Exercise 1: transitivity over equals

Theorem eq_trans : forall (T:Type) (x y z : T),
  x = y -> y = z -> x = z.
Proof.
  intros T x y z eq1 eq2.
  rewrite -> eq1.

yields

1 subgoal
T : Type
x, y, z : T
eq1 : x = y
eq2 : y = z
______________________________________(1/1)
y = z

How do we conclude this proof?

Exercise 1: introducing apply

Apply takes an hypothesis/lemma to conclude the goal.

  apply eq2.
Qed.

apply takes ?X to conclude a goal ?X (resolves foralls in the hypothesis).

1 subgoal
T : Type
x, y, z : T
eq1 : x = y
eq2 : y = z
______________________________________(1/1)
y = z

Applying conditional hypothesis

apply uses an hypothesis/theorem of format H1 -> ... -> Hn -> G, then solves goal G, and produces new goals H1, …, Hn.

Theorem eq_trans_2 : forall (T:Type) (x y z: T),
  (x = y -> y = z -> x = z) -> (* eq1 *)
  x = y ->                     (* eq2 *)
  y = z ->                     (* eq3 *)
  x = z.
Proof.
  intros T x y z eq1 eq2 eq3.
  apply eq1.   (* x = y -> y = z -> x = z *)

(Done in class.)

Rewriting conditional hypothesis

apply uses an hypothesis/theorem of format H1 -> ... -> Hn -> G, then solves goal G, and produces new goals H1, …, Hn.

Theorem eq_trans_3 : forall (T:Type) (x y z: T),
  (x = y -> y = z -> x = z) -> (* eq1 *)
  x = y ->                     (* eq2 *)
  y = z ->                     (* eq3 *)
  x = z.
Proof.
  intros T x y z eq1 eq2 eq3.
  rewrite -> eq1.   (* x = y -> y = z -> x = z *)

(Done in class.)

Notice that there are 2 conditions in eq1, so we get 3 goals to solve.

Recap

What's the difference between reflexivity, rewrite, and apply?

1. reflexivity solves goals that can be simplified as an equality like ?X = ?X
2. rewrite -> H takes an hypothesis H of type H1 -> ... -> Hn -> ?X = ?Y, finds any sub-term of the goal that matches ?X and replaces it by ?Y; it also produces goals H1,…, Hn. rewrite does not care about what your goal is, just that the goal must contain a pattern ?X.
3. apply H takes an hypothesis H of type H1 -> ... -> Hn -> G and solves goal G; it creates goals H1, …, Hn.

Apply with/Rewrite with

Theorem eq_trans_nat : forall (x y z: nat),
  x = 1 ->
  x = y ->
  y = z ->
  z = 1.
Proof.
  intros x y z eq1 eq2 eq3.
  assert (eq4: x = z). {
    apply eq_trans.

outputs

Unable to find an instance for the variable y.

We can supply the missing arguments using the keyword with: apply eq_trans with (y:=y).

Can we solve the same theorem but use rewrite instead?

Symmetry

What about this exercise?

Theorem eq_trans_nat : forall (x y z: nat),
  x = 1 ->
  x = y ->
  y = z ->
  1 = z.
Proof.
  intros x y z eq1 eq2 eq3.
  assert (eq4: x = z). {

Apply in example

Theorem silly3' : forall (n : nat),
  (Nat.eqb n 5 = true -> Nat.eqb (S (S n)) 7 = true) ->
  true = Nat.eqb n 5  ->
  true = Nat.eqb (S (S n)) 7.
Proof.
  intros n eq H.
  symmetry in H.
  apply eq in H.

(Done in class.)

Targetting hypothesis

• rewrite -> H1 in H2
• symmetry in H
• apply H1 in H2

Forward vs backward reasoning

If we have a theorem L: C1 -> C2 -> G:

• Goal takes last: apply to goal of type G and replaces G by C1 and C2
• Assumption takes first: apply to hypothesis L to an hypothesis H: C1 and rewrites H:C2 -> G

Proof styles:

• Forward reasoning: (apply in hypothesis) manipulate the hypothesis until we reach a goal.
  Standard in math textbooks.
• Backward reasoning: (apply to goal) manipulate the goal until you reach a state where you can apply the hypothesis.
  Idiomatic in Coq.

Recall our encoding of natural numbers

Inductive nat : Type :=
  | O : nat
  | S : nat -> nat.

1. Does the equation S n = 0 hold? Why?

Proving that S is injective (1/2)

Theorem S_injective : forall (n m : nat),
  S n = S m ->
  n = m.
Proof.
  intros n m eq1.
  injection eq1 as eq2.

If we run injection, we get:

1 subgoal
n, m : nat
eq1 : S n = S m
eq2 : n = m
______________________________________(1/1)
m = m

Disjoint constructors

Theorem Nat.eqb_0_l : forall n,
   Nat.eqb 0 n = true -> n = 0.
Proof.
  intros n eq1.
  destruct n.

(To do in class.)

Principle of explosion

Ex falso (sequitur) quodlibet

discriminate concludes absurd hypothesis, where there is an equality between different constructors. Use discriminate eq1 to conclude the proof below.

1 subgoal
n : nat
eq1 : false = true
______________________________________(1/1)
S n = 0

What we learned…

Tactics.v

• Exploding principle
• Forward and backward proof styles
• New tactics: apply H and apply H in
• Differences between apply and rewrite
• New tactics: symmetry
• New capability: rewrite ... in ...
• New capability: simpl in ...
• Constructors are disjoint and injective