InductionProof by Induction

Separate Compilation

Before getting started on this chapter, we need to import all of our definitions from the previous chapter:
From LF Require Export Basics.
For this Require Export command to work, Coq needs to be able to find a compiled version of Basics.v, called Basics.vo, in a directory associated with the prefix LF. This file is analogous to the .class files compiled from .java source files and the .o files compiled from .c files.
First create a file named _CoqProject containing the following line (if you obtained the whole volume "Logical Foundations" as a single archive, a _CoqProject should already exist and you can skip this step):
      -Q . LF This maps the current directory (".", which contains Basics.v, Induction.v, etc.) to the prefix (or "logical directory") "LF". Proof General and CoqIDE read _CoqProject automatically, so they know to where to look for the file Basics.vo corresponding to the library LF.Basics.
Once _CoqProject is thus created, there are various ways to build Basics.vo:
  • In Proof General or CoqIDE, the compilation should happen automatically when you submit the Require line above to PG.
  • If you want to compile from the command line, generate a Makefile using the coq_makefile utility, which comes installed with Coq (if you obtained the whole volume as a single archive, a Makefile should already exist and you can skip this step):
             coq_makefile -f _CoqProject *.v -o Makefile Note: You should rerun that command whenever you add or remove Coq files to the directory.
    Now you can compile Basics.v by running make with the corresponding .vo file as a target:
             make Basics.vo All files in the directory can be compiled by giving no arguments:
             make Under the hood, make uses the Coq compiler, coqc. You can also run coqc directly:
             coqc -Q . LF Basics.v But make also calculates dependencies between source files to compile them in the right order, so make should generally be preferred over explicit coqc.
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.
In particular, if you see a message like
        Compiled library Foo makes inconsistent assumptions over
        library Bar
check whether you have multiple installations of Coq on your machine. It may be that commands (like coqc) that you execute in a terminal window are getting a different version of Coq than commands executed by Proof General or CoqIDE.
  • Another common reason is that the library Bar was modified and recompiled without also recompiling Foo which depends on it. Recompile Foo, or everything if too many files are affected. (Using the third solution above: make clean; make.)
One more tip for CoqIDE users: If you see messages like Error: Unable to locate library Basics, a likely reason is inconsistencies between compiling things within CoqIDE vs using coqc from the command line. This typically happens when there are two incompatible versions of coqc installed on your system (one associated with CoqIDE, and one associated with coqc from the terminal). The workaround for this situation is compiling using CoqIDE only (i.e. choosing "make" from the menu), and avoiding using coqc directly at all.

Proof by Induction

We can prove that 0 is a neutral element for + on the left using just reflexivity. But the proof that it is also a neutral element on the right ...
Theorem add_0_r_firsttry : n:nat,
  n + 0 = n.
... 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.
  intros n.
  simpl. (* Does nothing! *)
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 add_0_r_secondtry : n:nat,
  n + 0 = n.
  intros n. destruct n as [| n'] eqn:E.
  - (* n = 0 *)
    reflexivity. (* so far so good... *)
  - (* n = S n' *)
    simpl. (* ...but here we are stuck again *)
We could use destruct n' to get one step further, but, since n can be arbitrarily large, we'll never get all the there if we just go on like this.
To prove interesting facts about numbers, lists, and other inductively defined sets, we often 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 add_0_r : n:nat, n + 0 = n.
  intros n. induction n as [| n' IHn'].
  - (* n = 0 *) reflexivity.
  - (* n = S n' *) simpl. rewriteIHn'. 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_n_n : n,
  minus n n = 0.
  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, standard, especially useful (basic_induction)

Prove the following using induction. You might need previously proven results.
Theorem mul_0_r : n:nat,
  n × 0 = 0.
  (* FILL IN HERE *) Admitted.

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

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

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

Exercise: 2 stars, standard (double_plus)

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

Exercise: 2 stars, standard (eqb_refl)

The following theorem relates the computational equality =? on nat with the definitional equality = on bool.
Theorem eqb_refl : n : nat,
  (n =? n) = true.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (even_S)

One inconvenient aspect of our definition of even n is the recursive call on n - 2. This makes proofs about even 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 even (S n) that works better with induction:
Theorem even_S : n : nat,
  even (S n) = negb (even n).
  (* FILL IN HERE *) Admitted.

Exercise: 1 star, standard, optional (destruct_induction)

Briefly explain the difference between the tactics destruct and induction.

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 involve 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.
Theorem mult_0_plus' : n m : nat,
  (n + 0 + 0) × m = n × m.
  intros n m.
  assert (H: n + 0 + 0 = n).
    { rewrite add_comm. simpl. rewrite add_comm. reflexivity. }
  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 (n + 0 + 0 = 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 n + 0 + 0 = 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.
As another 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 (add_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 add_comm will affect only the outer one...
Theorem plus_rearrange_firsttry : n m p q : nat,
  (n + m) + (p + q) = (m + n) + (p + q).
  intros n m p q.
  (* We just need to swap (n + m) for (m + n)... seems
     like add_comm should do the trick! *)

  rewrite add_comm.
  (* Doesn't work... Coq rewrites the wrong plus! :-( *)
To use add_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 add_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).
  intros n m p q.
  assert (H: n + m = m + n).
  { rewrite add_comm. reflexivity. }
  rewrite H. 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 add_assoc' : n m p : nat,
  n + (m + p) = (n + m) + p.
Proof. intros n m p. induction n as [| n' IHn']. reflexivity.
  simpl. rewrite IHn'. 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 add_assoc'' : n m p : nat,
  n + (m + p) = (n + m) + p.
  intros n m p. induction n as [| n' IHn'].
  - (* n = 0 *)
  - (* n = S n' *)
    simpl. rewrite IHn'. reflexivity. Qed.
... and if you're used to Coq you might 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 that
              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 now show that
              (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, especially useful (add_comm_informal)

Translate your solution for add_comm into an informal proof:
Theorem: Addition is commutative.
Proof: (* FILL IN HERE *)
(* Do not modify the following line: *)
Definition manual_grade_for_add_comm_informal : option (nat×string) := None.

Exercise: 2 stars, standard, optional (eqb_refl_informal)

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

More Exercises

Exercise: 3 stars, standard, especially useful (mul_comm)

Use assert to help prove add_shuffle3. You don't need to use induction yet.
Theorem add_shuffle3 : n m p : nat,
  n + (m + p) = m + (n + p).
  (* FILL IN HERE *) Admitted.
Now prove commutativity of multiplication. You will probably want to look for (or define and prove) a "helper" theorem to be used in the proof of this one. Hint: what is n × (1 + k)?
Theorem mul_comm : m n : nat,
  m × n = n × m.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (plus_leb_compat_l)

If a hypothesis has the form H: P a = b, then rewrite H will rewrite a to b in the goal, and add P as a new subgoal. Use that in the inductive step of this exercise.
Check leb.

Theorem plus_leb_compat_l : n m p : nat,
  n <=? m = true (p + n) <=? (p + m) = true.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars, standard, optional (more_exercises)

Take a piece of paper. For each of the following theorems, first think about whether (a) it can be proved using only simplification and rewriting, (b) it also requires 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 reflect before you hack!)
Theorem leb_refl : n:nat,
  (n <=? n) = true.
  (* FILL IN HERE *) Admitted.

Theorem zero_neqb_S : n:nat,
  0 =? (S n) = false.
  (* FILL IN HERE *) Admitted.

Theorem andb_false_r : b : bool,
  andb b false = false.
  (* FILL IN HERE *) Admitted.

Theorem S_neqb_0 : n:nat,
  (S n) =? 0 = false.
  (* FILL IN HERE *) Admitted.

Theorem mult_1_l : n:nat, 1 × n = n.
  (* FILL IN HERE *) Admitted.

Theorem all3_spec : b c : bool,
    (andb b c)
    (orb (negb b)
         (negb c))
  = true.
  (* FILL IN HERE *) Admitted.

Theorem mult_plus_distr_r : n m p : nat,
  (n + m) × p = (n × p) + (m × p).
  (* FILL IN HERE *) Admitted.

Theorem mult_assoc : n m p : nat,
  n × (m × p) = (n × m) × p.
  (* FILL IN HERE *) Admitted.

Exercise: 2 stars, standard, optional (add_shuffle3')

The replace tactic allows you to specify a particular subterm to rewrite and what you want it rewritten to: replace (t) with (u) replaces (all copies of) expression t in the goal by expression u, and generates t = u as an additional subgoal. This is often useful when a plain rewrite acts on the wrong part of the goal.
Use the replace tactic to do a proof of add_shuffle3', just like add_shuffle3 but without needing assert.
Theorem add_shuffle3' : n m p : nat,
  n + (m + p) = m + (n + p).
  (* FILL IN HERE *) Admitted.

Nat to Bin and Back to Nat

Recall the bin type we defined in Basics:
Inductive bin : Type :=
  | Z
  | B0 (n : bin)
  | B1 (n : bin)
Before you start working on the next exercise, replace the stub definitions of incr and bin_to_nat, below, with your solution from Basics. That will make it possible for this file to be graded on its own.
Fixpoint incr (m:bin) : bin
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Fixpoint bin_to_nat (m:bin) : nat
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
In Basics, we did some unit testing of bin_to_nat, but we didn't prove its correctness. Now we'll do so.

Exercise: 3 stars, standard, especially useful (binary_commute)

Prove that the following diagram commutes:
              bin ----------------------> bin
               |                           |
    bin_to_nat |                           |  bin_to_nat
               |                           |
               v                           v
              nat ----------------------> nat
That is, incrementing a binary number and then converting it to a (unary) natural number yields the same result as first converting it to a natural number and then incrementing.
If you want to change your previous definitions of incr or bin_to_nat to make the property easier to prove, feel free to do so!
Theorem bin_to_nat_pres_incr : b : bin,
  bin_to_nat (incr b) = 1 + bin_to_nat b.
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars, standard (nat_bin_nat)

Write a function to convert natural numbers to binary numbers.
Fixpoint nat_to_bin (n:nat) : bin
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Prove that, if we start with any nat, convert it to bin, and convert it back, we get the same nat which we started with.
Hint: This proof should go through smoothly using the previous exercise about incr as a lemma. If not, revisit your definitions of the functions involved and consider whether they are more complicated than necessary: the shape of a proof by induction will match the recursive structure of the program being verified, so make the recursions as simple as possible.
Theorem nat_bin_nat : n, bin_to_nat (nat_to_bin n) = n.
  (* FILL IN HERE *) Admitted.

Bin to Nat and Back to Bin (Advanced)

The opposite direction -- starting with a bin, converting to nat, then converting back to bin -- turns out to be problematic. That is, the following theorem does not hold.
Theorem bin_nat_bin_fails : b, nat_to_bin (bin_to_nat b) = b.
Let's explore why that theorem fails, and how to prove a modified version of it. We'll start with some lemmas that might seem unrelated, but will turn out to be relevant.

Exercise: 2 stars, advanced (double_bin)

Prove this lemma about double, which we defined earlier in the chapter.
Lemma double_incr : n : nat, double (S n) = S (S (double n)).
  (* FILL IN HERE *) Admitted.
Now define a similar doubling function for bin.
Definition double_bin (b:bin) : bin
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
Check that your function correctly doubles zero.
Example double_bin_zero : double_bin Z = Z.
(* FILL IN HERE *) Admitted.
Prove this lemma, which corresponds to double_incr.
Lemma double_incr_bin : b,
    double_bin (incr b) = incr (incr (double_bin b)).
  (* FILL IN HERE *) Admitted.
Let's return to our desired theorem:
Theorem bin_nat_bin_fails : b, nat_to_bin (bin_to_nat b) = b.
The theorem fails because there are some bin such that we won't necessarily get back to the original bin, but instead to an "equivalent" bin. (We deliberately leave that notion undefined here for you to think about.)
Explain in a comment, below, why this failure occurs. Your explanation will not be graded, but it's important that you get it clear in your mind before going on to the next part. If you're stuck on this, think about alternative implementations of double_bin that might have failed to satisfy double_bin_zero yet otherwise seem correct.
To solve that problem, we can introduce a normalization function that selects the simplest bin out of all the equivalent bin. Then we can prove that the conversion from bin to nat and back again produces that normalized, simplest bin.

Exercise: 4 stars, advanced (bin_nat_bin)

Define normalize. You will need to keep its definition as simple as possible for later proofs to go smoothly. Do not use bin_to_nat or nat_to_bin, but do use double_bin.
Hint: Structure the recursion such that it always reaches the end of the bin and process each bit only once. Do not try to "look ahead" at future bits.
Fixpoint normalize (b:bin) : bin
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.
It would be wise to do some Example proofs to check that your definition of normalize works the way you intend before you proceed. They won't be graded, but fill them in below.
Finally, prove the main theorem. The inductive cases could be a bit tricky.
Hint 1: Start by trying to prove the main statement, see where you get stuck, and see if you can find a lemma -- perhaps requiring its own inductive proof -- that will allow the main proof to make progress. You might end up with a couple of these.
Hint 2: Lemma double_incr_bin that you proved above will be helpful, too.
Theorem bin_nat_bin : b, nat_to_bin (bin_to_nat b) = normalize b.
  (* FILL IN HERE *) Admitted.
(* 2022-05-16 18:01 *)