# InductionProof by Induction

To prove the following theorem, which tactics will we need besides
intros and reflexivity? (1) none (2) rewrite, (3)
destruct, (4) both rewrite and destruct, or (5) can't be
done with the tactics we've seen.

Theorem review1: (orb true false) = true.

What about the next one?

Theorem review2: ∀b, (orb true b) = true.

Which tactics do we need besides intros and reflexivity? (1)
none (2) rewrite, (3) destruct, (4) both rewrite and
destruct, or (5) can't be done with the tactics we've seen.
What if we change the order of the arguments of orb?

Theorem review3: ∀b, (orb b true) = true.

Which tactics do we need besides intros and reflexivity? (1)
none (2) rewrite, (3) destruct, (4) both rewrite and
destruct, or (5) can't be done with the tactics we've seen.
What about this one?

Theorem review4 : ∀n : nat, 0 + n = n.

(1) none (2) rewrite, (3) destruct, (4) both rewrite and
destruct, or (5) can't be done with the tactics we've seen.
What about this?

Theorem review5 : ∀n : nat, n + 0 = n.

(1) none (2) rewrite, (3) destruct, (4) both rewrite and
destruct, or (5) can't be done with the tactics we've seen.
# Separate Compilation

Require Export Basics.

From now on, we'll use the Case tactic to mark points in
the proof script where we're dealing with different subgoals
generated by destruct (and other tactics like induction, which
we'll see shortly):

Theorem andb_true_elim1 : ∀b c : bool,

andb b c = true → b = true.

Proof.

intros b c H.

destruct b.

Case "b = true". (* <----- here *)

reflexivity.

Case "b = false". (* <---- and here *)

rewrite ← H.

reflexivity.

Qed.

# Proof by Induction

*right*is a little harder to show:

Theorem plus_0_r_firsttry : ∀n:nat,

n + 0 = n.

Proof.

intros n.

simpl. (* Does nothing! *)

Abort.

And reasoning by cases using destruct n doesn't get us much
further: the branch of the case analysis where we assume n = 0
goes through, but in the branch where n = S n' for some n' we
get stuck in exactly the same way. We could use destruct n' to
get one step further, but since n can be arbitrarily large, if we
try to keep on like this we'll never be done.

Theorem plus_0_r_secondtry : ∀n:nat,

n + 0 = n.

Proof.

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

Case "n = 0".

reflexivity. (* so far so good... *)

Case "n = S n'".

simpl. (* ...but here we are stuck again *)

Abort.

We need a bigger hammer: the
In Coq's "goal-directed" style of reasoning, we begin with the
goal of proving P(n) for all n and break it down (by applying
the induction tactic) into two separate subgoals: first showing
P(O) and then showing P(n') → P(S n'). For example...

*principle of induction*over natural numbers...- If P(n) is some proposition involving a natural number n,
and we want to show that P holds for
*all*numbers, we can reason like this:- show that P(O) holds
- show that, if P(n') holds, then so does P(S n')
- conclude that P(n) holds for all n.

Theorem plus_0_r : ∀n:nat, n + 0 = n.

Proof.

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

Case "n = 0". reflexivity.

Case "n = S n'". simpl. rewrite → IHn'. reflexivity. Qed.

Let's try this one together:

Theorem minus_diag : ∀n,

minus n n = 0.

Proof.

(* WORK IN CLASS *) Admitted.

Another result, which we'll need in a moment,
that can be proved by induction (left as an exercise).

Theorem plus_comm : ∀n m : nat,

n + m = m + n.

Proof.

(* FILL IN HERE *) Admitted.

We've seen that there are goals that destruct can't solve but
induction can. What about the other way around? Are there steps
in a proof that can be solved by pure case analysis (destruct)
but not using induction?
(1) No
(2) Yes

# Proofs Within Proofs

Theorem mult_0_plus' : ∀n m : nat,

(0 + n) × m = n × m.

Proof.

intros n m.

assert (H: 0 + n = n).

Case "Proof of assertion". reflexivity.

rewrite → H.

reflexivity. Qed.

Actually, assert will turn out to be handy in many sorts of
situations. For example, suppose we want to prove that (n + m)
+ (p + q) = (m + n) + (p + q). The only difference between the
two sides of the = is that the arguments m and n to the
first inner + are swapped, so it seems we should be able to
use the commutativity of addition (plus_comm) to rewrite one
into the other. However, the rewrite tactic is a little stupid
about

*where*it applies the rewrite. There are three uses of + here, and it turns out that doing rewrite → plus_comm will affect only the*outer*one.Theorem plus_rearrange_firsttry : ∀n m p q : nat,

(n + m) + (p + q) = (m + n) + (p + q).

Proof.

intros n m p q.

(* We just need to swap (n + m) for (m + n)...

it seems like plus_comm should do the trick! *)

rewrite → plus_comm.

(* Doesn't work...Coq rewrote the wrong plus! *)

Abort.

To get plus_comm to apply at the point where we want it, we can
introduce a local lemma stating that n + m = m + n (for
the particular m and n that we are talking about here), prove
this lemma using plus_comm, and then use this lemma to do the
desired rewrite.

Theorem plus_rearrange : ∀n m p q : nat,

(n + m) + (p + q) = (m + n) + (p + q).

Proof.

intros n m p q.

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

Case "Proof of assertion".

rewrite → plus_comm. reflexivity.

rewrite → H. reflexivity. Qed.

(* $Date: 2013-07-17 16:19:11 -0400 (Wed, 17 Jul 2013) $ *)