# InductionProof by Induction

The next line imports all of our definitions from the previous chapter.

Require Export Basics.

For it to work, you 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.)
Here are two ways to compile your code:
• CoqIDE:
Open Basics.v. In the "Compile" menu, click on "Compile Buffer".
• Command line:
Run coqc Basics.v

# 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 trying to remember that the first five subgoals belong to the inner case analysis and the remaining seven cases are what remains of the outer one...) Disciplined use of indentation and comments can help, but a better way is to use the Case tactic.

Here's an example of how Case is used. Step through the following proof and observe how the context changes.

Theorem andb_true_elim1 : b c : bool,
andb b c = true b = true.
Proof.
intros b c H.
destruct b.
Case "b = true". (* <----- here *)
reflexivity.
Case "b = false". (* <---- and here *)
rewrite H.
reflexivity.
Qed.

Case does something very straightforward: 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 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 while the string left by the previous one is still in the context, we get a nice clear error message.
For nested case analyses (e.g., when we want to use a destruct to solve a goal that has itself been generated by a destruct), there is an SCase ("subcase") tactic.

#### Exercise: 2 stars (andb_true_elim2)

Prove andb_true_elim2, marking cases (and subcases) when you use destruct.

Theorem andb_true_elim2 : b c : bool,
andb b c = true c = true.
Proof.
(* FILL IN HERE *) Admitted.
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) advice 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 reasonable convention is to limit yourself to 80-character lines. Lines longer than this are hard to read and can be inconvenient to display and print. Many editors have features that help enforce this.

# Proof by Induction

We proved in the last chapter that 0 is a neutral element for + on the left using a simple argument. The fact that it is also a neutral element on the right...

Theorem plus_0_r_firsttry : n:nat,
n + 0 = n.

... cannot be proved in the same simple way. Just applying reflexivity doesn't work: 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, but in the branch where n = S n' for some n' we get stuck in exactly the same way. We could use destruct n' to get one step further, but since n can be arbitrarily large, if we try to keep on like this we'll never be done.

Theorem plus_0_r_secondtry : n:nat,
n + 0 = n.
Proof.
intros n. destruct n as [| n'].
Case "n = 0".
reflexivity. (* so far so good... *)
Case "n = S n'".
simpl. (* ...but here we are stuck again *)
Abort.

To prove such facts — indeed, to prove most interesting facts about numbers, lists, and other inductively defined sets — we need a more powerful reasoning principle: induction.
Recall (from high school) 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 but the order is backwards: 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: first showing P(O) and then showing P(n') P(S n'). Here's how this works for the theorem we are trying to prove at the moment:

Theorem plus_0_r : n:nat, n + 0 = n.
Proof.
intros n. induction n as [| n'].
Case "n = 0". reflexivity.
Case "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. In the first branch, n is replaced by 0 and the goal becomes 0 + 0 = 0, which follows by simplification. In the second, 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'). The goal in this case becomes (S n') + 0 = S n', which simplifies to S (n' + 0) = S n', which in turn follows from the induction hypothesis.

Theorem minus_diag : n,
minus n n = 0.
Proof.
(* WORKED IN CLASS *)
intros n. induction n as [| n'].
Case "n = 0".
simpl. reflexivity.
Case "n = S n'".
simpl. rewrite IHn'. reflexivity. Qed.

#### Exercise: 2 stars (basic_induction)

Prove the following lemmas 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: 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 very often broken into a sequence of theorems, with later proofs referring to earlier theorems. Occasionally, 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 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 can also 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).
Case "Proof of assertion". reflexivity.
rewrite H.
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. (Note that we could also name the assertion with as just as we did above with destruct and induction, i.e., assert (0 + n = n) as H. Also note that we mark the proof of this assertion with a Case, both for readability and so that, when using Coq interactively, we can see when we're finished proving the assertion by observing when the "Proof of assertion" string disappears from the context.) The second goal is the same as the one at the point where we invoke assert, except that, in the context, we 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.
Actually, assert will turn out to be handy in many sorts of situations. 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 a little stupid 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)...
it seems like plus_comm should do the trick! *)

rewrite plus_comm.
(* Doesn't work...Coq rewrote the wrong plus! *)
Abort.

To get plus_comm to apply at the point where we want 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 this lemma 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).
Case "Proof of assertion".
rewrite plus_comm. reflexivity.
rewrite H. reflexivity. Qed.

#### Exercise: 4 stars (mult_comm)

Use assert to help prove this theorem. You shouldn't need to use induction.

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

Now prove commutativity of multiplication. (You will probably need to define and prove a separate subsidiary theorem to be used in the proof of this one.) You may find that plus_swap comes in handy.

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

#### Exercise: 2 stars, optional (evenb_n__oddb_Sn)

Prove the following simple fact:

Theorem evenb_n__oddb_Sn : n : nat,
evenb n = negb (evenb (S n)).
Proof.
(* FILL IN HERE *) Admitted.

# More Exercises

#### Exercise: 3 stars, 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 hacking!)

Theorem ble_nat_refl : n:nat,
true = ble_nat n n.
Proof.
(* FILL IN HERE *) Admitted.

Theorem zero_nbeq_S : n:nat,
beq_nat 0 (S n) = false.
Proof.
(* FILL IN HERE *) Admitted.

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

Theorem plus_ble_compat_l : n m p : nat,
ble_nat n m = true ble_nat (p + n) (p + m) = true.
Proof.
(* FILL IN HERE *) Admitted.

Theorem S_nbeq_0 : n:nat,
beq_nat (S n) 0 = false.
Proof.
(* FILL IN HERE *) Admitted.

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

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

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

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

#### Exercise: 2 stars, optional (beq_nat_refl)

Prove the following theorem. Putting true on the left-hand side of the equality may seem odd, but this is how the theorem is stated in the standard library, so we follow suit. Since rewriting works equally well in either direction, we will have no problem using the theorem no matter which way we state it.

Theorem beq_nat_refl : n : nat,
true = beq_nat n n.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars, optional (plus_swap')

The replace tactic allows you to specify a particular subterm to rewrite and what you want it rewritten to. More precisely, 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 plus_swap', just like plus_swap but without needing assert (n + m = m + n).

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

#### Exercise: 3 stars (binary_commute)

Recall the increment and binary-to-unary functions that you wrote for the binary exercise in the Basics chapter. Prove that these functions commute — that is, incrementing a binary number and then converting it to unary yields the same result as first converting it to unary and then incrementing.
(Before you start working on this exercise, please copy the definitions from your solution to the binary exercise here so that this file can be graded on its own. If you find yourself wanting to change your original definitions to make the property easier to prove, feel free to do so.)

(* FILL IN HERE *)

#### Exercise: 5 stars, advanced (binary_inverse)

This exercise is a continuation of the previous exercise about binary numbers. You will need your definitions and theorems from the previous exercise to complete this one.
(a) First, write a function to convert natural numbers to binary numbers. Then prove that starting with any natural number, converting to binary, then converting back yields the same natural number you started with.
(b) You might naturally think that we should also prove the opposite direction: that starting with a binary number, converting to a natural, and then back to binary yields the same number we started with. However, it is not true! Explain what the problem is.
(c) Define a function normalize from binary numbers to binary numbers such that for any binary number b, converting to a natural and then back to binary yields (normalize b). Prove it.
Again, feel free to change your earlier definitions if this helps here.

(* FILL IN HERE *)

## Formal vs. Informal Proof

"Informal proofs are algorithms; formal proofs are code."
The question of what, exactly, constitutes a "proof" of a mathematical claim has challenged philosophers for millenia. A rough and ready definition, though, 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. 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 recipes are formal proofs.
Because we are 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.
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']. reflexivity.
simpl. rewrite IHn'. reflexivity. Qed.

Coq is perfectly happy with this as a proof. For a human, however, it is difficult to make much sense of it. 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 might write it 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.
The overall form of the proof is basically similar. 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).
Here is a formal proof that shows the structure more clearly:

Theorem plus_assoc'' : n m p : nat,
n + (m + p) = (n + m) + p.
Proof.
intros n m p. induction n as [| n'].
Case "n = 0".
reflexivity.
Case "n = S n'".
simpl. rewrite IHn'. reflexivity. Qed.

#### Exercise: 2 stars, advanced (plus_comm_informal)

Translate your solution for plus_comm into an informal proof.