# TacticsMore Basic Tactics

- how to use auxiliary lemmas in both "forward-style" and "backward-style" proofs;
- how to reason about data constructors (in particular, how to use the fact that they are injective and disjoint);
- how to create a strong induction hypothesis (and when such strengthening is required); and
- more details on how to reason by case analysis.

# The apply Tactic

Theorem silly1 : ∀(n m o p : nat),

n = m →

[n;o] = [n;p] →

[n;o] = [m;p].

Proof.

intros n m o p eq

_{1}eq

_{2}.

rewrite ← eq

_{1}.

At this point, we could finish with "rewrite → eq

_{2}. reflexivity." as we have done several times before. We can achieve the same effect in a single step by using the apply tactic instead:apply eq

_{2}. 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 eq

_{1}eq

_{2}.

apply eq

_{2}. apply eq

_{1}. 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 ∀ that binds 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 eq_{2}in the following proof, the universal variable q in eq_{2}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 eq

_{1}eq

_{2}.

apply eq

_{2}. apply eq

_{1}. Qed.

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.

☐
(* FILL IN HERE *)

☐

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

Briefly explain the difference between the tactics apply and rewrite. What are the situations where both can usefully be applied?☐

# The apply ... with ... 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 eq

_{1}eq

_{2}.

rewrite → eq

_{1}. rewrite → eq

_{2}. reflexivity. Qed.

Since this is a common pattern, we might like to pull 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 eq

_{1}eq

_{2}. rewrite → eq

_{1}. rewrite → eq

_{2}.

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 eq

_{1}eq

_{2}.

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.

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.

m = (minustwo o) →

(n + p) = m →

(n + p) = (minustwo o).

Proof.

(* FILL IN HERE *) Admitted.

☐

# The inversion Tactic

Inductive nat : Type :=

| O : nat

| S : nat → nat.

It is obvious 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 more facts:
| O : nat

| S : nat → nat.

- The constructor S is
*injective*. That is, if S n = S m, it must be the case that n = m. - The constructors O and S are
*disjoint*. That is, O is not equal to S n for any n.

By writing inversion H at this point, we ask Coq to
generate all equations that it can infer from H as additional
hypotheses, replacing variables in the goal as it goes. In the
present example, this amounts to adding a new hypothesis H

_{1}: n = m and replacing n by m in the goal.inversion H. reflexivity. Qed.

Here's a more interesting example that shows how multiple
equations can be derived at once.

Theorem inversion_ex

_{1}: ∀(n m o : nat),

[n; m] = [o; o] →

[n] = [m].

Proof.

intros n m o H. inversion H. reflexivity. Qed.

It is possible to name the equations that inversion
generates with an as ... clause:

Theorem inversion_ex

_{2}: ∀(n m : nat),

[n] = [m] →

n = m.

Proof.

intros n o H. inversion H as [Hno]. reflexivity. Qed.

Example inversion_ex

x :: y :: l = z :: j →

y :: l = x :: j →

x = y.

Proof.

(* FILL IN HERE *) Admitted.

_{3}: ∀(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.

☐
While the injectivity of constructors allows us to reason
that ∀ (n m : nat), S n = S m → n = m, the converse of
this implication is an instance of a more general fact about
constructors and functions, which we will find useful below:

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.

When used on a hypothesis involving an equality between

*different*constructors (e.g., S n = O), inversion solves the goal immediately. To see why this makes sense, consider the following proof:
We can proceed by case analysis on n. The first case is
trivial.

destruct n as [| n'].

- (* n = 0 *)

intros H. reflexivity.

However, the second one doesn't look so simple: assuming
beq_nat 0 (S n') = true, we must show S n' = 0, but the latter
clearly contradictory! The way forward lies in the assumption.
After simplifying the goal state, we see that beq_nat 0 (S n') =
true has become false = true:

- (* n = S n' *)

simpl.

If we use inversion on this hypothesis, Coq notices that
the subgoal we are working on is impossible, and therefore removes
it from further consideration.

intros H. inversion H. Qed.

This is an instance of a general logical principle known as
the

*principle of explosion*, which asserts that a contradiction entails anything, even false things. For instance:Theorem inversion_ex

_{4}: ∀(n : nat),

S n = O →

2 + 2 = 5.

Proof.

intros n contra. inversion contra. Qed.

Theorem inversion_ex

_{5}: ∀(n m : nat),

false = true →

[n] = [m].

Proof.

intros n m contra. inversion contra. Qed.

If you find the principle of explosion confusing, remember
that these proofs are not actually showing that the conclusion of
the statement holds. Rather, they are arguing that the situation
described by the premise can never arise, so the implication is
vacuous. We'll explore the principle of explosion of more detail
in the next chapter.
#### Exercise: 1 star (inversion_ex

#### Exercise: 1 star (inversion_ex_{6})

Example inversion_ex

(x y z : X) (l j : list X),

x :: y :: l = [] →

y :: l = z :: j →

x = z.

Proof.

(* FILL IN HERE *) Admitted.

_{6}: ∀(X : Type)(x y z : X) (l j : list X),

x :: y :: l = [] →

y :: l = z :: j →

x = z.

Proof.

(* FILL IN HERE *) Admitted.

☐
To summarize this discussion, suppose H is a hypothesis in the
context or a previously proven lemma of the form

c a

for some constructors c and d and arguments a_{1}a_{2}... an = d b_{1}b_{2}... bm_{1}... an and b_{1}... bm. Then inversion H has the following effect:- If c and d are the same constructor, then, by the
injectivity of this constructor, we know that a
_{1}= b_{1}, a_{2}= b_{2}, 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, and the current goal doesn't have to be considered. In this case, inversion H marks the current goal as completed and pops it off the goal stack.

# Using Tactics on Hypotheses

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, apply L in H matches some conditional statement
L (of the form L
In other words, apply L in H gives us a form of "forward
reasoning": from L
Here is a variant of a proof from above, using forward reasoning
throughout instead of backward reasoning.

_{1}→ L_{2}, say) against a hypothesis H in the context. However, unlike ordinary apply (which rewrites a goal matching L_{2}into a subgoal L_{1}), apply L in H matches H against L_{1}and, if successful, replaces it with L_{2}._{1}→ L_{2}and a hypothesis matching L_{1}, it produces a hypothesis matching L_{2}. By contrast, apply L is "backward reasoning": it says that if we know L_{1}→L_{2}and we are trying to prove L_{2}, it suffices to prove L_{1}.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, idiomatic use of Coq tends to favor backward reasoning, but in some situations the forward style can be easier to think about.#### Exercise: 3 stars, recommended (plus_n_n_injective)

Practice using "in" variants in this exercise. (Hint: use plus_n_Sm.)Theorem plus_n_n_injective : ∀n m,

n + n = m + m →

n = m.

Proof.

intros n. induction n as [| n'].

(* FILL IN HERE *) Admitted.

☐

# Varying the Induction Hypothesis

Theorem double_injective: ∀n m,

double n = double m → n = m.

The way we double n = double m → n = m.

*start*this proof is a bit delicate: if we begin 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'].

- (* n = O *) simpl. intros eq. destruct m as [| m'].

+ (* m = O *) reflexivity.

+ (* m = S m' *) inversion eq.

- (* n = S n' *) intros eq. destruct m as [| m'].

+ (* m = O *) inversion eq.

+ (* m = S m' *) apply f_equal.

At this point, 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
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, for
holds, by showing
If we look closely at the second statement, it is saying something
rather strange: it says that, for a
then we can prove
To see why this is strange, let's think of a particular m —
say, 5. The statement is then saying that, if we know
then we can prove
But knowing Q doesn't give us any help at all with proving
R! (If we tried to prove R from Q, we would start with
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.)
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
then trying to prove a relation involving
The good proof of double_injective leaves m in the goal
statement at the point where the induction tactic is invoked on
n:

*these particular*n and m, then n = m.*all*n, that the proposition- P n = "if double n = double m, then n = m"

- P O
- P n → P (S n)

*particular*m, if we know- "if double n = double m then n = m"

- "if double (S n) = double m then S n = m".

- Q = "if double n = 10 then n = 5"

- R = "if double (S n) = 10 then S n = 5".

*every*n but just a*single*m.Theorem double_injective : ∀n m,

double n = double m →

n = m.

Proof.

intros n. induction n as [| n'].

- (* n = O *) simpl. intros m eq. destruct m as [| m'].

+ (* m = O *) reflexivity.

+ (* m = S m' *) inversion eq.

- (* n = S n' *) simpl.

Notice that both the goal and the induction hypothesis are
different this time: 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've chosen a particular m and introduced the assumption
that double n = double m. Since we are doing a case analysis on
n, we also need a case analysis on m to keep the two "in sync."

destruct m as [| m'].

+ (* m = O *) simpl.

The 0 case is trivial:

At this point, since we are in the second branch of the destruct
m, the m' mentioned in the context is the predecessor of the
m 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 current m' (this instantiation is
performed automatically by the apply in the next step), then
IHn' gives us exactly what we need to finish the proof.

apply IHn'. inversion eq. reflexivity. Qed.

What you should take away from all this 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 following exercise requires the same pattern.

#### Exercise: 2 stars (beq_nat_true)

☐

#### 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 to
obtain a more general IH doesn't always work by itself; 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'].

- (* m = O *) simpl. intros eq. destruct n as [| n'].

+ (* n = O *) reflexivity.

+ (* n = S n' *) inversion eq.

- (* m = S m' *) intros eq. destruct n as [| n'].

+ (* n = O *) inversion eq.

+ (* n = S n' *) apply f_equal.

(* Stuck again here, just like before. *)

Abort.

The problem here 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 twist 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, selectively taking variables 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'].

- (* m = O *) simpl. intros n eq. destruct n as [| n'].

+ (* n = O *) reflexivity.

+ (* n = S n' *) inversion eq.

- (* m = S m' *) intros n eq. destruct n as [| n'].

+ (* n = O *) inversion eq.

+ (* 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.
Before we close this section and move on to some exercises, let's
digress briefly and use beq_nat_true to prove a similar property
about identifiers that we'll need in later chapters:

*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.
- Second, 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'.

Theorem beq_id_true : ∀x y,

beq_id x y = true → x = y.

Proof.

intros [m] [n]. simpl. intros H.

assert (H' : m = n). { apply beq_nat_true. apply H. }

rewrite H'. reflexivity.

Qed.

Theorem nth_error_after_last: ∀(n : nat) (X : Type) (l : list X),

length l = n →

nth_error l n = None.

Proof.

(* FILL IN HERE *) Admitted.

☐

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

Prove this by induction on l_{1}, without using app_length from Lists.Theorem app_length_cons : ∀(X : Type) (l

_{1}l

_{2}: list X)

(x : X) (n : nat),

length (l

_{1}++ (x :: l

_{2})) = n →

S (length (l

_{1}++ l

_{2})) = n.

Proof.

(* FILL IN HERE *) Admitted.

☐

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

Prove this by induction on l, without using app_length from Lists.Theorem app_length_twice : ∀(X:Type) (n:nat) (l:list X),

length l = n →

length (l ++ l) = n + n.

Proof.

(* FILL IN HERE *) Admitted.

☐

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

Prove the following principle of induction over two naturals.Theorem double_induction: ∀(P : nat → nat → Prop),

P 0 0 →

(∀m, P m 0 → P (S m) 0) →

(∀n, P 0 n → P 0 (S n)) →

(∀m n, P m n → P (S m) (S n)) →

∀m n, P m n.

Proof.

(* FILL IN HERE *) Admitted.

☐

# Unfolding Definitions

... and try to prove a simple fact about square...

... we get stuck: simpl doesn't simplify anything at this point,
and since we haven't proved any other facts about square, there
is nothing we can apply or rewrite with.
To make progress, we can manually unfold the definition of
square:

Now we have plenty to work with: both sides of the equality are
expressions involving multiplication, and we have lots of facts
about multiplication at our disposal. In particular, we know that
it is commutative and associative, and from these facts it is not
hard to finish the proof.

rewrite mult_assoc.

assert (H : n * m * n = n * n * m).

{ rewrite mult_comm. apply mult_assoc. }

rewrite H. rewrite mult_assoc. reflexivity.

Qed.

At this point, a deeper discussion of unfolding and simplification
is in order.
You may already have observed that tactics like simpl,
reflexivity, and apply will often unfold the definitions of
functions automatically when it allows them to make progress. For
example, if we define foo m to be the constant 5,

then the simpl in the following proof (or the reflexivity, if
we omit the simpl) will unfold foo m to (fun x ⇒ 5) m and
then further simplify this expression to just 5.

However, this automatic unfolding is rather conservative. For
example, if we define a slightly more complicated function
involving a pattern match...

...then the analogous proof will get stuck:

Fact silly_fact_2_FAILED : ∀m, bar m + 1 = bar (m + 1) + 1.

Proof.

intros m.

simpl. (* Does nothing! *)

Abort.

The reason that simpl doesn't make progress here is that it
notices that, after tentatively unfolding bar m, it is left with
a match whose scrutinee, m, is a variable, so the match cannot
be simplified further. (It is not smart enough to notice that the
two branches of the match are identical.) So it gives up on
unfolding bar m and leaves it alone. Similarly, tentatively
unfolding bar (m+1) leaves a match whose scrutinee is a
function application (that, itself, cannot be simplified, even
after unfolding the definition of +), so simpl leaves it
alone.
At this point, there are two ways to make progress. One is to use
destruct m to break the proof into two cases, each focusing on a
more concrete choice of m (O vs S _). In each case, the
match inside of bar can now make progress, and the proof is
easy to complete.

Fact silly_fact_2 : ∀m, bar m + 1 = bar (m + 1) + 1.

Proof.

intros m.

destruct m.

- simpl. reflexivity.

- simpl. reflexivity.

Qed.

This approach works, but it depends on our recognizing that the
match hidden inside bar is what was preventing us from making
progress.
A more straightforward way to finish the proof is to explicitly
tell Coq to unfold bar.

Now it is apparent that we are stuck on the match expressions on
both sides of the =, and we can use destruct to finish the
proof without thinking too hard.

destruct m.

- reflexivity.

- reflexivity.

Qed.

# Using destruct on Compound Expressions

*expression*. We can also do this with destruct.

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).

- (* beq_nat n 3 = true *) reflexivity.

- (* beq_nat n 3 = false *) destruct (beq_nat n 5).

+ (* beq_nat n 5 = true *) reflexivity.

+ (* 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 .... But either
n is equal to 3 or it isn't, so we can 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.

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

Theorem combine_split : ∀X Y (l : list (X * Y)) l

split l = (l

combine l

Proof.

(* FILL IN HERE *) Admitted.

_{1}l_{2},split l = (l

_{1}, l_{2}) →combine l

_{1}l_{2}= l.Proof.

(* FILL IN HERE *) Admitted.

☐
However, destructing compound expressions requires a bit of
care, as such destructs can sometimes 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.

Now suppose that we want to convince Coq of the (rather
obvious) fact 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 performed 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 beq_nat n 3 = true in this
branch of the case analysis, 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, giving it a
name that 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. *)

- (* e

_{3}= true *) apply beq_nat_true in Heqe3.

rewrite → Heqe3. reflexivity.

- (* e

_{3}= 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.

+ (* e

_{5}= true *)

apply beq_nat_true in Heqe5.

rewrite → Heqe5. reflexivity.

+ (* e

_{5}= false *) inversion eq. Qed.

Theorem bool_fn_applied_thrice :

∀(f : bool → bool) (b : bool),

f (f (f b)) = f b.

Proof.

(* FILL IN HERE *) Admitted.

∀(f : bool → bool) (b : bool),

f (f (f b)) = f b.

Proof.

(* FILL IN HERE *) Admitted.

☐

# Review

*automation*tactics that make Coq help us with low-level details. But basically we've got what we need to get work done.

- 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

☐
Theorem: For any nats n m, beq_nat n m = beq_nat m n.
Proof:
(* FILL IN HERE *)

☐

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

Give an informal proof of this lemma that corresponds to your formal proof above:☐

#### 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.

beq_nat n m = true →

beq_nat m p = true →

beq_nat n p = true.

Proof.

(* FILL IN HERE *) Admitted.

☐
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 l

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

We proved, in an exercise above, 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? When is this property true?_{1}and l_{2}for split combine l_{1}l_{2}= (l_{1},l_{2}) to be true?)Definition split_combine_statement : Prop

(* REPLACE THIS LINE WITH := _your_definition_ . *) . Admitted.

Theorem split_combine : split_combine_statement.

Proof.

(* FILL IN HERE *) Admitted.

☐

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

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

☐
Finally, prove a theorem existsb_existsb' stating that
existsb' and existsb have the same behavior.

#### Exercise: 4 stars, advanced, recommended (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:
forallb negb [false;false] = true

forallb evenb [0;2;4;5] = false

forallb (beq_nat 5) [] = true

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

existsb oddb [1;0;0;0;0;3] = true

existsb evenb [] = false

*nonrecursive*version of existsb — call it existsb' — using forallb and negb.(* FILL IN HERE *)

☐