# TacticsMore Basic Tactics

- how to use auxiliary lemmas in both "forward-" 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 strengthen an induction hypothesis, and when such strengthening is required; and
- more details on how to reason by case analysis.

# The apply Tactic

*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 eq

_{1}eq

_{2}.

rewrite <- eq

_{1}.

Here, we could finish with "rewrite → eq

_{2}. reflexivity." as we have done several times before. We can finish this proof 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 →

(n = m → [n;o] = [m;p]) →

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

Proof.

intros n m o p eq

_{1}eq

_{2}.

apply eq

_{2}. apply eq

_{1}. Qed.

Typically, when we use apply H, the statement H will
begin with a ∀ that introduces some

*universally quantified 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.

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

Complete the following proof using only intros and apply.Theorem silly_ex :

(∀ n, evenb n = true → oddb (S n) = true) →

evenb 2 = true →

oddb 3 = true.

Proof.

(* FILL IN HERE *) Admitted.

☐

Here we cannot use apply directly, but we can use the symmetry
tactic, which switches the left and right sides of an equality in
the goal.

symmetry.

simpl.

(This simpl is optional, since apply will perform
simplification first, if needed.)

apply H. Qed.

#### Exercise: 3 stars, standard (apply_exercise1)

(*Hint*: You can use apply with previously defined lemmas, not just hypotheses in the context. Remember that Search is your friend.)

Theorem rev_exercise1 : ∀ (l l' : list nat),

l = rev l' →

l' = rev l.

Proof.

(* FILL IN HERE *) Admitted.

☐

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

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

☐

# 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 that records, 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
variable we are instantiating. We could instead write apply
trans_eq with [c;d].)
Coq also has a tactic transitivity that accomplishes the
same purpose as applying trans_eq. The tactic requires us to
state the instantiation we want, just like apply with does.

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}.

transitivity [c;d].

apply eq

_{1}. apply eq

_{2}. Qed.

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 injection and discriminate Tactics

Inductive nat : Type :=

| O

| S (n : nat).

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

Theorem S_injective : ∀ (n m : nat),

S n = S m →

n = m.

Proof.

intros n m H

_{1}.

assert (H

_{2}: n = pred (S n)). { reflexivity. }

rewrite H

_{2}. rewrite H

_{1}. reflexivity.

Qed.

This technique can be generalized to any constructor by
writing the equivalent of pred -- i.e., writing a function that
"undoes" one application of the constructor. As a more convenient
alternative, Coq provides a tactic called injection that allows
us to exploit the injectivity of any constructor. Here is an
alternate proof of the above theorem using injection:

By writing injection H as Hmn at this point, we are asking Coq
to generate all equations that it can infer from H using the
injectivity of constructors (in the present example, the equation
n = m). Each such equation is added as a hypothesis (with the
name Hmn in this case) into the context.

injection H as Hnm. apply Hnm.

Qed.

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

Theorem injection_ex

_{1}: ∀ (n m o : nat),

[n; m] = [o; o] →

[n] = [m].

Proof.

intros n m o H.

(* WORKED IN CLASS *)

injection H as H

_{1}H

_{2}.

rewrite H

_{1}. rewrite H

_{2}. reflexivity.

Qed.

Alternatively, if you just say injection H with no as clause,
then all the equations will be turned into hypotheses at the
beginning of the goal.

Theorem injection_ex

_{2}: ∀ (n m o : nat),

[n; m] = [o; o] →

[n] = [m].

Proof.

intros n m o H.

injection H.

(* WORKED IN CLASS *)

intros H

_{1}H

_{2}. rewrite H

_{1}. rewrite H

_{2}. reflexivity.

Qed.

Example injection_ex

x :: y :: l = z :: j →

j = z :: l →

x = y.

Proof.

(* FILL IN HERE *) Admitted.

☐

_{3}: ∀ (X : Type) (x y z : X) (l j : list X),x :: y :: l = z :: j →

j = z :: l →

x = y.

Proof.

(* FILL IN HERE *) Admitted.

☐

*assumed*that two such terms are equal, we are justified in concluding anything we want, since the assumption is nonsensical.

We can proceed by case analysis on n. The first case is
trivial.

destruct n as [| n'] eqn:E.

- (* n = 0 *)

intros H. reflexivity.

However, the second one doesn't look so simple: assuming 0
=? (S n') = true, we must show S n' = 0! The way forward is to
observe that the assumption itself is nonsensical:

- (* n = S n' *)

simpl.

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

intros H. discriminate H.

Qed.

This is an instance of a logical principle known as the

*principle of explosion*, which asserts that a contradictory hypothesis entails anything (even false things!).Theorem discriminate_ex

_{1}: ∀ (n : nat),

S n = O →

2 + 2 = 5.

Proof.

intros n contra. discriminate contra. Qed.

Theorem discriminate_ex

_{2}: ∀ (n m : nat),

false = true →

[n] = [m].

Proof.

intros n m contra. discriminate contra. Qed.

If you find the principle of explosion confusing, remember
that these proofs are
#### Exercise: 1 star, standard (discriminate_ex

*not*showing that the conclusion of the statement holds. Rather, they are showing that,*if*the nonsensical situation described by the premise did somehow arise,*then*the nonsensical conclusion would also follow, because we'd be living in an inconsistent universe where every statement is true. We'll explore the principle of explosion in more detail in the next chapter.#### Exercise: 1 star, standard (discriminate_ex_{3})

Example discriminate_ex

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

x :: y :: l = [] →

x = z.

Proof.

(* FILL IN HERE *) Admitted.

☐

_{3}:∀ (X : Type) (x y z : X) (l j : list X),

x :: y :: l = [] →

x = z.

Proof.

(* FILL IN HERE *) Admitted.

☐

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.

Theorem eq_implies_succ_equal : ∀ (n m : nat),

n = m → S n = S m.

Proof. intros n m H. apply f_equal. apply H. Qed.

There is also a tactic named `f_equal` that can prove such
theorems. Given a goal of the form f a

_{1}... an = g b_{1}... bn, the tactic f_equal will produce subgoals of the form f = g, a_{1}= b_{1}, ..., an = bn. At the same time, any of these subgoals that are simple enough (e.g., immediately provable by reflexivity) will be automatically discharged by f_equal.Theorem eq_implies_succ_equal' : ∀ (n m : nat),

n = m → S n = S m.

Proof. intros n m H. f_equal. apply H. Qed.

# Using Tactics on Hypotheses

Theorem S_inj : ∀ (n m : nat) (b : bool),

(S n) =? (S m) = b →

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

Theorem silly3' : ∀ (n : nat),

(n =? 5 = true → (S (S n)) =? 7 = true) →

true = (n =? 5) →

true = ((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
The informal proofs that you've seen in math or computer science
classes probably tended to use forward reasoning. In general,
idiomatic use of Coq favors backward reasoning, but in some
situations the forward style can be easier to think about.

*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.# Varying the Induction Hypothesis

Theorem double_injective: ∀ n m,

double n = double m → n = m.

intros n. induction n.

intros n m. induction n.

Theorem double_injective_FAILED : ∀ n m,

double n = double m →

n = m.

Proof.

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

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

+ (* m = O *) reflexivity.

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

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

+ (* m = O *) discriminate 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: that, for a
then we can prove
To see why this is strange, let's think of a particular (arbitrary,
but fixed) 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
helpful about whether double n is 10 (indeed, it strongly
suggests that double n is
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 statement involving
A successful proof of double_injective leaves m in the goal
statement at the point where the induction tactic is invoked on
n:

*those 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".

*not*10!!), so Q is useless.*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' IHn'].

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

+ (* m = O *) reflexivity.

+ (* m = S m' *) simpl in eq. discriminate eq.

- (* n = S n' *)

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 whichever 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'] eqn:E.

+ (* m = O *)

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'. simpl in eq. injection eq as goal. apply goal. Qed.

What you should take away from all this is that we need to be
careful, when using induction, that we are not trying to prove
something too specific: When proving a property involving two
variables n and m by induction on n, it is sometimes
crucial to leave m generic.
The following exercise follows the same pattern.

#### Exercise: 2 stars, standard (eqb_true)

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

Give a careful informal proof of eqb_true, being as explicit as possible about quantifiers.(* FILL IN HERE *)

(* Do not modify the following line: *)

Definition manual_grade_for_informal_proof : option (nat×string) := None.

☐

#### Exercise: 3 stars, standard, recommended (plus_n_n_injective)

In addition to being careful about how you use intros, practice using "in" variants in this proof. (Hint: use plus_n_Sm.)*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' IHm'].

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

+ (* n = O *) reflexivity.

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

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

+ (* n = O *) discriminate eq.

+ (* 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. (And 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
works, 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! Rather we want to state them in the clearest and
most 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' IHm'].

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

+ (* n = O *) reflexivity.

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

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

+ (* n = O *) discriminate eq.

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

apply IHm'. injection eq as goal. apply goal. 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.
- 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'.

#### Exercise: 3 stars, standard, recommended (gen_dep_practice)

Prove this by induction on l.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.

☐

# Unfolding Definitions

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

... we appear to be stuck: simpl doesn't simplify anything, 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 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, some discussion of unfolding and simplification is
in order.
We already have observed that tactics like simpl, reflexivity,
and apply will often unfold the definitions of functions
automatically when this 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 somewhat 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 cannot itself 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 eqn:E.

- 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 forward 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 eqn:E.

- reflexivity.

- reflexivity.

Qed.

# Using destruct on Compound Expressions

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

Definition sillyfun (n : nat) : bool :=

if n =? 3 then false

else if n =? 5 then false

else false.

Theorem sillyfun_false : ∀ (n : nat),

sillyfun n = false.

Proof.

intros n. unfold sillyfun.

destruct (n =? 3) eqn:E

_{1}.

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

- (* n =? 3 = false *) destruct (n =? 5) eqn:E

_{2}.

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

+ (* n =? 5 = false *) reflexivity. Qed.

After unfolding sillyfun in the above proof, we find that
we are stuck on if (n =? 3) then ... else .... But either
n is equal to 3 or it isn't, so we can use destruct (eqb
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, standard (combine_split)

Here is an implementation of the split function mentioned in chapter Poly:Fixpoint split {X Y : Type} (l : list (X×Y))

: (list X) × (list Y) :=

match l with

| [] ⇒ ([], [])

| (x, y) :: t ⇒

match split t with

| (lx, ly) ⇒ (x :: lx, y :: ly)

end

end.

Prove that split and combine are inverses in the following
sense:

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

_{1}l

_{2},

split l = (l

_{1}, l

_{2}) →

combine l

_{1}l

_{2}= l.

Proof.

(* FILL IN HERE *) Admitted.

☐

Now suppose that we want to convince Coq that sillyfun1 n
yields true only when n is odd. If we start the proof like
this (with no eqn: on the destruct)...

Theorem sillyfun1_odd_FAILED : ∀ (n : nat),

sillyfun1 n = true →

oddb n = true.

Proof.

intros n eq. unfold sillyfun1 in eq.

destruct (n =? 3).

(* stuck... *)

Abort.

... then we are 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 quite brutal --
in this case, it thows away every occurrence of 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 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 want here is to substitute away all existing occurences of
n =? 3, but at the same time add an equation to the context that
records which case we are in. This is precisely what the eqn:
qualifier does.

Theorem sillyfun1_odd : ∀ (n : nat),

sillyfun1 n = true →

oddb n = true.

Proof.

intros n eq. unfold sillyfun1 in eq.

destruct (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 eqb_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, allowing us to finish the

proof. *)

destruct (n =? 5) eqn:Heqe5.

+ (* e

_{5}= true *)

apply eqb_true in Heqe5.

rewrite → Heqe5. reflexivity.

+ (* e

_{5}= false *) discriminate 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
- transitivity y: prove a goal x=z by proving two new subgoals,
x=y and y=z
- 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
- injection: reason by injectivity on equalities
between values of inductively defined types
- discriminate: reason by disjointness of constructors on
equalities between values of inductively defined types
- assert (H: e) (or 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
- f_equal: change a goal of the form f x = f y into x = y

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

Give an informal proof of this lemma that corresponds to your formal proof above:
(* FILL IN HERE *)

☐

☐

Theorem eqb_trans : ∀ n m p,

n =? m = true →

m =? p = true →

n =? p = true.

Proof.

(* FILL IN HERE *) Admitted.

☐

n =? m = true →

m =? p = true →

n =? p = true.

Proof.

(* FILL IN HERE *) Admitted.

☐

#### 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

(* (": Prop" means that we are giving a name to a

logical proposition here.) *)

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

Theorem split_combine : split_combine_statement.

Proof.

(* FILL IN HERE *) Admitted.

(* Do not modify the following line: *)

Definition manual_grade_for_split_combine : option (nat×string) := None.

☐

#### 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.

☐

#### 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 (eqb 5) [] = true

existsb (eqb 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

*nonrecursive*version of existsb -- call it existsb' -- using forallb and negb.

Fixpoint forallb {X : Type} (test : X → bool) (l : list X) : bool

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

Example test_forallb_1 : forallb oddb [1;3;5;7;9] = true.

Proof. (* FILL IN HERE *) Admitted.

Example test_forallb_2 : forallb negb [false;false] = true.

Proof. (* FILL IN HERE *) Admitted.

Example test_forallb_3 : forallb evenb [0;2;4;5] = false.

Proof. (* FILL IN HERE *) Admitted.

Example test_forallb_4 : forallb (eqb 5) [] = true.

Proof. (* FILL IN HERE *) Admitted.

Fixpoint existsb {X : Type} (test : X → bool) (l : list X) : bool

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

Example test_existsb_1 : existsb (eqb 5) [0;2;3;6] = false.

Proof. (* FILL IN HERE *) Admitted.

Example test_existsb_2 : existsb (andb true) [true;true;false] = true.

Proof. (* FILL IN HERE *) Admitted.

Example test_existsb_3 : existsb oddb [1;0;0;0;0;3] = true.

Proof. (* FILL IN HERE *) Admitted.

Example test_existsb_4 : existsb evenb [] = false.

Proof. (* FILL IN HERE *) Admitted.

Definition existsb' {X : Type} (test : X → bool) (l : list X) : bool

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

Theorem existsb_existsb' : ∀ (X : Type) (test : X → bool) (l : list X),

existsb test l = existsb' test l.

Proof. (* FILL IN HERE *) Admitted.

☐

(* 30 Apr 2020 *)