# The apply Tactic

We can use apply when some hypothesis or an earlier lemma exactly matches the goal:

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.
Here, we could finish with "rewrite eq2. reflexivity." as we have done several times before. We can finish this proof in a single step by using the apply tactic instead:

apply eq2. Qed.
apply also works with conditional hypotheses:

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 eq1 eq2.
apply eq2. apply eq1. Qed.
Observe how Coq picks appropriate values for the -quantified variables of the hypothesis:

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.
The goal must match the hypothesis exactly for apply to work:

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 applywith 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.
rewriteeq1. rewriteeq2. 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 = mm = on = o.
Proof.
intros X n m o eq1 eq2. rewriteeq1. rewriteeq2.
reflexivity. Qed.
Applying this lemma to the example above requires a slight variation on apply:

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.
Doing apply trans_eq doesn't work! But...
apply trans_eq with (m:=[c;d]).
does.

apply eq1. apply eq2. 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 eq1 eq2.
transitivity [c;d].
apply eq1. apply eq2. Qed.

# The injection and discriminate Tactics

In Coq, the constructors of inductive types are injective and disjoint.
E.g., for nat...
• 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 H1.
assert (H2: n = pred (S n)). { reflexivity. }
rewrite H2. rewrite H1. reflexivity.
Qed.
As a convenience, the injection tactic allows us to exploit injectivity of any constructor.

Theorem S_injective' : (n m : nat),
S n = S m
n = m.
Proof.
intros n m H.
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_ex1 : (n m o : nat),
[n; m] = [o; o] →
[n] = [m].
Proof.
intros n m o H.
(* WORK IN CLASS *) Admitted.
So much for injectivity of constructors. What about disjointness?
The principle of disjointness says that two terms beginning with different constructors (like O and S, or true and false) can never be equal. This means that, any time we find ourselves in a context where we've 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 = truen = 0.
Proof.
intros n.

destruct n as [| n'] eqn:E.
- (* n = 0 *)
intros H. reflexivity.

- (* n = S n' *)
simpl.

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_ex1 : (n : nat),
S n = O
2 + 2 = 5.
Proof.
intros n contra. discriminate contra. Qed.

Theorem discriminate_ex2 : (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.

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.

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.

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: AB) (x y: A),
x = yf x = f y.
Proof. intros A B f x y eq. rewrite eq. reflexivity. Qed.

Theorem eq_implies_succ_equal : (n m : nat),
n = mS n = S m.
Proof. intros n m H. apply f_equal. apply H. Qed.
Or use the f_equal tactic.

Theorem eq_implies_succ_equal' : (n m : nat),
n = mS n = S m.
Proof. intros n m H. f_equal. apply H. Qed.

# Using Tactics on Hypotheses

Many tactics come with "... in ..." variants that work on hypotheses instead of goals.

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."
By contrast, the variant apply... in... is "forward reasoning": it says "We know Y and we know Y X, so we also know X."

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

Recall this function for doubling a natural number (from the Induction chapter):

Fixpoint double (n:nat) :=
match n with
| OO
| 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.
What went wrong?
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 every n but just a single m.
A successful 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' 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.
The following theorem illustrates the same point:

Theorem eqb_true : n m,
n =? m = truen = 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.
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 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.

# Unfolding Definitions

It sometimes happens that we need to manually unfold a name that has been introduced by a Definition so that we can manipulate the expression it denotes. For example, if we define...

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.
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...

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

The destruct tactic can be used on expressions as well as variables:

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:E1.
- (* n =? 3 = true *) reflexivity.
- (* n =? 3 = false *) destruct (n =? 5) eqn:E2.
+ (* 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.
However, when destructing compound expressions, the information recorded by the eqn: can actually be critical: if we leave it out, then destruct can erase information we need to complete a proof.

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.
Adding the eqn: qualifier saves this information so we can use it.

Theorem sillyfun1_odd : (n : nat),
sillyfun1 n = true
oddb n = true.
Proof.
intros n eq. unfold sillyfun1 in eq.
destruct (n =? 3) eqn:Heqe3.
- (* e3 = true *) apply eqb_true in Heqe3.
rewriteHeqe3. reflexivity.
- (* e3 = false *)
destruct (n =? 5) eqn:Heqe5.
+ (* e5 = true *)
apply eqb_true in Heqe5.
rewriteHeqe5. reflexivity.
+ (* e5 = false *) discriminate eq. Qed.

# Micro Sermon

Mindless proof-hacking is a terrible temptation...
Try to resist!