# TacticsMore Basic Tactics

# 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}.

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.

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.

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 silly3_firsttry : ∀ (n : nat),

true = (n =? 5) →

(S (S n)) =? 7 = true.

Proof.

intros n H.

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.

# 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.

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}.

Doing apply trans_eq doesn't work! But...

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

does.

apply eq

_{1}. apply eq

_{2}. Qed.

transitivity is also a 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}.

transitivity [c;d].

apply eq

_{1}. apply eq

_{2}. Qed.

# The injection and discriminate Tactics

*injective*and

*disjoint*.

- if S n = S m then it must hold that n = m (that is, S is
*injective*aka*one-to-one*) - O is not equal to S n for any n (that is, O and S are
*disjoint*)

### For example, we can prove the injectivity of S by using the pred function defined in Basics.v.

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.

Theorem S_injective' : ∀ (n m : nat),

S n = S m →

n = m.

Proof.

intros n m H.

injection H as Hnm. apply Hnm.

Qed.

Theorem injection_ex

_{1}: ∀ (n m o : nat),

[n; m] = [o; o] →

[n] = [m].

Proof.

intros n m o H.

(* WORK IN CLASS *) Admitted.

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

### The discriminate tactic embodies this principle: It is used on a hypothesis involving an equality between different constructors (e.g., S n = O), and it solves the current goal immediately. Here is an example:

Theorem eqb_0_l : ∀ n,

0 =? n = true → n = 0.

Proof.

intros n.

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

- (* n = 0 *)

intros H. reflexivity.

- (* n = S n' *)

simpl.

intros H. discriminate H.

Qed.

*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.

Suppose Coq's proof state looks like

x : bool

y : bool

H : negb x = negb y

============================

y = x
and we apply the tactic injection H. What will happen?
(1) "No more subgoals."
(2) The tactic fails.
(3) The goal becomes x = y → y = x .
(4) None of the above.

x : bool

y : bool

H : negb x = negb y

============================

y = x

Now suppose Coq's proof state looks like

x : nat

y : nat

H : x + 1 = y + 1

============================

y = x
and we apply the tactic injection H. What will happen?
(1) "No more subgoals."
(2) The tactic fails.
(3) The goal becomes x = y → y = x .
(4) None of the above.

x : nat

y : nat

H : x + 1 = y + 1

============================

y = x

Finally, suppose Coq's proof state looks like

x : nat

y : nat

H : 1 + x = 1 + y

============================

y = x
and we apply the tactic injection H. What will happen?
(1) "No more subgoals."
(2) The tactic fails.
(3) The goal becomes x = y → y = x .
(4) None of the above.

x : nat

y : nat

H : 1 + x = 1 + y

============================

y = x

### 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 both constructors and functions, which we will find convenient in a few places 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.

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.

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.

### The ordinary apply tactic is a form of "backward reasoning": it says "We're trying to prove X and we know Y → X, so if we can prove Y we'll be done."

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.

# Varying the Induction Hypothesis

Fixpoint double (n:nat) :=

match n with

| O ⇒ O

| S n' ⇒ S (S (double n'))

end.

### Suppose we want to show that double is injective -- i.e., that it maps different arguments to different results. The way we

*start*this proof is a little bit delicate:

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.

*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' *)

intros m eq.

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

+ (* m = O *)

simpl in eq. discriminate eq.

+ (* m = S m' *)

apply f_equal.

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.

Theorem eqb_true : ∀ n m,

n =? m = true → n = m.

Proof.

(* WORK IN CLASS *) Admitted.

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

Theorem plus_n_n_injective : ∀ n m,

n + n = m + m →

n = m.

Proof.

(* WORK IN CLASS *) Admitted.

n + n = m + m →

n = m.

Proof.

(* WORK IN CLASS *) Admitted.

### The strategy of doing fewer intros before an induction to obtain a more general IH doesn't always work by itself; sometimes some

*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 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.

# Unfolding Definitions

Definition square n := n × n.

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

Lemma square_mult : ∀ n m, square (n × m) = square n × square m.

Proof.

intros n m.

simpl.

... 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:

unfold 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.

Definition foo (x: nat) := 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.

Fact silly_fact_1 : ∀ m, foo m + 1 = foo (m + 1) + 1.

Proof.

intros m.

simpl.

reflexivity.

Qed.

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

Definition bar x :=

match x with

| O ⇒ 5

| S _ ⇒ 5

end.

...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.

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

Proof.

intros m.

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

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.

### The eqn: part of the destruct tactic is optional: So far, we've chosen to include it most of the time, just for the sake of documentation.

Definition sillyfun1 (n : nat) : bool :=

if n =? 3 then true

else if n =? 5 then true

else false.

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.

Theorem sillyfun1_odd : ∀ (n : nat),

sillyfun1 n = true →

oddb n = true.

Proof.

intros n eq. unfold sillyfun1 in eq.

destruct (n =? 3) eqn:Heqe3.

- (* e

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

rewrite → Heqe3. reflexivity.

- (* e

_{3}= false *)

destruct (n =? 5) eqn:Heqe5.

+ (* e

_{5}= true *)

apply eqb_true in Heqe5.

rewrite → Heqe5. reflexivity.

+ (* e

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