Require Export Poly.

This chapter introduces several more Coq tactics that, together, allow us to prove many more theorems about the functional programs we are writing.

# The apply Tactic

We often encounter situations where the goal to be proved is exactly the same as some hypothesis in the context or some previously proved lemma.

Theorem silly1 : (n m o p : nat),
n = m
[n;o] = [n;p]
[n;o] = [m;p].
Proof.
intros n m o p eq1 eq2.
rewrite eq1.
(* At this point, we could finish with
"rewrite eq2. reflexivity." as we have
done several times above. But we can achieve the
same effect in a single step by using the

apply eq2. Qed.

The apply tactic also works with conditional hypotheses and lemmas: if the statement being applied is an implication, then the premises of this implication will be added to the list of subgoals needing to be proved.

Theorem silly2 : (n m o p : nat),
n = m
((q r : nat), q = r [q;o] = [r;p])
[n;o] = [m;p].
Proof.
intros n m o p eq1 eq2.
apply eq2. apply eq1. Qed.

You may find it instructive to experiment with this proof and see if there is a way to complete it using just rewrite instead of apply.
Typically, when we use apply H, the statement H will begin with a binding some universal variables. When Coq matches the current goal against the conclusion of H, it will try to find appropriate values for these variables. For example, when we do apply eq2 in the following proof, the universal variable q in eq2 gets instantiated with n and r gets instantiated with m.

Theorem silly2a : (n m : nat),
(n,n) = (m,m)
((q r : nat), (q,q) = (r,r) [q] = [r])
[n] = [m].
Proof.
intros n m eq1 eq2.
apply eq2. apply eq1. Qed.

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

Complete the following proof without using simpl.

Theorem silly_ex :
(n, evenb n = true oddb (S n) = true)
evenb 3 = true
oddb 4 = true.
Proof.
(* FILL IN HERE *) Admitted.
To use the apply tactic, the (conclusion of the) fact being applied must match the goal exactly — for example, apply will not work if the left and right sides of the equality are swapped.

Theorem silly3_firsttry : (n : nat),
true = beq_nat n 5
beq_nat (S (S n)) 7 = true.
Proof.
intros n H.
simpl.
(* Here we cannot use apply directly *)
Abort.

In this case we can use the symmetry tactic, which switches the left and right sides of an equality in the goal.

Theorem silly3 : (n : nat),
true = beq_nat n 5
beq_nat (S (S n)) 7 = true.
Proof.
intros n H.
symmetry.
simpl. (* Actually, this simpl is unnecessary, since
apply will perform simplification first. *)

apply H. Qed.

#### Exercise: 3 stars (apply_exercise1)

Hint: you can use apply with previously defined lemmas, not just hypotheses in the context. Remember that SearchAbout is your friend.

Theorem rev_exercise1 : (l l' : list nat),
l = rev l'
l' = rev l.
Proof.
(* FILL IN HERE *) Admitted.

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

Briefly explain the difference between the tactics apply and rewrite. Are there situations where both can usefully be applied? (* FILL IN HERE *)

# The apply...with... Tactic

The following silly example uses two rewrites in a row to get from [a,b] to [e,f].

Example trans_eq_example : (a b c d e f : nat),
[a;b] = [c;d]
[c;d] = [e;f]
[a;b] = [e;f].
Proof.
intros a b c d e f eq1 eq2.
rewrite eq1. rewrite eq2. reflexivity. Qed.

Since this is a common pattern, we might abstract it out as a lemma recording once and for all the fact that equality is transitive.

Theorem trans_eq : (X:Type) (n m o : X),
n = m m = o n = o.
Proof.
intros X n m o eq1 eq2. rewrite eq1. rewrite eq2.
reflexivity. Qed.

Now, we should be able to use trans_eq to prove the above example. However, to do this we need a slight refinement of the apply tactic.

Example trans_eq_example' : (a b c d e f : nat),
[a;b] = [c;d]
[c;d] = [e;f]
[a;b] = [e;f].
Proof.
intros a b c d e f eq1 eq2.
(* If we simply tell Coq apply trans_eq at this point,
it can tell (by matching the goal against the
conclusion of the lemma) that it should instantiate X
with [nat]n with [a,b], and o with [e,f].
However, the matching process doesn't determine an
instantiation for m: we have to supply one explicitly
by adding with (m:=[c,d]) to the invocation of
apply. *)

apply trans_eq with (m:=[c;d]). apply eq1. apply eq2. Qed.

Actually, we usually don't have to include the name m in the with clause; Coq is often smart enough to figure out which instantiation we're giving. We could instead write: apply trans_eq with [c,d].

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

Example trans_eq_exercise : (n m o p : nat),
m = (minustwo o)
(n + p) = m
(n + p) = (minustwo o).
Proof.
(* FILL IN HERE *) Admitted.

# The inversion tactic

Recall the definition of natural numbers:
Inductive nat : Type :=
| O : nat
| S : nat  nat.
It is clear from this definition that every number has one of two forms: either it is the constructor O or it is built by applying the constructor S to another number. But there is more here than meets the eye: implicit in the definition (and in our informal understanding of how datatype declarations work in other programming languages) are two other facts:
• The constructor S is injective. That is, the only way we can have S n = S m is if n = m.
• The constructors O and S are disjoint. That is, O is not equal to S n for any n.
Similar principles apply to all inductively defined types: all constructors are injective, and the values built from distinct constructors are never equal. For lists, the cons constructor is injective and nil is different from every non-empty list. For booleans, true and false are unequal. (Since neither true nor false take any arguments, their injectivity is not an issue.)
Coq provides a tactic called inversion that allows us to exploit these principles in proofs.
The inversion tactic is used like this. Suppose H is a hypothesis in the context (or a previously proven lemma) of the form
c a1 a2 ... an = d b1 b2 ... bm
for some constructors c and d and arguments a1 ... an and b1 ... bm. Then inversion H instructs Coq to "invert" this equality to extract the information it contains about these terms:
• If c and d are the same constructor, then we know, by the injectivity of this constructor, that a1 = b1, a2 = b2, etc.; inversion H adds these facts to the context, and tries to use them to rewrite the goal.
• If c and d are different constructors, then the hypothesis H is contradictory. That is, a false assumption has crept into the context, and this means that any goal whatsoever is provable! In this case, inversion H marks the current goal as completed and pops it off the goal stack.
The inversion tactic is probably easier to understand by seeing it in action than from general descriptions like the above. Below you will find example theorems that demonstrate the use of inversion and exercises to test your understanding.

Theorem eq_add_S : (n m : nat),
S n = S m
n = m.
Proof.
intros n m eq. inversion eq. reflexivity. Qed.

Theorem silly4 : (n m : nat),
[n] = [m]
n = m.
Proof.
intros n o eq. inversion eq. reflexivity. Qed.

As a convenience, the inversion tactic can also destruct equalities between complex values, binding multiple variables as it goes.

Theorem silly5 : (n m o : nat),
[n;m] = [o;o]
[n] = [m].
Proof.
intros n m o eq. inversion eq. reflexivity. Qed.

#### Exercise: 1 star (sillyex1)

Example sillyex1 : (X : Type) (x y z : X) (l j : list X),
x :: y :: l = z :: j
y :: l = x :: j
x = y.
Proof.
(* FILL IN HERE *) Admitted.

Theorem silly6 : (n : nat),
S n = O
2 + 2 = 5.
Proof.
intros n contra. inversion contra. Qed.

Theorem silly7 : (n m : nat),
false = true
[n] = [m].
Proof.
intros n m contra. inversion contra. Qed.

#### Exercise: 1 star (sillyex2)

Example sillyex2 : (X : Type) (x y z : X) (l j : list X),
x :: y :: l = []
y :: l = z :: j
x = z.
Proof.
(* FILL IN HERE *) Admitted.
While the injectivity of constructors allows us to reason (n m : nat), S n = S m n = m, the reverse direction of the implication is an instance of a more general fact about constructors and functions, which we will often find useful:

Theorem f_equal : (A B : Type) (f: A B) (x y: A),
x = y f x = f y.
Proof. intros A B f x y eq. rewrite eq. reflexivity. Qed.

Here's another illustration of inversion. This is a slightly roundabout way of stating a fact that we have already proved above. The extra equalities force us to do a little more equational reasoning and exercise some of the tactics we've seen recently.

Theorem length_snoc' : (X : Type) (v : X)
(l : list X) (n : nat),
length l = n
length (snoc l v) = S n.
Proof.
intros X v l. induction l as [| v' l'].
Case "l = []". intros n eq. rewrite eq. reflexivity.
Case "l = v' :: l'". intros n eq. simpl. destruct n as [| n'].
SCase "n = 0". inversion eq.
SCase "n = S n'".
apply f_equal. apply IHl'. inversion eq. reflexivity. Qed.

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

A couple more nontrivial but not-too-complicated proofs to work together in class, or for you to work as exercises.

Theorem beq_nat_0_l : n,
beq_nat 0 n = true n = 0.
Proof.
(* FILL IN HERE *) Admitted.

Theorem beq_nat_0_r : n,
beq_nat n 0 = true n = 0.
Proof.
(* FILL IN HERE *) Admitted.

# Using Tactics on Hypotheses

By default, most tactics work on the goal formula and leave the context unchanged. However, most tactics also have a variant that performs a similar operation on a statement in the context.
For example, the tactic simpl in H performs simplification in the hypothesis named H in the context.

Theorem S_inj : (n m : nat) (b : bool),
beq_nat (S n) (S m) = b
beq_nat n m = b.
Proof.
intros n m b H. simpl in H. apply H. Qed.

Similarly, the tactic apply L in H matches some conditional statement L (of the form L1 L2, say) against a hypothesis H in the context. However, unlike ordinary apply (which rewrites a goal matching L2 into a subgoal L1), apply L in H matches H against L1 and, if successful, replaces it with L2.
In other words, apply L in H gives us a form of "forward reasoning" — from L1 L2 and a hypothesis matching L1, it gives us a hypothesis matching L2. By contrast, apply L is "backward reasoning" — it says that if we know L1L2 and we are trying to prove L2, it suffices to prove L1.
Here is a variant of a proof from above, using forward reasoning throughout instead of backward reasoning.

Theorem silly3' : (n : nat),
(beq_nat n 5 = true beq_nat (S (S n)) 7 = true)
true = beq_nat n 5
true = beq_nat (S (S n)) 7.
Proof.
intros n eq H.
symmetry in H. apply eq in H. symmetry in H.
apply H. Qed.

Forward reasoning starts from what is given (premises, previously proven theorems) and iteratively draws conclusions from them until the goal is reached. Backward reasoning starts from the goal, and iteratively reasons about what would imply the goal, until premises or previously proven theorems are reached. If you've seen informal proofs before (for example, in a math or computer science class), they probably used forward reasoning. In general, Coq tends to favor backward reasoning, but in some situations the forward style can be easier to use or to think about.

#### Exercise: 3 stars (plus_n_n_injective)

Practice using "in" variants in this exercise.

Theorem plus_n_n_injective : n m,
n + n = m + m
n = m.
Proof.
intros n. induction n as [| n'].
(* Hint: use the plus_n_Sm lemma *)
(* FILL IN HERE *) Admitted.

# Varying the Induction Hypothesis

Sometimes it is important to control the exact form of the induction hypothesis when carrying out inductive proofs in Coq. In particular, we need to be careful about which of the assumptions we move (using intros) from the goal to the context before invoking the induction tactic. For example, suppose we want to show that the double function is injective — i.e., that it always maps different arguments to different results:
Theorem double_injectiven mdouble n = double m  n = m.
The way we start this proof is a little bit delicate: if we begin it with
intros n. induction n.
all is well. But if we begin it with
intros n m. induction n.
we get stuck in the middle of the inductive case...

Theorem double_injective_FAILED : n m,
double n = double m
n = m.
Proof.
intros n m. induction n as [| n'].
Case "n = O". simpl. intros eq. destruct m as [| m'].
SCase "m = O". reflexivity.
SCase "m = S m'". inversion eq.
Case "n = S n'". intros eq. destruct m as [| m'].
SCase "m = O". inversion eq.
SCase "m = S m'". apply f_equal.
(* Here we are stuck.  The induction hypothesis, IHn', does
not give us n' = m' -- there is an extra S in the
way -- so the goal is not provable. *)

Abort.

What went wrong?
The problem is that, at the point we invoke the induction hypothesis, we have already introduced m into the context — intuitively, we have told Coq, "Let's consider some particular n and m..." and we now have to prove that, if double n = double m for this particular n and m, then n = m.
The next tactic, induction n says to Coq: We are going to show the goal by induction on n. That is, we are going to prove that the proposition
• P n = "if double n = double m, then n = m"
holds for all n by showing
• P O
(i.e., "if double O = double m then O = m")
• P n P (S n)
(i.e., "if double n = double m then n = m" implies "if double (S n) = double m then S n = m").
If we look closely at the second statement, it is saying something rather strange: it says that, for a particular m, if we know
• "if double n = double m then n = m"
then we can prove
• "if double (S n) = double m then S n = m".
To see why this is strange, let's think of a particular m — say, 5. The statement is then saying that, if we know
• Q = "if double n = 10 then n = 5"
then we can prove
• R = "if double (S n) = 10 then S n = 5".
But knowing Q doesn't give us any help with proving R! (If we tried to prove R from Q, we would say something like "Suppose double (S n) = 10..." but then we'd be stuck: knowing that double (S n) is 10 tells us nothing about whether double n is 10, so Q is useless at this point.)
To summarize: Trying to carry out this proof by induction on n when m is already in the context doesn't work because we are trying to prove a relation involving every n but just a single m.
The good proof of double_injective leaves m in the goal statement at the point where the induction tactic is invoked on n:

Theorem double_injective : n m,
double n = double m
n = m.
Proof.
intros n. induction n as [| n'].
Case "n = O". simpl. intros m eq. destruct m as [| m'].
SCase "m = O". reflexivity.
SCase "m = S m'". inversion eq.
Case "n = S n'".
(* Notice that both the goal and the induction
hypothesis have changed: the goal asks us to prove
something more general (i.e., to prove the
statement for _every_ m), but the IH is
correspondingly more flexible, allowing us to
choose any m we like when we apply the IH.  *)

intros m eq.
(* Now we choose a particular m and introduce the
assumption that double n = double m.  Since we
are doing a case analysis on n, we need a case
analysis on m to keep the two "in sync." *)

destruct m as [| m'].
SCase "m = O".
(* The 0 case is trivial *)
inversion eq.
SCase "m = S m'".
apply f_equal.
(* At this point, since we are in the second
branch of the destruct m, the m' mentioned
in the context at this point is actually the
predecessor of the one we started out talking
about.  Since we are also in the S branch of
the induction, this is perfect: if we
instantiate the generic m in the IH with the
m' that we are talking about right now (this
instantiation is performed automatically by
apply), then IHn' gives us exactly what we
need to finish the proof. *)

apply IHn'. inversion eq. reflexivity. Qed.

What this teaches us is that we need to be careful about using induction to try to prove something too specific: If we're proving a property of n and m by induction on n, we may need to leave m generic.
The proof of this theorem has to be treated similarly:

#### Exercise: 2 stars (beq_nat_true)

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

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

Give a careful informal proof of beq_nat_true, being as explicit as possible about quantifiers.

(* FILL IN HERE *)
The strategy of doing fewer intros before an induction doesn't always work directly; sometimes a little rearrangement of quantified variables is needed. Suppose, for example, that we wanted to prove double_injective by induction on m instead of n.

Theorem double_injective_take2_FAILED : n m,
double n = double m
n = m.
Proof.
intros n m. induction m as [| m'].
Case "m = O". simpl. intros eq. destruct n as [| n'].
SCase "n = O". reflexivity.
SCase "n = S n'". inversion eq.
Case "m = S m'". intros eq. destruct n as [| n'].
SCase "n = O". inversion eq.
SCase "n = S n'". apply f_equal.
(* Stuck again here, just like before. *)
Abort.

The problem is that, to do induction on m, we must first introduce n. (If we simply say induction m without introducing anything first, Coq will automatically introduce n for us!)
What can we do about this? One possibility is to rewrite the statement of the lemma so that m is quantified before n. This will work, but it's not nice: We don't want to have to mangle the statements of lemmas to fit the needs of a particular strategy for proving them — we want to state them in the most clear and natural way.
What we can do instead is to first introduce all the quantified variables and then re-generalize one or more of them, taking them out of the context and putting them back at the beginning of the goal. The generalize dependent tactic does this.

Theorem double_injective_take2 : n m,
double n = double m
n = m.
Proof.
intros n m.
(* n and m are both in the context *)
generalize dependent n.
(* Now n is back in the goal and we can do induction on
m and get a sufficiently general IH. *)

induction m as [| m'].
Case "m = O". simpl. intros n eq. destruct n as [| n'].
SCase "n = O". reflexivity.
SCase "n = S n'". inversion eq.
Case "m = S m'". intros n eq. destruct n as [| n'].
SCase "n = O". inversion eq.
SCase "n = S n'". apply f_equal.
apply IHm'. inversion eq. reflexivity. Qed.

Let's look at an informal proof of this theorem. Note that the proposition we prove by induction leaves n quantified, corresponding to the use of generalize dependent in our formal proof.
Theorem: For any nats n and m, if double n = double m, then n = m.
Proof: Let m be a nat. We prove by induction on m that, for any n, if double n = double m then n = m.
• First, suppose m = 0, and suppose n is a number such that double n = double m. We must show that n = 0.
Since m = 0, by the definition of double we have double n = 0. There are two cases to consider for n. If n = 0 we are done, since this is what we wanted to show. Otherwise, if n = S n' for some n', we derive a contradiction: by the definition of double we would have double n = S (S (double n')), but this contradicts the assumption that double n = 0.
• Otherwise, suppose m = S m' and that n is again a number such that double n = double m. We must show that n = S m', with the induction hypothesis that for every number s, if double s = double m' then s = m'.
By the fact that m = S m' and the definition of double, we have double n = S (S (double m')). There are two cases to consider for n.
If n = 0, then by definition double n = 0, a contradiction. Thus, we may assume that n = S n' for some n', and again by the definition of double we have S (S (double n')) = S (S (double m')), which implies by inversion that double n' = double m'.
Instantiating the induction hypothesis with n' thus allows us to conclude that n' = m', and it follows immediately that S n' = S m'. Since S n' = n and S m' = m, this is just what we wanted to show.

#### Exercise: 3 stars (gen_dep_practice)

Prove this by induction on l.

Theorem index_after_last: (n : nat) (X : Type) (l : list X),
length l = n
index n l = None.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, advanced, optional (index_after_last_informal)

Write an informal proof corresponding to your Coq proof of index_after_last:
Theorem: For all sets X, lists l : list X, and numbers n, if length l = n then index n l = None.
Proof: (* FILL IN HERE *)

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

Prove this by induction on l.

Theorem length_snoc''' : (n : nat) (X : Type)
(v : X) (l : list X),
length l = n
length (snoc l v) = S n.
Proof.
(* FILL IN HERE *) Admitted.

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

Prove this by induction on l1, without using app_length.

Theorem app_length_cons : (X : Type) (l1 l2 : list X)
(x : X) (n : nat),
length (l1 ++ (x :: l2)) = n
S (length (l1 ++ l2)) = n.
Proof.
(* FILL IN HERE *) Admitted.

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

Prove this by induction on l, without using app_length.

Theorem app_length_twice : (X:Type) (n:nat) (l:list X),
length l = n
length (l ++ l) = n + n.
Proof.
(* FILL IN HERE *) Admitted.

# Using destruct on Compound Expressions

We have seen many examples where the destruct tactic is used to perform case analysis of the value of some variable. But sometimes we need to reason by cases on the result of some expression. We can also do this with destruct.
Here are some examples:

Definition sillyfun (n : nat) : bool :=
if beq_nat n 3 then false
else if beq_nat n 5 then false
else false.

Theorem sillyfun_false : (n : nat),
sillyfun n = false.
Proof.
intros n. unfold sillyfun.
destruct (beq_nat n 3).
Case "beq_nat n 3 = true". reflexivity.
Case "beq_nat n 3 = false". destruct (beq_nat n 5).
SCase "beq_nat n 5 = true". reflexivity.
SCase "beq_nat n 5 = false". reflexivity. Qed.

After unfolding sillyfun in the above proof, we find that we are stuck on if (beq_nat n 3) then ... else .... Well, either n is equal to 3 or it isn't, so we use destruct (beq_nat n 3) to let us reason about the two cases.
In general, the destruct tactic can be used to perform case analysis of the results of arbitrary computations. If e is an expression whose type is some inductively defined type T, then, for each constructor c of T, destruct e generates a subgoal in which all occurrences of e (in the goal and in the context) are replaced by c.

Theorem override_shadow : (X:Type) x1 x2 k1 k2 (f : natX),
(override (override f k1 x2) k1 x1) k2 = (override f k1 x1) k2.
Proof.
(* FILL IN HERE *) Admitted.

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

Complete the proof below

Theorem combine_split : X Y (l : list (X × Y)) l1 l2,
split l = (l1, l2)
combine l1 l2 = l.
Proof.
(* FILL IN HERE *) Admitted.
Sometimes, doing a destruct on a compound expression (a non-variable) will erase information we need to complete a proof. For example, suppose we define a function sillyfun1 like this:

Definition sillyfun1 (n : nat) : bool :=
if beq_nat n 3 then true
else if beq_nat n 5 then true
else false.

And suppose that we want to convince Coq of the rather obvious observation that sillyfun1 n yields true only when n is odd. By analogy with the proofs we did with sillyfun above, it is natural to start the proof like this:

Theorem sillyfun1_odd_FAILED : (n : nat),
sillyfun1 n = true
oddb n = true.
Proof.
intros n eq. unfold sillyfun1 in eq.
destruct (beq_nat n 3).
(* stuck... *)
Abort.

We get stuck at this point because the context does not contain enough information to prove the goal! The problem is that the substitution peformed by destruct is too brutal — it threw away every occurrence of beq_nat n 3, but we need to keep some memory of this expression and how it was destructed, because we need to be able to reason that since, in this branch of the case analysis, beq_nat n 3 = true, it must be that n = 3, from which it follows that n is odd.
What we would really like is to substitute away all existing occurences of beq_nat n 3, but at the same time add an equation to the context that records which case we are in. The eqn: qualifier allows us to introduce such an equation (with whatever name we choose).

Theorem sillyfun1_odd : (n : nat),
sillyfun1 n = true
oddb n = true.
Proof.
intros n eq. unfold sillyfun1 in eq.
destruct (beq_nat n 3) eqn:Heqe3.
(* Now we have the same state as at the point where we got stuck
above, except that the context contains an extra equality
assumption, which is exactly what we need to make progress. *)

Case "e3 = true". apply beq_nat_true in Heqe3.
rewrite Heqe3. reflexivity.
Case "e3 = false".
(* When we come to the second equality test in the body of the
function we are reasoning about, we can use eqn: again in the
same way, allow us to finish the proof. *)

destruct (beq_nat n 5) eqn:Heqe5.
SCase "e5 = true".
apply beq_nat_true in Heqe5.
rewrite Heqe5. reflexivity.
SCase "e5 = false". inversion eq. Qed.

#### Exercise: 2 stars (destruct_eqn_practice)

Theorem bool_fn_applied_thrice :
(f : bool bool) (b : bool),
f (f (f b)) = f b.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars (override_same)

Theorem override_same : (X:Type) x1 k1 k2 (f : natX),
f k1 = x1
(override f k1 x1) k2 = f k2.
Proof.
(* FILL IN HERE *) Admitted.

# Review

We've now seen a bunch of Coq's fundamental tactics. We'll introduce a few more as we go along through the coming lectures, and later in the course we'll introduce some more powerful automation tactics that make Coq do more of the low-level work in many cases. But basically we've got what we need to get work done.
Here are the ones we've seen:
• intros: move hypotheses/variables from goal to context
• reflexivity: finish the proof (when the goal looks like e = e)
• apply: prove goal using a hypothesis, lemma, or constructor
• apply... in H: apply a hypothesis, lemma, or constructor to a hypothesis in the context (forward reasoning)
• apply... with...: explicitly specify values for variables that cannot be determined by pattern matching
• simpl: simplify computations in the goal
• simpl in H: ... or a hypothesis
• rewrite: use an equality hypothesis (or lemma) to rewrite the goal
• rewrite ... in H: ... or a hypothesis
• symmetry: changes a goal of the form t=u into u=t
• symmetry in H: changes a hypothesis of the form t=u into u=t
• unfold: replace a defined constant by its right-hand side in the goal
• unfold... in H: ... or a hypothesis
• destruct... as...: case analysis on values of inductively defined types
• destruct... eqn:...: specify the name of an equation to be added to the context, recording the result of the case analysis
• induction... as...: induction on values of inductively defined types
• inversion: reason by injectivity and distinctness of constructors
• assert (e) as H: introduce a "local lemma" e and call it H
• generalize dependent x: move the variable x (and anything else that depends on it) from the context back to an explicit hypothesis in the goal formula

#### Exercise: 3 stars (beq_nat_sym)

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

#### Exercise: 3 stars, advanced, optional (beq_nat_sym_informal)

Give an informal proof of this lemma that corresponds to your formal proof above:
Theorem: For any nats n m, beq_nat n m = beq_nat m n.
Proof: (* FILL IN HERE *)

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

Theorem beq_nat_trans : n m p,
beq_nat n m = true
beq_nat m p = true
beq_nat n p = true.
Proof.
(* FILL IN HERE *) Admitted.

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

We have just proven that for all lists of pairs, combine is the inverse of split. How would you formalize the statement that split is the inverse of combine?
Complete the definition of split_combine_statement below with a property that states that split is the inverse of combine. Then, prove that the property holds. (Be sure to leave your induction hypothesis general by not doing intros on more things than necessary. Hint: what property do you need of l1 and l2 for split combine l1 l2 = (l1,l2) to be true?)

Definition split_combine_statement : Prop :=
(* FILL IN HERE *) admit.

Theorem split_combine : split_combine_statement.
Proof.
(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars (override_permute)

Theorem override_permute : (X:Type) x1 x2 k1 k2 k3 (f : natX),
beq_nat k2 k1 = false
(override (override f k2 x2) k1 x1) k3 = (override (override f k1 x1) k2 x2) k3.
Proof.
(* FILL IN HERE *) Admitted.

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

This one is a bit challenging. Pay attention to the form of your IH.

Theorem filter_exercise : (X : Type) (test : X bool)
(x : X) (l lf : list X),
filter test l = x :: lf
test x = true.
Proof.
(* FILL IN HERE *) Admitted.

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

Define two recursive Fixpoints, forallb and existsb. The first checks whether every element in a list satisfies a given predicate:
forallb oddb [1;3;5;7;9] = true

forallb negb [false;false] = true

forallb evenb [0;2;4;5] = false

forallb (beq_nat 5) [] = true
The second checks whether there exists an element in the list that satisfies a given predicate:
existsb (beq_nat 5) [0;2;3;6] = false

existsb (andb true) [true;true;false] = true

existsb oddb [1;0;0;0;0;3] = true

existsb evenb [] = false
Next, define a nonrecursive version of existsb — call it existsb' — using forallb and negb.
Prove that existsb' and existsb have the same behavior.

(* FILL IN HERE *)

(* \$Date: 2013-07-30 12:24:33 -0400 (Tue, 30 Jul 2013) \$ *)