InductionProof by Induction

Before getting started, we need to import all of our definitions from the previous chapter:
Require Export Basics.
For the Require Export to work, you first need to use coqc to compile Basics.v into Basics.vo. This is like making a .class file from a .java file, or a .o file from a .c file. There are two ways to do it:
  • In CoqIDE:
    Open Basics.v. In the "Compile" menu, click on "Compile Buffer".
  • From the command line: Either
    make Basics.vo
    (assuming you've downloaded the whole LF directory and have a working 'make' command) or
    coqc Basics.v
    (which should work from any terminal window).
If you have trouble (e.g., if you get complaints about missing identifiers later in the file), it may be because the "load path" for Coq is not set up correctly. The Print LoadPath. command may be helpful in sorting out such issues.

Proof by Induction

We proved in the last chapter that 0 is a neutral element for + on the left, using an easy argument based on simplification. We also observed that proving the fact that it is also a neutral element on the right...
Theorem plus_n_O_firsttry : ∀ n:nat,
  n = n + 0.
... can't be done in the same simple way. Just applying reflexivity doesn't work, since the n in n + 0 is an arbitrary unknown number, so the match in the definition of + can't be simplified.
Proof.
  intros n.
  simpl. (* Does nothing! *)
Abort.
And reasoning by cases using destruct n doesn't get us much further: the branch of the case analysis where we assume n = 0 goes through fine, but in the branch where n = S n' for some n' we get stuck in exactly the same way.
Theorem plus_n_O_secondtry : ∀ n:nat,
  n = n + 0.
Proof.
  intros n. destruct n as [| n'].
  - (* n = 0 *)
    reflexivity. (* so far so good... *)
  - (* n = S n' *)
    simpl. (* ...but here we are stuck again *)
Abort.
We could use destruct n' to get one step further, but, since n can be arbitrarily large, if we just go on like this we'll never finish.
To prove interesting facts about numbers, lists, and other inductively defined sets, we usually need a more powerful reasoning principle: induction.
Recall (from high school, a discrete math course, etc.) the principle of induction over natural numbers: If P(n) is some proposition involving a natural number n and we want to show that P holds for all numbers n, we can reason like this:
  • show that P(O) holds;
  • show that, for any n', if P(n') holds, then so does P(S n');
  • conclude that P(n) holds for all n.
In Coq, the steps are the same: we begin with the goal of proving P(n) for all n and break it down (by applying the induction tactic) into two separate subgoals: one where we must show P(O) and another where we must show P(n') P(S n'). Here's how this works for the theorem at hand:
Theorem plus_n_O : ∀ n:nat, n = n + 0.
Proof.
  intros n. induction n as [| n' IHn'].
  - (* n = 0 *) reflexivity.
  - (* n = S n' *) simpl. rewrite <- IHn'. reflexivity. Qed.
Like destruct, the induction tactic takes an as... clause that specifies the names of the variables to be introduced in the subgoals. Since there are two subgoals, the as... clause has two parts, separated by |. (Strictly speaking, we can omit the as... clause and Coq will choose names for us. In practice, this is a bad idea, as Coq's automatic choices tend to be confusing.)
In the first subgoal, n is replaced by 0. No new variables are introduced (so the first part of the as... is empty), and the goal becomes 0 = 0 + 0, which follows by simplification.
In the second subgoal, n is replaced by S n', and the assumption n' + 0 = n' is added to the context with the name IHn' (i.e., the Induction Hypothesis for n'). These two names are specified in the second part of the as... clause. The goal in this case becomes S n' = (S n') + 0, which simplifies to S n' = S (n' + 0), which in turn follows from IHn'.
Theorem minus_diag : ∀ n,
  minus n n = 0.
Proof.
  (* WORKED IN CLASS *)
  intros n. induction n as [| n' IHn'].
  - (* n = 0 *)
    simpl. reflexivity.
  - (* n = S n' *)
    simpl. rewriteIHn'. reflexivity. Qed.
(The use of the intros tactic in these proofs is actually redundant. When applied to a goal that contains quantified variables, the induction tactic will automatically move them into the context as needed.)

Exercise: 2 stars, recommended (basic_induction)

Prove the following using induction. You might need previously proven results.
Theorem mult_0_r : ∀ n:nat,
  n * 0 = 0.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem plus_n_Sm : ∀ n m : nat,
  S (n + m) = n + (S m).
Proof.
  (* FILL IN HERE *) Admitted.

Theorem plus_comm : ∀ n m : nat,
  n + m = m + n.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem plus_assoc : ∀ n m p : nat,
  n + (m + p) = (n + m) + p.
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars (double_plus)

Consider the following function, which doubles its argument:
Fixpoint double (n:nat) :=
  match n with
  | OO
  | S n'S (S (double n'))
  end.
Use induction to prove this simple fact about double:
Lemma double_plus : ∀ n, double n = n + n .
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, optional (evenb_S)

One inconvenient aspect of our definition of evenb n is the recursive call on n - 2. This makes proofs about evenb n harder when done by induction on n, since we may need an induction hypothesis about n - 2. The following lemma gives an alternative characterization of evenb (S n) that works better with induction:
Theorem evenb_S : ∀ n : nat,
  evenb (S n) = negb (evenb n).
Proof.
  (* FILL IN HERE *) Admitted.

Exercise: 1 star (destruct_induction)

Briefly explain the difference between the tactics destruct and induction.
(* FILL IN HERE *)

Proofs Within Proofs

In Coq, as in informal mathematics, large proofs are often broken into a sequence of theorems, with later proofs referring to earlier theorems. But sometimes a proof will require some miscellaneous fact that is too trivial and of too little general interest to bother giving it its own top-level name. In such cases, it is convenient to be able to simply state and prove the needed "sub-theorem" right at the point where it is used. The assert tactic allows us to do this. For example, our earlier proof of the mult_0_plus theorem referred to a previous theorem named plus_O_n. We could instead use assert to state and prove plus_O_n in-line:
Theorem mult_0_plus' : ∀ n m : nat,
  (0 + n) * m = n * m.
Proof.
  intros n m.
  assert (H: 0 + n = n). { reflexivity. }
  rewriteH.
  reflexivity. Qed.
The assert tactic introduces two sub-goals. The first is the assertion itself; by prefixing it with H: we name the assertion H. (We can also name the assertion with as just as we did above with destruct and induction, i.e., assert (0 + n = n) as H.) Note that we surround the proof of this assertion with curly braces { ... }, both for readability and so that, when using Coq interactively, we can see more easily when we have finished this sub-proof. The second goal is the same as the one at the point where we invoke assert except that, in the context, we now have the assumption H that 0 + n = n. That is, assert generates one subgoal where we must prove the asserted fact and a second subgoal where we can use the asserted fact to make progress on whatever we were trying to prove in the first place.
Another example of assert...
For example, suppose we want to prove that (n + m) + (p + q) = (m + n) + (p + q). The only difference between the two sides of the = is that the arguments m and n to the first inner + are swapped, so it seems we should be able to use the commutativity of addition (plus_comm) to rewrite one into the other. However, the rewrite tactic is not very smart about where it applies the rewrite. There are three uses of + here, and it turns out that doing rewrite plus_comm will affect only the outer one...
Theorem plus_rearrange_firsttry : ∀ n m p q : nat,
  (n + m) + (p + q) = (m + n) + (p + q).
Proof.
  intros n m p q.
  (* We just need to swap (n + m) for (m + n)... seems
     like plus_comm should do the trick! *)

  rewriteplus_comm.
  (* Doesn't work...Coq rewrote the wrong plus! *)
Abort.
To use plus_comm at the point where we need it, we can introduce a local lemma stating that n + m = m + n (for the particular m and n that we are talking about here), prove this lemma using plus_comm, and then use it to do the desired rewrite.
Theorem plus_rearrange : ∀ n m p q : nat,
  (n + m) + (p + q) = (m + n) + (p + q).
Proof.
  intros n m p q.
  assert (H: n + m = m + n).
  { rewriteplus_comm. reflexivity. }
  rewriteH. reflexivity. Qed.

Formal vs. Informal Proof

"Informal proofs are algorithms; formal proofs are code."
What constitutes a successful proof of a mathematical claim? The question has challenged philosophers for millennia, but a rough and ready definition could be this: A proof of a mathematical proposition P is a written (or spoken) text that instills in the reader or hearer the certainty that P is true — an unassailable argument for the truth of P. That is, a proof is an act of communication.
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 that P can be mechanically derived from a certain set of formal logical rules, and the proof is a recipe that guides the program in checking this fact. Such recipes are formal proofs.
Alternatively, the reader can be a human being, in which case the proof will be written in English or some other natural language, and will thus necessarily be informal. Here, the criteria for success are less clearly specified. A "valid" proof is one that makes the reader believe P. But the same proof may be read by many different readers, some of whom may be convinced by a particular way of phrasing the argument, while others may not be. Some readers may be particularly pedantic, inexperienced, or just plain thick-headed; the only way to convince them will be to make the argument in painstaking detail. But other readers, 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, since it is easier for them to fill in the details for themselves than to wade through a written presentation of them. Ultimately, there is no universal standard, because there is no single way of writing an informal proof that is guaranteed to convince every conceivable reader.
In practice, however, mathematicians have developed a rich set of conventions and idioms for writing about complex mathematical objects that — at least within a certain community — make communication fairly reliable. The conventions of this stylized form of communication give a fairly clear standard for judging proofs good or bad.
Because we are using Coq in this course, we will be working heavily with formal proofs. But this doesn't mean we can completely forget about informal ones! Formal proofs are useful in many ways, but they are not very efficient ways of communicating ideas between human beings.
For example, here is a proof that addition is associative:
Theorem plus_assoc' : ∀ n m p : nat,
  n + (m + p) = (n + m) + p.
Proof. intros n m p. induction n as [| n' IHn']. reflexivity.
  simpl. rewriteIHn'. reflexivity. Qed.
Coq is perfectly happy with this. For a human, however, it is difficult to make much sense of it. We can use comments and bullets to show the structure a little more clearly...
Theorem plus_assoc'' : ∀ n m p : nat,
  n + (m + p) = (n + m) + p.
Proof.
  intros n m p. induction n as [| n' IHn'].
  - (* n = 0 *)
    reflexivity.
  - (* n = S n' *)
    simpl. rewriteIHn'. reflexivity. Qed.
... and if you're used to Coq you may be able to 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.
A (pedantic) mathematician might write the proof something like this:
  • Theorem: For any n, m and p,
       n + (m + p) = (n + m) + p.
    Proof: By induction on n.
    • First, suppose n = 0. We must show
        0 + (m + p) = (0 + m) + p.
      This follows directly from the definition of +.
    • Next, suppose n = S n', where
        n' + (m + p) = (n' + m) + p.
      We must show
        (S n') + (m + p) = ((S n') + m) + p.
      By the definition of +, this follows from
        S (n' + (m + p)) = S ((n' + m) + p),
      which is immediate from the induction hypothesis. Qed.
The overall form of the proof is basically similar, and of course this is no accident: 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).

Exercise: 2 stars, advanced, recommended (plus_comm_informal)

Translate your solution for plus_comm into an informal proof:
Theorem: Addition is commutative.
Proof: (* FILL IN HERE *)

Exercise: 2 stars, optional (beq_nat_refl_informal)

Write an informal proof of the following theorem, using the informal proof of plus_assoc as a model. Don't just paraphrase the Coq tactics into English!
Theorem: true = beq_nat n n for any n.
Proof: (* FILL IN HERE *)