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2 Basics
Our laboratory for this course is the Coq proof assistant. Coq can be seen as a combination of two things:
- a simple and slightly idiosyncratic (but, in its way, extremely expressive) programming language, and
- a set of tools for stating logical assertions (including assertions about the behavior of programs) and assembling evidence of their truth. We will be investigating both aspects together.
2.1 Enumerated Types
In Coq’s programming language, almost nothing is built in—not even booleans or numbers! Instead, it provides powerful tools for defining new types of data and functions that process and transform them. Let’s start with a very simple example. The following definition tells Coq that we are defining a new set of data values. The set is called day and its members are monday, tuesday, etc. The lines of the definition can be read “monday is a day, tuesday is a day, etc.” Inductive day : Set := | monday : day | tuesday : day | wednesday : day | thursday : day | friday : day | saturday : day | sunday : day. Having defined this set, we can write functions that operate on its members.
6 2 Basics Definition next_weekday (d:day) : day := match d with | monday => tuesday | tuesday => wednesday | wednesday => thursday | thursday => friday | friday => monday | saturday => monday | sunday => monday end. One thing to note is that the argument and return types of this function are explicitly declared. Like most functional programming languages, Coq can often work out these types even if they are not given explicitly, but we’ll always include them to make reading easier. Having defined a function, we might like to check how it works on some examples. There are actually three different ways to do this in Coq.
- We can use the command Eval simpl to evaluate a compound expres- sion involving next_weekday. For example, if we give Coq the following input Eval simpl in (next_weekday (next_weekday saturday)). it will print the simplified result and its type: ! (^) = tuesday : day The keyword simpl (for “simplify”) tells Coq precisely how to evaluate the expression we give it. For the moment, simpl is the only one we’ll need; later on we’ll see some alternatives that are sometimes useful. If you have a computer handy, now would be an excellent moment to fire up the Coq interpreter under your favorite IDE—either CoqIde or Proof General—and try this for yourself. Load the file Basics.v from the book’s accompanying Coq sources, find the above example a little ways from the top, send it to Coq, and observe the result.
- We can record what we expect the result to be in the form of a Coq Example: Example test_next_weekday: (next_weekday (next_weekday saturday)) = tuesday.
8 2 Basics end. Definition ifb (b1 b2 b3:bool) : bool := match b1 with | true => b | false => b end. Definition andb (b1:bool) (b2:bool) : bool := ifb b1 b2 false. Definition orb (b1:bool) (b2:bool) : bool := ifb b1 true b2. The last three illustrate the syntax for multi-argument function definitions. The following four “unit tests” constitute a complete specification—a truth table—for the orb function: Example test_orb1: (orb true false) = true. Example test_orb2: (orb false false) = false. Example test_orb3: (orb false true ) = true. Example test_orb4: (orb true true ) = true. The proofs of these properties are precisely the same as what we saw above. From now on, proofs will generally be omitted, unless they are particularly relevant to the discussion. They can always be found in full in the accompa- nying Coq sources. 2.2.1 EXERCISE [!]: In the Coq source file Basics.v, you will find a com- ment containing incomplete implementations of two more boolean func- tions, nandb and and3b. Uncomment and finish them, making sure that Coq can verify the provided unit tests. The Check command causes Coq to print the type of an expression. For example, when presented with Check (negb true). Coq prints ! (^) negb true : bool Functions like negb itself are also data values, just like true and false. Their types are called function types. Check negb. ! (^) negb : bool → bool
2.3 Numbers 9 The type of negb is pronounced “bool arrow bool” and can be read, “Given an input of type bool, this function produces an output of type bool.” Simi- larly, the type of andb, written bool→bool→bool, can be read, “Given two inputs, both of type bool, this function produces an output of type bool.”
2.3 Numbers
The types we have defined so far are examples of “enumerated types”: their definitions explicitly enumerate a finite collection of elements. A more inter- esting way of defining a type is to give a collection of inductive rules describ- ing its elements. For example, we can define the natural numbers as follows: Inductive nat : Set := | O : nat | S : nat → nat. The clauses of this definition can be read:
- O is a natural number (note that this is the letter “O,” not the numeral “ 0 ”).
- S is a constructor that takes a natural number and yields another one—that is, if n is a natural number, then S n is too. We can write simple functions that pattern match on natural numbers just as we did above: Definition pred (n : nat) : nat := match n with | O => O | S n′^ => n′ end. Definition minustwo (n : nat) : nat := match n with | O => O | S O => O | S (S n′) => n′ end. Because natural numbers are such a pervasive form of data, Coq provides a tiny bit of special built-in magic for parsing and printing them: ordinary arabic numerals can be used as an alternative to the “unary” notation defined by the constructors S and O. Coq prints numbers in arabic form by default,
2.3 Numbers 11 Fixpoint evenb (n:nat) {struct n} : bool := match n with | O => true | S O => false | S (S n′) => evenb n′ end. The most important thing to note about this definition is the annotation {struct n} on the first line. This instructs Coq to check that we are per- forming a “structural recursion” over the argument n—i.e., that we make recursive calls only on strictly smaller values of n. This implies that all calls to evenb will eventually terminate. We can define oddb by a similar Fixpoint declaration, but here is a sim- pler definition that will be easier to work with later: Definition oddb (n:nat) : bool := negb (evenb n). Naturally, we can also define multi-argument functions by recursion. Fixpoint plus (n : nat) (m : nat) {struct n} : nat := match n with | O => m | S n′^ => S (plus n′^ m) end. Adding three to two now gives us five, as we’d expect. Eval simpl in (plus (S (S (S O))) (S (S O))). ! (^) = 5 : nat The simplification that Coq performs to reach this conclusion can be visual- ized as follows: plus (S (S (S O))) (S (S O)) = S (plus (S (S O)) (S (S O))) by the second clause of the match = S (S (plus (S O) (S (S O)))) by the second clause of the match = S (S (S (plus O (S (S O))))) by the second clause of the match = S (S (S (S (S O)))) by the first clause of the match. As a notational convenience, if two or more arguments have the same type, they can be written together. In the following definition, (n m : nat) means just the same as if we had written (n : nat) (m : nat).
12 2 Basics Fixpoint mult (n m : nat) {struct n} : nat := match n with | O => O | S n′^ => plus m (mult n′^ m) end. Other arithmetic functions like minus and exp can be defined similarly (see Basics.v). 2.3.1 EXERCISE [!]: Recall that the factorial function is defined like this in con- ventional mathematical notation: factorial( 0 ) = 1 factorial(n) = n ∗ (factorial(n − 1 )) if n > 0 Translate this into Coq’s notation. (An incomplete definition can be found in a comment in Basics.v.) When we say that Coq comes with nothing built-in, we really mean it: even equality testing for numbers is a user-defined operation! Fixpoint beq_nat (n m : nat) {struct n} : bool := match n with | O => match m with | O => true | S m′^ => false end | S n′^ => match m with | O => false | S m′^ => beq_nat n′^ m′ end end. 2.3.2 EXERCISE [!]: Complete the definition (in Basics.v) of the comparison function blt_nat.
2.4 Proof by Simplification
Now that we’ve defined a few datatypes and functions, let’s turn to the ques- tion of how to state and prove properties of their behavior. Actually, in a sense, we’ve already started doing this: each Example in the previous sections makes a precise claim about the behavior of some function on some particular inputs. The proofs of these claims were always the same:
14 2 Basics The intros tactic permits us to do this, by moving one or more quantifiers or hypotheses from the goal to a context of current assumptions. For example, here is a similar theorem with a slightly different proof. Theorem plus_1_l : forall n:nat, plus 1 n = S n. Proof. intros n. reflexivity.! Step through this proof in Coq and notice how the goal and context change at each point.
2.6 Proof by Rewriting
Here is a slightly more interesting theorem: Theorem plus_id_example : forall n m:nat, n = m → plus n n = plus m m. Instead of making a completely universal claim about all numbers n and m, this theorem talks about a more specialized property that only holds when n = m. The arrow symbol, written → in typeset code and -> in ascii .v files, is pronounced implies. Since n and m are arbitrary numbers, we can’t just use simplification to prove this theorem. Instead, we prove it by observing that, if we are assum- ing n = m, then we can replace n with m in the goal statement and obtain an equality with the same expression on both sides. The tactic that tells Coq to perform this replacement is called rewrite. Proof. intros n m. (* move both quantifiers into the context (^) ) intros H. ( move the hypothesis into the context (^) ) rewrite → H. ( Rewrite the goal using the hypothesis (^) *) reflexivity.! The first line of the proof moves the universally quantified variables n and m into the context. The second moves the hypothesis n = m into the con- text and gives it the name H. The third tells Coq to rewrite the current goal (plus n n = plus m m) by replacing the left side of the equality hypothesis H with the right side. (The arrow symbol in the rewrite has nothing to do with implication: it tells Coq to apply the rewrite from left to right. To rewrite from right to left, you can use rewrite ←. Try making this change in the above proof and see what difference it makes in Coq’s behavior.)
2.7 Case Analysis 15 2.6.1 EXERCISE [!]: Prove plus_id_exercise, found in Basics.v. Notice that in the file, the current proof consists of the keyword Admitted. The Admitted command tells Coq that we want to give up trying to prove this theorem and just accept it as a given. This can be useful for developing longer proofs, since we can state subsidiary facts that we believe will be useful for making some larger argument, use Admitted to accept them on faith for the moment, and continue thinking about the larger argument until we are sure it makes sense; then we can go back and fill in the proofs we skipped. Be careful, though: every time you say Admitted you are leaving a door open for total nonsense to enter Coq’s nice, rigorous, formally checked world! We can also use the rewrite tactic with a previously proved theorem instead of a hypothesis from the context. Theorem mult_0_plus : forall n m : nat, mult (plus 0 n) m = mult n m. Proof. intros n m. rewrite → plus_0_l. reflexivity.! 2.6.2 EXERCISE [!]: Prove the theorem mult_1_plus in Basics.v.
2.7 Case Analysis
Of course, not everything can be proved by simple calculation: In general, unknown, hypothetical values (arbitrary numbers, booleans, lists, etc.) can show up in the "head position" of functions that we want to reason about, blocking simplification. For example, if we try to prove the following fact using the simpl tactic as above, we get stuck. Theorem plus_1_neq_0_firsttry : forall n, beq_nat (plus n 1) 0 = false. The reason for this is that the definitions of both beq_nat and plus begin by performing a match on their first argument. But here, the first argument to plus is the unknown number n and the argument to neq_nat is the com- pound expression plus n 1; neither can be simplified. What we need is to be able to consider the possible forms of n separately. If n is O, then we can calculate the final result of beq_nat (plus n 1) 0 and check that it is, indeed, false. And if n = S n′^ for some n′, then, although we don’t know exactly what number plus n 1 yields, we can calculate that,
2.8 Naming Cases 17 Fixpoint plus (n : nat) (m : nat) {struct n} : nat := match n with | O => m | S n′^ => S (plus n′^ m) end. What will Coq print in response to this query? Eval simpl in (forall n, plus n 0 = n). What will Coq print in response to this query? Eval simpl in (forall n, plus 0 n = n). Briefly explain the difference.
2.8 Naming Cases
The fact that there is no explicit command for moving from one branch of a case analysis to the next can make proof scripts rather hard to read. In larger proofs, with nested case analyses, it can even become hard to stay oriented when you’re sitting with Coq and stepping through the proof. (Imagine try- ing to remember that the first five subgoals belong to the inner case analysis and the remaining seven are the cases that are left of the outer one...) Disci- plined use of indentation and comments can help, but a better way is to use the Case tactic. Theorem andb_true_l : forall b c, andb b c = true → b = true. Proof. intros b c H. destruct b. Case "b = true". reflexivity. Case "b = false". rewrite ← H. reflexivity.! Case does something very trivial: It simply adds a string that we choose (tagged with the identifier “Case”) to the context for the current goal. When subgoals are generated, this string is carried over into their contexts. When the last of these subgoals is finally proved and the next top-level goal (a sib- ling of the current one) becomes active, this string will no longer appear in the context and we will be able to see that the case where we introduced it is complete. Also, as a sanity check, if we try to execute a new Case tactic
18 2 Basics while the string left by the previous one is still in the context, we get a nice clear error message. For nested case analyses (i.e., when we want to use a destruct to solve a goal that has itself been generated by a destruct), there is an SCase (“sub- case”) tactic. For deeper nesting there are SSCase, SSSCase, etc. Case and its friends are not actually built-in facilities of Coq: they can be programmed using “Ltac,” Coq’s language for writing user-defined tactics. You can see the actual definitions at this point in the Basics.v file, if you’re curious, but there’s no need to understand any of the details of how they work. 2.8.1 EXERCISE [!]: Prove the theorem andb_true_r in Basics.v, using Case (or SCase) to mark the branches of each destruct. There are no hard and fast rules for how proofs should be formatted in Coq—in particular, where lines should be broken and how sections of the proof should be indented to indicate their nested structure. However, if the places where multiple subgoals are generated are marked with explicit Case tactics placed at the beginning of lines, then the proof will be readable almost no matter what choices are made about other aspects of layout. This is a good place to mention one other piece of (possibly obvious) ad- vice about line lengths. Beginning Coq users sometimes tend to the extremes, either writing each tactic on its own line or entire proofs on one line. Good style lies somewhere in the middle. In particular, one convention (not just for Coq proofs, but for all programming) is to limit yourself to 80 character lines. Lines longer than this are hard to read and can be inconvenient to dis- play and print. Many editors have features that help enforce this. Of course, there is no need to make every line this long, but it’s a simple and reasonable upper bound.
2.9 Induction
We proved above that 0 is a neutral element for plus on the left using a simple partial evaluation argument. The fact that it is also a neutral element on the right... Theorem plus_0_r : forall n:nat, plus n 0 = n. ... cannot be proved in the same simple way. Just applying reflexivity doesn’t work: the n in plus 0 n is an arbitrary unknown number, so the match in the definition of plus can’t be simplified. And reasoning by cases using destruct n doesn’t get us much further: the branch of the case anal- ysis where we assume n = 0 goes through, but in the branch where n = S n′
20 2 Basics Theorem plus_n_Sm : forall n m : nat, S (plus n m) = plus n (S m). Theorem plus_comm : forall n m : nat, plus n m = plus m n.
2.10 Formal vs. Informal Proofs
The question of what, exactly, constitutes a proof of a mathematical claim has challenged philosophers throughout the ages. A rough and ready definition, though, could be this: a proof of a mathematical proposition P is a written (or, sometimes, spoken) text that instills in the reader or hearer the certainty that P is true. That is, a proof is an act of communication. Now, acts of communication may involve different sorts of readers. On one hand, the “reader” can be a program like Coq, in which case the “belief” that is instilled is a simple mechanical check that P can be derived from a certain set of formal logical rules, and the proof is a recipe that guides the program in performing this check. Such recipies are called formal proofs. Alternatively, the reader can be a human being, in which case the proof will be written in English or some other natural language, thus necessarily informal. Here, the criteria for success are less clearly specified. A “good” proof is one that makes the reader believe P. But the same proof may be read by many different readers, some of whom may be convinced by a par- ticular way of phrasing the argument, while others may not be. One reader may be particularly pedantic, inexperienced, or just plain thick-headed; the only way to convince them will be to make the argument in painstaking de- tail. But another reader, more familiar in the area, may find all this detail so overwhelming that they lose the overall thread. All they want is to be told the main ideas, because it is easier to fill in the details for themselves. Ulti- mately, there is no universal standard, because is no single way of writing an informal proof that is guaranteed to convince every conceivable reader. In practice, however, mathematicians have developed a rich set of conventions and idioms for writing about complex mathematical objects that, within a certain community, make communication fairly reliable. The conventions of this stylized form of communication give a fairly clear standard for judging proofs good or bad. Because we will be using Coq in this course, we will be working heavily with formal proofs. But this doesn’t mean we can ignore the informal ones! Formal proofs are useful in many ways, but they are not very efficient ways of communicating ideas between human beings.
2.11 Proofs Within Proofs 21 For example, consider this statement: Theorem plus_assoc : forall n m p : nat, plus n (plus m p) = plus (plus n m) p. Coq is perfectly happy with this as a proof: Proof. intros n m p. induction n as [| n′]. Case "n = 0". reflexivity. Case "n = S n′". simpl. rewrite → IHn′. reflexivity.! For a human, however, it is difficult to make much sense of this. If you’re used to Coq you can probably step through the tactics one after the other in your mind and imagine the state of the context and goal stack at each point, but if the proof were even a little bit more complicated this would be next to impossible. Instead, a mathematician would write it as in Figure 2-
- The overall form of the proof is basically similar. (This is no accident, of course: Coq has been designed so that its induction tactic generates the same sub-goals, in the same order, as the bullet points that a mathematician would write.) But there are significant differences of detail: the formal proof is much more explicit in some ways (e.g., the use of reflexivity) but much less explicit in others; in particular, the “proof state” at any given point in the Coq proof is completely implicit, whereas the informal proof reminds the reader several times where things stand. 2.10.1 EXERCISE [!!]: Write an informal proof corresponding to your formal proof of plus_comm (from exercise 2.9.1). Use the informal proof of plus_assoc as a model. Don’t just paraphrase the Coq tactics into English! 2.10.2 EXERCISE [!!]: In Basics.v, you will find a careful informal proof of a theorem called beq_nat_refl. Write a corresponding Coq proof.
2.11 Proofs Within Proofs
In Coq, as in informal mathematics, large proofs are very often broken into sequence of theorems, with later proofs referring to earlier theorems. Occa- sionally, however, a proof will need some miscellaneous fact that is too trivial (and of too little general interest) to bother giving it its own top-level name. In such cases, it is convenient to be able to simply state and prove the needed “sub-theorem” right at the point where it is used. The assert tactic allows
2.11 Proofs Within Proofs 23 the asserted fact and a second subgoal where we can use the asserted fact to make progress on whatever we were trying to prove in the first place. Actually, assert will turn out to be handy in many sorts of situations. For example, suppose we want to prove that plus (plus n m) (plus p q) = plus (plus m n) (plus p q). The only difference between the two sides of the = is that the arguments m and n to the first inner plus are swapped, so it seems we should be able to use the commutativity of addition (plus_comm) to rewrite one into the other. However, the rewrite tactic is a little stupid about where it applies the rewrite. There are three uses of plus here, and it turns out that do- ing rewrite → plus_comm will affect only the outer one. (Try it!) To get plus_comm to apply at the point where we want it, we can introduce a local lemma stating that plus n m = plus m n (for the particular m and n that we are talking about here), prove this lemma using plus_comm, and then use this lemma to do the desired rewrite. Proof. intros n m p q. assert (plus n m = plus m n). Case "Proof of assertion". rewrite → plus_comm. reflexivity. rewrite → H. reflexivity.! 2.11.1 EXERCISE [!!]: Use assert to help prove the theorem plus_swap in Basics.v. Then use plus_swap to prove mult_comm. 2.11.2 EXERCISE [!!]: The theorem evenb_n__oddb_Sn in Basics.v states that, if n is even, then its successor is odd. Prove it. 2.11.3 EXERCISE [!!]: Take a piece of paper. Find the section marked “More exer- cises” in Basics.v. For each of the theorems there, first think about whether (a) it can be proved using only simplification and rewriting, (b) it also re- quires case analysis (destruct), or (c) it also requires induction. Write down your prediction. Then fill in the proof. (There is no need to turn in your piece of paper; this is just to encourage you to think before hacking!)
