# PropPropositions and Evidence

Require Export MoreCoq.

In previous chapters, we have seen many examples of factual claims (propositions) and ways of presenting evidence of their truth (proofs). In particular, we have worked extensively with equality propositions of the form e1 = e2, with implications (P Q), and with quantified propositions ( x, P).
In this chapter we take a deeper look at the way propositions are expressed in Coq and at the structure of the logical evidence that we construct when we carry out proofs.
Some of the concepts in this chapter may seem a bit abstract on a first encounter. We've included a lot of exercises, most of which should be quite approachable even if you're still working on understanding the details of the text. Try to work as many of them as you can, especially the one-starred exercises.

# Inductively Defined Propositions

This chapter will take us on a first tour of the propositional (logical) side of Coq. As a running example, let's define a simple property of natural numbers — we'll call it "beautiful."
Informally, a number is beautiful if it is 0, 3, 5, or the sum of two beautiful numbers.
More pedantically, we can define beautiful numbers by giving four rules:
• Rule b_0: The number 0 is beautiful.
• Rule b_3: The number 3 is beautiful.
• Rule b_5: The number 5 is beautiful.
• Rule b_sum: If n and m are both beautiful, then so is their sum.
We will see many definitions like this one during the rest of the course, and for purposes of informal discussions, it is helpful to have a lightweight notation that makes them easy to read and write. Inference rules are one such notation:
 (b_0) beautiful 0
 (b_3) beautiful 3
 (b_5) beautiful 5
 beautiful n     beautiful m (b_sum) beautiful (n+m)
Each of the textual rules above is reformatted here as an inference rule; the intended reading is that, if the premises above the line all hold, then the conclusion below the line follows. For example, the rule b_sum says that, if n and m are both beautiful numbers, then it follows that n+m is beautiful too. The rules with no premises above the line are called axioms.
These rules define the property beautiful. That is, if we want to convince someone that some particular number is beautiful, our argument must be based on these rules. For a simple example, suppose we claim that the number 5 is beautiful. To support this claim, we just need to point out that rule b_5 says so. Or, if we want to claim that 8 is beautiful, we can support our claim by first observing that 3 and 5 are both beautiful (by rules b_3 and b_5) and then pointing out that their sum, 8, is therefore beautiful by rule b_sum. This argument can be expressed graphically with the following proof tree:
----------- (b_3)   ----------- (b_5)
beautiful 3         beautiful 5
------------------------------- (b_sum)
beautiful 8
Of course, there are other ways of using these rules to argue that 8 is beautiful, for instance:
----------- (b_5)   ----------- (b_3)
beautiful 5         beautiful 3
------------------------------- (b_sum)
beautiful 8

#### Exercise: 1 star (varieties_of_beauty)

How many different ways are there to show that 8 is beautiful?

(* FILL IN HERE *)
In Coq, we can express the definition of beautiful as follows:

Inductive beautiful : nat Prop :=
b_0 : beautiful 0
| b_3 : beautiful 3
| b_5 : beautiful 5
| b_sum : n m, beautiful n beautiful m beautiful (n+m).

The first line declares that beautiful is a proposition — or, more formally, a family of propositions "indexed by" natural numbers. (That is, for each number n, the claim that "n is beautiful" is a proposition.) Such a family of propositions is often called a property of numbers. Each of the remaining lines embodies one of the rules for beautiful numbers.
We can use Coq's tactic scripting facility to assemble proofs that particular numbers are beautiful.

Theorem three_is_beautiful: beautiful 3.
Proof.
(* This simply follows from the axiom b_3. *)
apply b_3.
Qed.

Theorem eight_is_beautiful: beautiful 8.
Proof.
(* First we use the rule b_sum, telling Coq how to
instantiate n and m. *)

apply b_sum with (n:=3) (m:=5).
(* To solve the subgoals generated by b_sum, we must provide
evidence of beautiful 3 and beautiful 5. Fortunately we
have axioms for both. *)

apply b_3.
apply b_5.
Qed.

# Proof Objects

Look again at the formal definition of the beautiful property. The opening keyword, Inductive, has been used up to this point to declare new types of data, such as numbers and lists. Does this interpretation also make sense for the Inductive definition of beautiful? That is, can we view evidence of beauty as some kind of data structure? Yes, we can!
The trick is to introduce an alternative pronunciation of ":". Instead of "has type," we can also say "is a proof of." For example, the second line in the definition of beautiful declares that b_0 : beautiful 0. Instead of "b_0 has type beautiful 0," we can say that "b_0 is a proof of beautiful 0." Similarly for b_3 and b_5.
This pun between types and propositions (between : as "has type" and : as "is a proof of" or "is evidence for") is called the Curry-Howard correspondence. It proposes a deep connection between the world of logic and the world of computation.
```                 propositions  ~  types
proofs        ~  data values
```
Many useful insights follow from this connection. To begin with, it gives us a natural interpretation of the type of b_sum constructor:

Check b_sum.
(* ===> b_sum : forall n m,
beautiful n ->
beautiful m ->
beautiful (n+m) *)

This can be read "b_sum is a constructor that takes four arguments — two numbers, n and m, and two values, of types beautiful n and beautiful m — and yields evidence for the proposition beautiful (n+m)."
In view of this, we might wonder whether we can write an expression of type beautiful 8 by applying b_sum to appropriate arguments. Indeed, we can:

Check (b_sum 3 5 b_3 b_5).
(* ===> beautiful (3 + 5) *)

The expression b_sum 3 5 b_3 b_5 can be thought of as instantiating the parameterized constructor b_sum with the specific arguments 3 5 and the corresponding proof objects for its premises beautiful 3 and beautiful 5 (Coq is smart enough to figure out that 3+5=8). Alternatively, we can think of b_sum as a primitive "evidence constructor" that, when applied to two particular numbers, wants to be further applied to evidence that those two numbers are beautiful; its type, [ n m, beautiful n beautiful m beautiful (n+m), ] expresses this functionality, in the same way that the polymorphic type X, list X in the previous chapter expressed the fact that the constructor nil can be thought of as a function from types to empty lists with elements of that type.
This gives us an alternative way to write the proof that 8 is beautiful:

Theorem eight_is_beautiful': beautiful 8.
Proof.
apply (b_sum 3 5 b_3 b_5).
Qed.

Notice that we're using apply here in a new way: instead of just supplying the name of a hypothesis or previously proved theorem whose type matches the current goal, we are supplying an expression that directly builds evidence with the required type.

## Proof Scripts and Proof Objects

These proof objects lie at the core of how Coq operates.
When Coq is following a proof script, what is happening internally is that it is gradually constructing a proof object — a term whose type is the proposition being proved. The tactics between the Proof command and the Qed instruct Coq how to build up a term of the required type. To see this process in action, let's use the Show Proof command to display the current state of the proof tree at various points in the following tactic proof.

Theorem eight_is_beautiful'': beautiful 8.
Proof.
Show Proof.
apply b_sum with (n:=3) (m:=5).
Show Proof.
apply b_3.
Show Proof.
apply b_5.
Show Proof.
Qed.

At any given moment, Coq has constructed a term with some "holes" (indicated by ?1, ?2, and so on), and it knows what type of evidence is needed at each hole. In the Show Proof output, lines of the form ?1 beautiful n record these requirements. (The here has nothing to do with either implication or function types — it is just an unfortunate choice of concrete syntax for the output!)
Each of the holes corresponds to a subgoal, and the proof is finished when there are no more subgoals. At this point, the Theorem command gives a name to the evidence we've built and stores it in the global context.
Tactic proofs are useful and convenient, but they are not essential: in principle, we can always construct the required evidence by hand. Indeed, we don't even need the Theorem command: we can instead use Definition to directly give a global name to a piece of evidence.

Definition eight_is_beautiful''' : beautiful 8 :=
b_sum 3 5 b_3 b_5.

All these different ways of building the proof lead to exactly the same evidence being saved in the global environment.

Print eight_is_beautiful.
(* ===> eight_is_beautiful    = b_sum 3 5 b_3 b_5 : beautiful 8 *)
Print eight_is_beautiful'.
(* ===> eight_is_beautiful'   = b_sum 3 5 b_3 b_5 : beautiful 8 *)
Print eight_is_beautiful''.
(* ===> eight_is_beautiful''  = b_sum 3 5 b_3 b_5 : beautiful 8 *)
Print eight_is_beautiful'''.
(* ===> eight_is_beautiful''' = b_sum 3 5 b_3 b_5 : beautiful 8 *)

#### Exercise: 1 star (six_is_beautiful)

Give a tactic proof and a proof object showing that 6 is beautiful.

Theorem six_is_beautiful :
beautiful 6.
Proof.
(* FILL IN HERE *) Admitted.

Definition six_is_beautiful' : beautiful 6 :=
(* FILL IN HERE *) admit.

#### Exercise: 1 star (nine_is_beautiful)

Give a tactic proof and a proof object showing that 9 is beautiful.

Theorem nine_is_beautiful :
beautiful 9.
Proof.
(* FILL IN HERE *) Admitted.

Definition nine_is_beautiful' : beautiful 9 :=
(* FILL IN HERE *) admit.

## Implications and Functions

In Coq's computational universe (where we've mostly been living until this chapter), there are two sorts of values with arrows in their types: constructors introduced by Inductive-ly defined data types, and functions.
Similarly, in Coq's logical universe, there are two ways of giving evidence for an implication: constructors introduced by Inductive-ly defined propositions, and... functions!
For example, consider this statement:

Theorem b_plus3: n, beautiful n beautiful (3+n).
Proof.
intros n H.
apply b_sum.
apply b_3.
apply H.
Qed.

What is the proof object corresponding to b_plus3?
We're looking for an expression whose type is n, beautiful n beautiful (3+n) — that is, a function that takes two arguments (one number and a piece of evidence) and returns a piece of evidence! Here it is:

Definition b_plus3' : n, beautiful n beautiful (3+n) :=
fun n => fun H : beautiful n =>
b_sum 3 n b_3 H.

Check b_plus3'.
(* ===> b_plus3' : forall n, beautiful n -> beautiful (3+n) *)

Recall that fun n => blah means "the function that, given n, yields blah." Another equivalent way to write this definition is:

Definition b_plus3'' (n : nat) (H : beautiful n) : beautiful (3+n) :=
b_sum 3 n b_3 H.

Check b_plus3''.
(* ===> b_plus3'' : forall n, beautiful n -> beautiful (3+n) *)

#### Exercise: 2 stars (b_times2)

Theorem b_times2: n, beautiful n beautiful (2*n).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, optional (b_times2')

Write a proof object corresponding to b_times2 above

Definition b_times2': n, beautiful n beautiful (2*n) :=
(* FILL IN HERE *) admit.

#### Exercise: 2 stars (b_timesm)

Theorem b_timesm: n m, beautiful n beautiful (m*n).
Proof.
(* FILL IN HERE *) Admitted.

## Induction Over Proof Objects

Since we use the keyword Induction to define primitive propositions together with their evidence, we might wonder whether there are some sort of induction principles associated with these definitions. Indeed there are, and in this section we'll take a look at how they can be used.
Besides constructing evidence that numbers are beautiful, we can also reason about such evidence.
The fact that we introduced beautiful with an Inductive declaration tells us not only that the constructors b_0, b_3, b_5 and b_sum are ways to build evidence, but also that these two constructors are the only ways to build evidence that numbers are beautiful.
In other words, if someone gives us evidence E for the assertion beautiful n, then we know that E must have one of four shapes:
• E is b_0 (and n is O),
• E is b_3 (and n is 3),
• E is b_5 (and n is 5), or
• E is b_sum n1 n2 E1 E2 (and n is n1+n2, where E1 is evidence that n1 is beautiful and E2 is evidence that n2 is beautiful).
This gives rise to an induction principle for proofs — i.e., we can use the induction tactic that we have already seen for reasoning about inductively defined data to reason about inductively defined evidence.
To illustrate this, let's define another property of numbers:

Inductive gorgeous : nat Prop :=
g_0 : gorgeous 0
| g_plus3 : n, gorgeous n gorgeous (3+n)
| g_plus5 : n, gorgeous n gorgeous (5+n).

#### Exercise: 1 star (gorgeous_tree)

Write out the definition of gorgeous numbers using inference rule notation.
(* FILL IN HERE *)
It seems intuitively obvious that, although gorgeous and beautiful are presented using slightly different rules, they are actually the same property in the sense that they are true of the same numbers. Indeed, we can prove this.

Theorem gorgeous__beautiful : n,
gorgeous n beautiful n.
Proof.
intros n H.
induction H as [|n'|n'].
Case "g_0".
apply b_0.
Case "g_plus3".
apply b_sum. apply b_3.
apply IHgorgeous.
Case "g_plus5".
apply b_sum. apply b_5. apply IHgorgeous.
Qed.

Notice that the argument proceeds by induction on the evidence H!
Let's see what happens if we try to prove this by induction on n instead of induction on the evidence H.

Theorem gorgeous__beautiful_FAILED : n,
gorgeous n beautiful n.
Proof.
intros. induction n as [| n'].
Case "n = 0". apply b_0.
Case "n = S n'". (* We are stuck! *)
Abort.

The problem here is that doing induction on n doesn't yield a useful induction hypothesis. Knowing how the property we are interested in behaves on the predecessor of n doesn't help us prove that it holds for n. Instead, we would like to be able to have induction hypotheses that mention other numbers, such as n - 3 and n - 5. This is given precisely by the shape of the constructors for gorgeous.

#### Exercise: 1 star (gorgeous_plus13)

Theorem gorgeous_plus13: n,
gorgeous n gorgeous (13+n).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars, optional (gorgeous_plus13_po):

Give the proof object for theorem gorgeous_plus13 above.

Definition gorgeous_plus13_po: n, gorgeous n gorgeous (13+n):=
(* FILL IN HERE *) admit.

#### Exercise: 2 stars (gorgeous_sum)

Theorem gorgeous_sum : n m,
gorgeous n gorgeous m gorgeous (n + m).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, advanced (beautiful__gorgeous)

Theorem beautiful__gorgeous : n, beautiful n gorgeous n.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, optional (b_times2)

Prove the g_times2 theorem below without using gorgeous__beautiful. You might find the following helper lemma useful.

Lemma helper_g_times2 : x y z, x + (z + y)= z + x + y.
Proof.
(* FILL IN HERE *) Admitted.

Theorem g_times2: n, gorgeous n gorgeous (2*n).
Proof.
intros n H. simpl.
induction H.
(* FILL IN HERE *) Admitted.

## From Boolean Functions to Propositions

In chapter Basics we defined a function evenb that tests a number for evenness, yielding true if so. We can use this function to define the proposition that some number n is even:

Definition even (n:nat) : Prop :=
evenb n = true.

That is, we can define "n is even" to mean "the function evenb returns true when applied to n."
Another alternative is to define the concept of evenness directly. Instead of going via the evenb function ("a number is even if a certain computation yields true"), we can say what the concept of evenness means by giving two different ways of presenting evidence that a number is even.

Inductive ev : nat Prop :=
| ev_0 : ev O
| ev_SS : n:nat, ev n ev (S (S n)).

This definition says that there are two ways to give evidence that a number m is even. First, 0 is even, and ev_0 is evidence for this. Second, if m = S (S n) for some n and we can give evidence e that n is even, then m is also even, and ev_SS n e is the evidence.

#### Exercise: 1 star (double_even)

Construct a tactic proof of the following proposition.

Theorem double_even : n,
ev (double n).
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 4 stars, optional (double_even_pfobj)

Try to predict what proof object is constructed by the above tactic proof. (Before checking your answer, you'll want to strip out any uses of Case, as these will make the proof object look a bit cluttered.)

### Discussion: Computational vs. Inductive Definitions

We have seen that the proposition "n is even" can be phrased in two different ways — indirectly, via a boolean testing function evenb, or directly, by inductively describing what constitutes evidence for evenness. These two ways of defining evenness are about equally easy to state and work with. Which we choose is basically a question of taste.
However, for many other properties of interest, the direct inductive definition is preferable, since writing a testing function may be awkward or even impossible.
One such property is beautiful. This is a perfectly sensible definition of a set of numbers, but we cannot translate its definition directly into a Coq Fixpoint (or into a recursive function in any other common programming language). We might be able to find a clever way of testing this property using a Fixpoint (indeed, it is not too hard to find one in this case), but in general this could require arbitrarily deep thinking. In fact, if the property we are interested in is uncomputable, then we cannot define it as a Fixpoint no matter how hard we try, because Coq requires that all Fixpoints correspond to terminating computations.
On the other hand, writing an inductive definition of what it means to give evidence for the property beautiful is straightforward.

## Inversion on Proof Objects

Besides induction, we can use the other tactics in our toolkit to reason about evidence. For example, this proof uses destruct on evidence.

Theorem ev_minus2: n,
ev n ev (pred (pred n)).
Proof.
intros n E.
destruct E as [| n' E'].
Case "E = ev_0". simpl. apply ev_0.
Case "E = ev_SS n' E'". simpl. apply E'. Qed.

#### Exercise: 1 star, optional (ev_minus2_n)

What happens if we try to destruct on n instead of E?

(* FILL IN HERE *)

#### Exercise: 1 star (ev__even)

Here is a proof that the inductive definition of evenness implies the computational one.

Theorem ev__even : n,
ev n even n.
Proof.
intros n E. induction E as [| n' E'].
Case "E = ev_0".
unfold even. reflexivity.
Case "E = ev_SS n' E'".
unfold even. apply IHE'.
Qed.

Could this proof also be carried out by induction on n instead of E? If not, why not?

(* FILL IN HERE *)
The induction principle for inductively defined propositions does not follow quite the same form as that of inductively defined sets. For now, you can take the intuitive view that induction on evidence ev n is similar to induction on n, but restricts our attention to only those numbers for which evidence ev n could be generated. We'll look at the induction principle of ev in more depth below, to explain what's really going on.

#### Exercise: 1 star (l_fails)

The following proof attempt will not succeed.
Theorem l : n,
ev n.
Proof.
intros n. induction n.
Case "O". simpl. apply ev_0.
Case "S".
...
Intuitively, we expect the proof to fail because not every number is even. However, what exactly causes the proof to fail?
(* FILL IN HERE *)

#### Exercise: 2 stars (ev_sum)

Here's another exercise requiring induction.

Theorem ev_sum : n m,
ev n ev m ev (n+m).
Proof.
(* FILL IN HERE *) Admitted.
Another situation where we want to analyze evidence for evenness is when proving that, if n+2 is even, then n is.
Our first idea might be to use destruct for this kind of case analysis:

Theorem SSev_ev_firsttry : n,
ev (S (S n)) ev n.
Proof.
intros n E.
destruct E as [| n' E'].
(* Stuck: destruct gives us an unprovable subgoal here! *)
Abort.

In the first sub-goal, we've lost the information that n is 0. We could have used remember, but then we still need inversion on both cases.

Theorem SSev_ev_secondtry : n,
ev (S (S n)) ev n.
Proof.
intros n E. remember (S (S n)) as n2.
destruct E as [| n' E'].
Case "n = 0". inversion Heqn2.
Case "n = S n'". inversion Heqn2. rewrite H0. apply E'.
Qed.

There is a much simpler way to do this. We can use inversion directly on the inductively defined proposition ev (S (S n)).

Theorem SSev__even : n,
ev (S (S n)) ev n.
Proof.
intros n E. inversion E as [| n' E']. apply E'. Qed.

This use of inversion may seem a bit mysterious at first. Until now, we've only used inversion on equality propositions, to utilize injectivity of constructors or to discriminate between different constructors. But we see here that inversion can also be applied to analyzing evidence for inductively defined propositions.
Here's how inversion works in general. Suppose the name I refers to an assumption P in the current context, where P has been defined by an Inductive declaration. Then, for each of the constructors of P, inversion I generates a subgoal in which I has been replaced by the exact, specific conditions under which this constructor could have been used to prove P. Some of these subgoals will be self-contradictory; inversion throws these away. The ones that are left represent the cases that must be proved to establish the original goal.
In this particular case, the inversion analyzed the construction ev (S (S n)), determined that this could only have been constructed using ev_SS, and generated a new subgoal with the arguments of that constructor as new hypotheses. (It also produced an auxiliary equality, which happens to be useless here.) We'll begin exploring this more general behavior of inversion in what follows.

#### Exercise: 1 star (inversion_practice)

Theorem SSSSev__even : n,
ev (S (S (S (S n)))) ev n.
Proof.
(* FILL IN HERE *) Admitted.

The inversion tactic can also be used to derive goals by showing the absurdity of a hypothesis.

Theorem even5_nonsense :
ev 5 2 + 2 = 9.
Proof.
(* FILL IN HERE *) Admitted.
We can generally use inversion on inductive propositions. This illustrates that in general, we get one case for each possible constructor. Again, we also get some auxiliary equalities that are rewritten in the goal but not in the other hypotheses.

Theorem ev_minus2': n,
ev n ev (pred (pred n)).
Proof.
intros n E. inversion E as [| n' E'].
Case "E = ev_0". simpl. apply ev_0.
Case "E = ev_SS n' E'". simpl. apply E'. Qed.

#### Exercise: 3 stars, advanced (ev_ev__ev)

Finding the appropriate thing to do induction on is a bit tricky here:

Theorem ev_ev__ev : n m,
ev (n+m) ev n ev m.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, optional (ev_plus_plus)

Here's an exercise that just requires applying existing lemmas. No induction or even case analysis is needed, but some of the rewriting may be tedious. You'll want the replace tactic used for plus_swap' in Basics.v

Theorem ev_plus_plus : n m p,
ev (n+m) ev (n+p) ev (m+p).
Proof.
(* FILL IN HERE *) Admitted.

## Building Proof Objects Incrementally (Optional)

As you probably noticed while solving the exercises earlier in the chapter, constructing proof objects is more involved than constructing the corresponding tactic proofs. Fortunately, there is a bit of syntactic sugar that we've already introduced to help in the construction: the admit term, which we've sometimes used to force Coq into accepting incomplete exercies. As an example, let's walk through the process of constructing a proof object demonstrating the beauty of 16.

Definition b_16_atmpt_1 : beautiful 16 := admit.

Maybe we can use b_sum to construct a term of type beautiful 16? Recall that b_sum is of type
n m : natbeautiful n  beautiful m  beautiful (n + m)
If we can demonstrate the beauty of 5 and 11, we should be done.

In the attempt above, we've omitted the proofs of the propositions that 5 and 11 are beautiful. But the first of these is already axiomatized in b_5:

Definition b_16_atmpt_3 : beautiful 16 := b_sum 5 11 b_5 admit.

What remains is to show that 11 is beautiful. We repeat the procedure:

Definition b_16_atmpt_4 : beautiful 16 :=

Definition b_16_atmpt_5 : beautiful 16 :=
b_sum 5 11 b_5 (b_sum 5 6 b_5 admit).

Definition b_16_atmpt_6 : beautiful 16 :=
b_sum 5 11 b_5 (b_sum 5 6 b_5 (b_sum 3 3 admit admit)).

And finally, we can complete the proof object:

Definition b_16 : beautiful 16 :=
b_sum 5 11 b_5 (b_sum 5 6 b_5 (b_sum 3 3 b_3 b_3)).

To recap, we've been guided by an informal proof that we have in our minds, and we check the high level details before completing the intricacies of the proof. The admit term allows us to do this.

#### Exercise: 4 stars (palindromes)

A palindrome is a sequence that reads the same backwards as forwards.
• Define an inductive proposition pal on list X that captures what it means to be a palindrome. (Hint: You'll need three cases. Your definition should be based on the structure of the list; just having a single constructor
c : ll = rev l  pal l
may seem obvious, but will not work very well.)
• Prove that
lpal (l ++ rev l).
• Prove that
lpal l  l = rev l.

(* FILL IN HERE *)

#### Exercise: 5 stars, optional (palindrome_converse)

Using your definition of pal from the previous exercise, prove that
ll = rev l  pal l.

(* FILL IN HERE *)

#### Exercise: 4 stars, advanced (subsequence)

A list is a subsequence of another list if all of the elements in the first list occur in the same order in the second list, possibly with some extra elements in between. For example,
[1,2,3]
is a subsequence of each of the lists
[1,2,3]
[1,1,1,2,2,3]
[1,2,7,3]
[5,6,1,9,9,2,7,3,8]
but it is not a subsequence of any of the lists
[1,2]
[1,3]
[5,6,2,1,7,3,8]
• Define an inductive proposition subseq on list nat that captures what it means to be a subsequence. (Hint: You'll need three cases.)
• Prove that subsequence is reflexive, that is, any list is a subsequence of itself.
• Prove that for any lists l1, l2, and l3, if l1 is a subsequence of l2, then l1 is also a subsequence of l2 ++ l3.
• (Optional, harder) Prove that subsequence is transitive — that is, if l1 is a subsequence of l2 and l2 is a subsequence of l3, then l1 is a subsequence of l3. Hint: choose your induction carefully!

(* FILL IN HERE *)

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

Suppose we give Coq the following definition:
Inductive R : nat  list nat  Prop :=
| c1 : R 0 []
| c2 : n lR n l  R (S n) (n :: l)
| c3 : n lR (S nl  R n l.
Which of the following propositions are provable?
• R 2 [1,0]
• R 1 [1,2,1,0]
• R 6 [3,2,1,0]

(* \$Date: 2013-03-10 13:20:46 -0400 (Sun, 10 Mar 2013) \$ *)