n) = 2 * (S n) + 2 * sum_to n. Proof. intros n. simpl. ring. Qed. (** The proof above simplifies the application of [sum_to (S n)], then invokes a new tactic called [ring]. That tactic is able to automatically search for proofs of equations involving addition and multiplication of natural numbers. Now that we have our helper lemma, we can use it to prove the theorem: *) Theorem sum_sq_no_div : forall n : nat, 2 * sum_to n = n * (n+1). Proof. intros n. induction n as [ | k IH]. - trivial. - rewrite -> sum_helper. rewrite -> IH. ring. Qed. (** Once more, after doing the rewriting with the lemma and the inductive hypothesis, we're left with algebraic equations that can be proved simply by invoking the [ring] tactic. Finally, we can use [sum_sq_no_div] to prove the original theorem involving division. To do that, we need to first prove another lemma that shows we can transform a multiplication into a division: *) Lemma div_helper : forall a b c : nat, c <> 0 -> c * a = b -> a = b / c. Proof. intros a b c neq eq. rewrite <- eq. rewrite Nat.mul_comm. rewrite Nat.div_mul. trivial. assumption. Qed. (** That lemma involves two library theorems, [mult_comm] and [Nat.div_mul]. How did we know to use these? Coq can help us search for useful theorems. Right after we [rewrite <- eq] in the above proof, our subgoal is [a = c * a / c]. It looks like we ought to be able to cancel the [c] term on the right-hand side. So we can search for a theorem that would help us do that. The [Search] command takes wildcards and reports all theorems that match the pattern we supply, for example: *) Search (_ * _ / _). (** Unfortunately, the search results currently seem to be broken in Visual Studio Code: we get only one matching library theorem, instead of all of them. Doing the search in [coqide] or [coqtop] (the Coq REPL), or just browsing through the web documentation, does reveal a useful theorem: << Nat.div_mul: forall a b : nat, b <> 0 -> a * b / b = a >> That would let us cancel a term from the numerator and denominator, but it requires the left-hand side of the equality to be of the form [a * b / b], whereas we have [c * a / b]. The problem is that the two sides of the multiplication are reversed. No worries; multiplication is commutative, and there is a library theorem that proves it. Again, we could find that theorem: *) Search (_ * _ = _ * _). (** One of the results is: << Nat.mul_comm: forall n m : nat, n * m = m * n >> Putting those two library theorems to use, we're able to prove the lemma as above. Finally, we can use that lemma to prove our classic theorem about summation. *) Theorem sum_sq : forall n : nat, sum_to n = n * (n+1) / 2. Proof. intros n. apply div_helper. - discriminate. - rewrite sum_sq_no_div. trivial. Qed. (** When we use [apply div_helper] in that proof, Coq generates two new subgoals---one for each of the propositions [c <> 0] and [c * a = b] in the type of [div_helper]. _Summary_: wow, that was a lot of work to prove that seemingly simple classic theorem! We had to figure out how to code [sum_to], and we had to deal with a lot of complications involving algebra. This situation is not uncommon: the theorems that we think of as easy with pencil-and-paper (like arithmetic) turn out to be hard to convince Coq of, whereas the theorems that we think of as challenging with pencil-and-paper (like induction) turn out to be easy. (**********************************************************************) ** Rings and fields In the proof we just did above about summation, we used a tactic called [ring] that we said searches for proofs about algebraic equations involving multiplication and addition. Let's look more closely at that tactic. When we studied OCaml modules, we looked at _rings_ as an example of a mathematical abstraction of addition, multiplication, and negation. Here is an OCaml signature for a ring: << module type Ring = sig type t val zero : t val one : t val add : t -> t -> t val mult : t -> t -> t val neg : t -> t end >> We could implement that signature with a representation type [t] that is [int], or [float], or even [bool]. The names given in [Ring] are suggestive of the operations they represent, but to really specify how those operations should behave, we need to write some equations that relate them. Below are the equations that (it turns out) fully specify [zero], [one], [add], and [mult]. Rather than use those identifiers, we use the more familiar notation of , , [+], and [*]. << 0 + x = 0 x + y = y + x x + (y + z) = (x + y) + z 0 * x = 0 1 * x = 1 x * y = y * x x * (y * z) = (x * y) * z (x + y) * z = (x * z) + (y * z) >> Technically these equations specify what is known as a _commutative semi-ring_. It's a _semi_-ring because we don't have equations specifying negation yet. It's a _commutative_ semi-ring because the [*] operation commutes. (The [+] operation commutes too, but that's always required of a semi-ring.) The first group of equations specifies how [+] behaves on its own. The second group specifies how [*] behaves on its own. The final equation specifies how [+] and [*] interact. If we extend the equations above with the following two, we get a specification for a _ring_: << x - y = x + (-y) x + (-x) = 0 >> It's a remarkable fact from the study of _abstract algebra_ that those equations completely specify a ring. Any theorem you want to prove about addition, multiplication, and negation follows from those equations. We call the equations the _axioms_ that specify a ring. Rings don't have a division operation. Let's introduce a new operator called [inv] (short for "inverse"), and let's write [1/x] as syntactic sugar for [inv x]. If we take all the the ring axioms and add the following axiom for [inv], we get what is called a _field_: << x * 1/x = 1 if x<>0 >> A field is an abstraction of addition, multiplication, negation, and division. Note that OCaml [int]s do not satisfy the [inv] axiom above. For example, [2 * (1/2)] equals  in OCaml, not . OCaml [float]s mostly do satisfy the field axioms, up to the limits of floating-point arithmetic. And in mathematics, the rational numbers and the real numbers are fields. Coq provides two tactics, [ring] and [field], that automatically search for proofs using the ring and field axioms. The [ring] tactic was already loaded for us when we wrote [Require Import Arith] earlier in this file. We can use the [ring] tactic to easily prove equalities that follow from the ring axioms. Here are two examples. *) Theorem plus_comm : forall a b, a + b = b + a. Proof. intros a b. ring. Qed. Theorem foil : forall a b c d, (a + b) * (c + d) = a*c + b*c + a*d + b*d. Proof. intros a b c d. ring. Qed. (** Coq infers the types of the variables above to be [nat], because the [+] and [*] operators are defined on [nat]. The proofs that the [ring] tactic finds can be quite complicated. For example, try looking at the output of the following command. It's so long that we won't put that output in this file! *) Print foil. (** Of course, [ring] won't find proofs of equations that don't actually hold. For example, if we had a typo in our statement of [foil], then [ring] would fail. *) Theorem broken_foil: forall a b c d, (a + b) * (c + d) = a*c + b*c + c*d + b*d. Proof. intros a b c d. try ring. Abort. (** Here's a theorem that [ring], perhaps surprisingly, cannot prove. *) Theorem sub_add_1 : forall a : nat, a - 1 + 1 = a. Proof. intros a. try ring. Abort. (** What's going wrong here is that [nat] is really only a semi-ring, not a ring. That is, [nat] doesn't satisfy the axioms about negation. Why? Remember that the natural numbers stop at ; we don't get any negative numbers. So if [a] is  in the above theorem, [a-1] actually evaluates to  rather than [-1]. *) Compute 0-1. (* 0 : nat *) (** If we want to reason about the integers instead of the natural numbers, we can use a library called [ZArith] for that. The name comes from the fact that [Z] is used in mathematics to denote the integers. *) Require Import ZArith. Open Scope Z_scope. (** The [Open Scope] command causes the [ZArith] library's scope to be used to resolve names, hence [+] becomes the operator on [Z] instead of on [nat], as does [-], etc. *) Compute 0-1. (* -1 : Z *) (** Now we can prove the theorem from before. *) Theorem sub_add_1 : forall a : Z, a - 1 + 1 = a. Proof. intros a. ring. Qed. (** Before going on, let's close the [Z] scope so that the operators go back to working on [nat], as usual. *) Close Scope Z_scope. Compute 0-1. (* 0 : nat *) (** Coq also provides implementations of the rational numbers as a field, as well as the real numbers as a field. To get the [field] tactic, we first need to load the [Field] library. *) Require Import Field. (** The rational numbers are provided in a couple different Coq libraries; the one we'll use here is [Qcanon]. In mathematics, [Q] denotes the rational numbers, and [canon] indicates that the numbers are stored in a _canonical form_---that is, as simplified fractions. For example, [Qcanon] would represent [2/4] as [1/2], eliminating the common factor of  from the numerator and the denominator. (The [QArith] library provides rational numbers that are not in canonical form.) *) Require Import Qcanon. Open Scope Qc_scope. Theorem frac_qc: forall x y z : Qc, z <> 0 -> (x + y) / z = x / z + y /z. Proof. intros x y z z_not_0. field. assumption. Qed. Close Scope Qc_scope. (** The real numbers are provided in the [Reals] library. Here's that same theorem again. *) Module RealExample. (** This code is in its own module for an annoying reason: [Reals] redefines its own [nil], which will interefere with the examples want to give further below in this file with lists. *) Require Import Reals. Open Scope R_scope. Theorem frac_r : forall x y z, z <> 0 -> (x + y) / z = x / z + y /z. Proof. intros x y z z_not_0. field. assumption. Qed. (** The assumption that [z <> 0] was needed in the above theorems to avoid division by zero. If we omitted that assumption, the [field] tactic would leave us with an unprovable subgoal, as in the proof below. *) Theorem frac_r_broken : forall x y z, (x + y) / z = x / z + y /z. Proof. intros x y z. field. Abort. Close Scope R_scope. End RealExample. (** (**********************************************************************) ** Induction principles When we studied the Curry-Howard correspondence, we learned that proofs correspond to programs. That correspondence applies to inductive proofs as well, and as it turns out, inductive proofs correspond to recursive programs. Intuitively, that's because an inductive proof involves an inductive hypothesis---which is an instance of the theorem that is being proved, but applied to a smaller value. Likewise, recursive programs involve recursive calls---which are like another instance of the function that is already being evaluated, but on a smaller value. To get a more concrete idea of what this means, let's look at the proof value (i.e., program) that Coq produces for our original inductive proof in these notes: *) Print app_nil. (** Coq responds: << app_nil = fun (A : Type) (lst : list A) => list_ind (fun lst0 : list A => lst0 ++ nil = lst0) eq_refl (fun (h : A) (t : list A) (IH : t ++ nil = t) => eq_ind_r (fun l : list A => h :: l = h :: t) eq_refl IH) lst : forall (A : Type) (lst : list A), lst ++ nil = lst >> That's dense, but let's start picking it apart. First, we see that [app_nil] is a function that takes in two arguments: [A] and [lst]. Then it immediately applies another function named [list_ind]. That function was defined for us in the standard library, and it's what "implements" induction on lists. Let's check it out: *) Check list_ind. (** Coq responds: << list_ind : forall (A : Type) (P : list A -> Prop), P nil -> (forall (a : A) (l : list A), P l -> P (a :: l)) -> forall l : list A, P l >> We call [list_ind] the _induction principle_ for lists. It is a proposition that says, intuitively, that induction is a valid reasoning principle for lists. In more detail, it takes these arguments: - [A], which is the type of the list elements. - [P], which is the property to be proved by induction. For example, the property being proved in [app_nil] is [fun (lst: list A) => lst ++ nil = lst]. - [P nil], which is a proof that [P] holds of the empty list. In other words, a proof of the base case. - A final argument of type [forall (a : A) (l : list A), P l -> P (a :: l)]. This is the proof of the inductive case. It takes an argument [a], which is the head of a list, [l], which is the tail of a list, and a proof [P l] that [P] holds of [l]. So, [P l] is the inductive hypothesis. The output is of type [P (a :: l)], which is a proof that [P] holds of [a::l]. Finally, [list_ind] returns a value of type [forall l : list A, P l], which is a proof that [P] holds of all lists. Ok, so that's the type of [list_ind]: a proposition asserting that if you have a proof of the base case, and a proof of the inductive case, you can assemble those to prove that a property holds of a list. Next, what's the _value_ of [list_ind]? In other words, what's the proof that [list_ind] itself is actually a true proposition? *) Print list_ind. (** Coq responds: << list_ind = fun (A : Type) (P : list A -> Prop) => list_rect P ... >> So [list_ind] takes in [A] and [P] and just immediately applies another function, [list_rect], to [P]. (The name [rect] is not especially helpful to understand, but alludes to [rec]ursion over a [t]ype.) Before we look at [list_rect]'s actual implementation, let's look at our own equivalent implementation that is easier to read: *) Fixpoint my_list_rect (A : Type) (P : list A -> Prop) (baseCase : P nil) (inductiveCase : forall (h : A) (t : list A), P t -> P (h::t)) (lst : list A) : P lst := match lst with | nil => baseCase | h::t => inductiveCase h t (my_list_rect A P baseCase inductiveCase t) end. (** The arguments to [my_list_rect] are the same as the arguments to [list_ind]: an element type, a property to be proved, a proof of the base case, and a proof of the inductive case. Then [my_list_rect] takes an argument [lst], which is the list for which we want to prove that [P] holds. Finally, [my_list_rect] returns that proof specifically for [lst]. The body of [my_list_rect] constructs the proof that [P] holds of [lst]. It does so by matching against [lst]: - If [lst] is empty, then [my_list_rect] returns the proof of the base case. - If [lst] is [h::t], then [my_list_rect] returns the proof of the inductive case. To construct that proof, it applies [inductiveCase] to [h] and [t] as the head and tail. But [inductiveCase] also requires a final argument, which is the proof that [P] holds of [t]. To construct that proof, [my_list_rect] calls itself recursively on [t]. That recursive call is exactly why we said that inductive proofs are recursive programs. The inductive proof needs evidence that the inductive hypothesis holds of the smaller list, and recursing on that smaller list produces the evidence. It's not immediately obvious, but [my_list_rect] is almost just [fold_right]. Here's how we could implement [fold_right] in Coq, with a slightly different argument order than the same function in OCaml: *) Fixpoint my_fold_right {A B : Type} (init : B) (f : A -> B -> B) (lst : list A) := match lst with | nil => init | h::t => f h (my_fold_right init f t) end. (** Now compare the body of [my_fold_right] with [my_list_rect]: << my_fold_right's body: match lst with | nil => init | h::t => f h (my_fold_right init f t) end. my_list_rect's body: match lst with | nil => baseCase | h::t => inductiveCase h t (my_list_rect A P baseCase inductiveCase t) end. >> Both match against [lst]. If [lst] is empty, both return an initial/base-case value. If [lst] is non-empty, both recurse on the tail, then pass the result of the recursive call to a function ([f] or [inductiveCase]) that combines that result with the head. The only essential difference is that [f] does not take [t] directly as an input, whereas [inductiveCase] does. So there you have it: induction over a list is really just folding over the list, eventually reaching the empty list and returning the proof of the base case for it, then working the way back up the call stack, assembling an ever-larger proof for each element of the list. ##An inductive proof is a recursive program.## Going back to the actual definition of [list_rect], here it is: *) Print list_rect. (** Coq responds: << list_rect = fun (A : Type) (P : list A -> Type) (f : P nil) (f0 : forall (a : A) (l : list A), P l -> P (a :: l)) => fix F (l : list A) : P l := match l as l0 return (P l0) with | nil => f | y :: l0 => f0 y l0 (F l0) end : forall (A : Type) (P : list A -> Type), P nil -> (forall (a : A) (l : list A), P l -> P (a :: l)) -> forall l : list A, P l >> That uses different syntax, but it ends up defining the same function as [my_list_rect]. Whenever you define an inductive type, Coq automatically generates the induction principle and recursive function that implements it for you. For example, we could define our own lists: *) Inductive mylist (A:Type) : Type := | mynil : mylist A | mycons : A -> mylist A -> mylist A. (** Coq automatically generates [mylist_rect] and [mylist_ind] for us: *) Print mylist_ind. Print mylist_rect. (** ** Summary Coq excels as a proof assistant when it comes to proof by induction. Whenever we define an inductive type, Coq generates an induction principle for us automatically. That principle is really a recursive program that knows how to assemble evidence for a proposition, given the constructors of the inductive type. The [induction] tactic manages the proof for us, automatically figuring out what the base case and the inductive case, and automatically generating the inductive hypothesis. ** Terms and concepts - append - base case - field - [fix] - [Fixpoint] - induction - induction principle - inductive case - inductive hypothesis - lemma - Peano natural numbers - [Prop] vs [bool] - ring - searching for library theorems - semi-ring - syntactically smaller restriction on recursive calls ** Tactics - [field] - [induction] - [rewrite] - [ring] - tacticals: [try] ** Further reading - _Software Foundations, Volume 1: Logical Foundations_. # Chapter 2 through 4: Induction, Lists, Poly#. - _Interactive Theorem Proving and Program Development_. Chapters 6 through 10. Available # online from the Cornell library#. *)