# LogicLogic in Coq

Set Warnings "-notation-overridden,-parsing".

From LF Require Export Tactics.

From LF Require Export Tactics.

In previous chapters, we have seen many examples of factual
claims (
Before diving into details, let's talk a bit about the status of
mathematical statements in Coq. Recall that Coq is a

*propositions*) and ways of presenting evidence of their truth (*proofs*). In particular, we have worked extensively with*equality propositions*of the form e_{1}= e_{2}, with implications (P → Q), and with quantified propositions (∀ x, P). In this chapter, we will see how Coq can be used to carry out other familiar forms of logical reasoning.*typed*language, which means that every sensible expression in its world has an associated type. Logical claims are no exception: any statement we might try to prove in Coq has a type, namely Prop, the type of*propositions*. We can see this with the Check command:
Check 3 = 3.

(* ===> Prop *)

Check ∀n m : nat, n + m = m + n.

(* ===> Prop *)

(* ===> Prop *)

Check ∀n m : nat, n + m = m + n.

(* ===> Prop *)

Note that
Simply

*all*syntactically well-formed propositions have type Prop in Coq, regardless of whether they are true.*being*a proposition is one thing; being*provable*is something else!
Check 2 = 2.

(* ===> Prop *)

Check ∀n : nat, n = 2.

(* ===> Prop *)

Check 3 = 4.

(* ===> Prop *)

(* ===> Prop *)

Check ∀n : nat, n = 2.

(* ===> Prop *)

Check 3 = 4.

(* ===> Prop *)

Indeed, propositions don't just have types: they are
So far, we've seen one primary place that propositions can appear:
in Theorem (and Lemma and Example) declarations.

*first-class objects*that can be manipulated in the same ways as the other entities in Coq's world.
Theorem plus_2_2_is_4 :

2 + 2 = 4.

Proof. reflexivity. Qed.

2 + 2 = 4.

Proof. reflexivity. Qed.

But propositions can be used in many other ways. For example, we
can give a name to a proposition using a Definition, just as we
have given names to expressions of other sorts.

Definition plus_fact : Prop := 2 + 2 = 4.

Check plus_fact.

(* ===> plus_fact : Prop *)

Check plus_fact.

(* ===> plus_fact : Prop *)

We can later use this name in any situation where a proposition is
expected — for example, as the claim in a Theorem declaration.

Theorem plus_fact_is_true :

plus_fact.

Proof. reflexivity. Qed.

plus_fact.

Proof. reflexivity. Qed.

We can also write
For instance, the following function takes a number
and returns a proposition asserting that this number is equal to
three:

*parameterized*propositions — that is, functions that take arguments of some type and return a proposition.
Definition is_three (n : nat) : Prop :=

n = 3.

Check is_three.

(* ===> nat -> Prop *)

n = 3.

Check is_three.

(* ===> nat -> Prop *)

In Coq, functions that return propositions are said to define
For instance, here's a (polymorphic) property defining the
familiar notion of an

*properties*of their arguments.*injective function*.
Definition injective {A B} (f : A → B) :=

∀x y : A, f x = f y → x = y.

Lemma succ_inj : injective S.

Proof.

intros n m H. injection H as H

Qed.

∀x y : A, f x = f y → x = y.

Lemma succ_inj : injective S.

Proof.

intros n m H. injection H as H

_{1}. apply H_{1}.Qed.

The equality operator = is also a function that returns a
Prop.
The expression n = m is syntactic sugar for eq n m (defined
using Coq's Notation mechanism). Because eq can be used with
elements of any type, it is also polymorphic:

Check @eq.

(* ===> forall A : Type, A -> A -> Prop *)

(* ===> forall A : Type, A -> A -> Prop *)

(Notice that we wrote @eq instead of eq: The type
argument A to eq is declared as implicit, so we need to turn
off implicit arguments to see the full type of eq.)

# Logical Connectives

## Conjunction

*conjunction*, or

*logical and*, of propositions A and B is written A ∧ B, representing the claim that both A and B are true.

Example and_example : 3 + 4 = 7 ∧ 2 * 2 = 4.

To prove a conjunction, use the split tactic. It will generate
two subgoals, one for each part of the statement:

Proof.

split.

- (* 3 + 4 = 7 *) reflexivity.

- (* 2 + 2 = 4 *) reflexivity.

Qed.

split.

- (* 3 + 4 = 7 *) reflexivity.

- (* 2 + 2 = 4 *) reflexivity.

Qed.

For any propositions A and B, if we assume that A is true
and we assume that B is true, we can conclude that A ∧ B is
also true.

Lemma and_intro : ∀A B : Prop, A → B → A ∧ B.

Proof.

intros A B HA HB. split.

- apply HA.

- apply HB.

Qed.

Proof.

intros A B HA HB. split.

- apply HA.

- apply HB.

Qed.

Since applying a theorem with hypotheses to some goal has the
effect of generating as many subgoals as there are hypotheses for
that theorem, we can apply and_intro to achieve the same effect
as split.

Example and_example' : 3 + 4 = 7 ∧ 2 * 2 = 4.

Proof.

apply and_intro.

- (* 3 + 4 = 7 *) reflexivity.

- (* 2 + 2 = 4 *) reflexivity.

Qed.

Proof.

apply and_intro.

- (* 3 + 4 = 7 *) reflexivity.

- (* 2 + 2 = 4 *) reflexivity.

Qed.

Example and_exercise :

∀n m : nat, n + m = 0 → n = 0 ∧ m = 0.

Proof.

(* FILL IN HERE *) Admitted.

☐
∀n m : nat, n + m = 0 → n = 0 ∧ m = 0.

Proof.

(* FILL IN HERE *) Admitted.

*use*a conjunctive hypothesis to help prove something else — we employ the destruct tactic.

Lemma and_example2 :

∀n m : nat, n = 0 ∧ m = 0 → n + m = 0.

Proof.

(* WORKED IN CLASS *)

intros n m H.

destruct H as [Hn Hm].

rewrite Hn. rewrite Hm.

reflexivity.

Qed.

∀n m : nat, n = 0 ∧ m = 0 → n + m = 0.

Proof.

(* WORKED IN CLASS *)

intros n m H.

destruct H as [Hn Hm].

rewrite Hn. rewrite Hm.

reflexivity.

Qed.

As usual, we can also destruct H right when we introduce it,
instead of introducing and then destructing it:

Lemma and_example2' :

∀n m : nat, n = 0 ∧ m = 0 → n + m = 0.

Proof.

intros n m [Hn Hm].

rewrite Hn. rewrite Hm.

reflexivity.

Qed.

∀n m : nat, n = 0 ∧ m = 0 → n + m = 0.

Proof.

intros n m [Hn Hm].

rewrite Hn. rewrite Hm.

reflexivity.

Qed.

You may wonder why we bothered packing the two hypotheses n = 0
and m = 0 into a single conjunction, since we could have also
stated the theorem with two separate premises:

Lemma and_example2'' :

∀n m : nat, n = 0 → m = 0 → n + m = 0.

Proof.

intros n m Hn Hm.

rewrite Hn. rewrite Hm.

reflexivity.

Qed.

∀n m : nat, n = 0 → m = 0 → n + m = 0.

Proof.

intros n m Hn Hm.

rewrite Hn. rewrite Hm.

reflexivity.

Qed.

For this theorem, both formulations are fine. But it's important
to understand how to work with conjunctive hypotheses because
conjunctions often arise from intermediate steps in proofs,
especially in bigger developments. Here's a simple example:

Lemma and_example3 :

∀n m : nat, n + m = 0 → n * m = 0.

Proof.

(* WORKED IN CLASS *)

intros n m H.

assert (H' : n = 0 ∧ m = 0).

{ apply and_exercise. apply H. }

destruct H' as [Hn Hm].

rewrite Hn. reflexivity.

Qed.

∀n m : nat, n + m = 0 → n * m = 0.

Proof.

(* WORKED IN CLASS *)

intros n m H.

assert (H' : n = 0 ∧ m = 0).

{ apply and_exercise. apply H. }

destruct H' as [Hn Hm].

rewrite Hn. reflexivity.

Qed.

Another common situation with conjunctions is that we know
A ∧ B but in some context we need just A (or just B).
The following lemmas are useful in such cases:

Lemma proj1 : ∀P Q : Prop,

P ∧ Q → P.

Proof.

intros P Q [HP HQ].

apply HP. Qed.

P ∧ Q → P.

Proof.

intros P Q [HP HQ].

apply HP. Qed.

Lemma proj2 : ∀P Q : Prop,

P ∧ Q → Q.

Proof.

(* FILL IN HERE *) Admitted.

☐
P ∧ Q → Q.

Proof.

(* FILL IN HERE *) Admitted.

Theorem and_commut : ∀P Q : Prop,

P ∧ Q → Q ∧ P.

Proof.

intros P Q [HP HQ].

split.

- (* left *) apply HQ.

- (* right *) apply HP. Qed.

P ∧ Q → Q ∧ P.

Proof.

intros P Q [HP HQ].

split.

- (* left *) apply HQ.

- (* right *) apply HP. Qed.

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

(In the following proof of associativity, notice how the*nested*intros pattern breaks the hypothesis H : P ∧ (Q ∧ R) down into HP : P, HQ : Q, and HR : R. Finish the proof from there.)

Theorem and_assoc : ∀P Q R : Prop,

P ∧ (Q ∧ R) → (P ∧ Q) ∧ R.

Proof.

intros P Q R [HP [HQ HR]].

(* FILL IN HERE *) Admitted.

☐
P ∧ (Q ∧ R) → (P ∧ Q) ∧ R.

Proof.

intros P Q R [HP [HQ HR]].

(* FILL IN HERE *) Admitted.

Check and.

(* ===> and : Prop -> Prop -> Prop *)

(* ===> and : Prop -> Prop -> Prop *)

## Disjunction

*disjunction*, or

*logical or*, of two propositions: A ∨ B is true when either A or B is. (This infix notation stands for or A B, where or : Prop → Prop → Prop.)

Lemma or_example :

∀n m : nat, n = 0 ∨ m = 0 → n * m = 0.

Proof.

(* This pattern implicitly does case analysis on

n = 0 ∨ m = 0 *)

intros n m [Hn | Hm].

- (* Here, n = 0 *)

rewrite Hn. reflexivity.

- (* Here, m = 0 *)

rewrite Hm. rewrite <- mult_n_O.

reflexivity.

Qed.

∀n m : nat, n = 0 ∨ m = 0 → n * m = 0.

Proof.

(* This pattern implicitly does case analysis on

n = 0 ∨ m = 0 *)

intros n m [Hn | Hm].

- (* Here, n = 0 *)

rewrite Hn. reflexivity.

- (* Here, m = 0 *)

rewrite Hm. rewrite <- mult_n_O.

reflexivity.

Qed.

Conversely, to show that a disjunction holds, we need to show that
one of its sides does. This is done via two tactics, left and
right. As their names imply, the first one requires
proving the left side of the disjunction, while the second
requires proving its right side. Here is a trivial use...

Lemma or_intro : ∀A B : Prop, A → A ∨ B.

Proof.

intros A B HA.

left.

apply HA.

Qed.

Proof.

intros A B HA.

left.

apply HA.

Qed.

... and here is a slightly more interesting example requiring both
left and right:

Lemma zero_or_succ :

∀n : nat, n = 0 ∨ n = S (pred n).

Proof.

(* WORKED IN CLASS *)

intros [|n].

- left. reflexivity.

- right. reflexivity.

Qed.

∀n : nat, n = 0 ∨ n = S (pred n).

Proof.

(* WORKED IN CLASS *)

intros [|n].

- left. reflexivity.

- right. reflexivity.

Qed.

## Falsehood and Negation

So far, we have mostly been concerned with proving that certain things are*true*— addition is commutative, appending lists is associative, etc. Of course, we may also be interested in negative results, showing that some given proposition is

*not*true. In Coq, such statements are expressed with the negation operator ¬.

*principle of explosion*from the Tactics chapter; it asserts that, if we assume a contradiction, then any other proposition can be derived.

Module MyNot.

Definition not (P:Prop) := P → False.

Notation "¬x" := (not x) : type_scope.

Check not.

(* ===> Prop -> Prop *)

End MyNot.

Definition not (P:Prop) := P → False.

Notation "¬x" := (not x) : type_scope.

Check not.

(* ===> Prop -> Prop *)

End MyNot.

Since False is a contradictory proposition, the principle of
explosion also applies to it. If we get False into the proof
context, we can use destruct on it to complete any goal:

Theorem ex_falso_quodlibet : ∀(P:Prop),

False → P.

Proof.

(* WORKED IN CLASS *)

intros P contra.

destruct contra. Qed.

False → P.

Proof.

(* WORKED IN CLASS *)

intros P contra.

destruct contra. Qed.

The Latin

*ex falso quodlibet*means, literally, "from falsehood follows whatever you like"; this is another common name for the principle of explosion.#### Exercise: 2 stars, standard, optional (not_implies_our_not)

Show that Coq's definition of negation implies the intuitive one mentioned above:
Fact not_implies_our_not : ∀(P:Prop),

¬P → (∀(Q:Prop), P → Q).

Proof.

(* FILL IN HERE *) Admitted.

☐
¬P → (∀(Q:Prop), P → Q).

Proof.

(* FILL IN HERE *) Admitted.

Notation "x ≠ y" := (~(x = y)).

Theorem zero_not_one : 0 ≠ 1.

Proof.

Proof.

The proposition 0 ≠ 1 is exactly the same as
~(0 = 1), that is not (0 = 1), which unfolds to
(0 = 1) → False. (We use unfold not explicitly here
to illustrate that point, but generally it can be omitted.)

unfold not.

To prove an inequality, we may assume the opposite
equality...

intros contra.

... and deduce a contradiction from it. Here, the
equality O = S O contradicts the disjointness of
constructors O and S, so discriminate takes care
of it.

discriminate contra.

Qed.

Qed.

It takes a little practice to get used to working with negation in
Coq. Even though you can see perfectly well why a statement
involving negation is true, it can be a little tricky at first to
get things into the right configuration so that Coq can understand
it! Here are proofs of a few familiar facts to get you warmed
up.

Theorem not_False :

¬False.

Proof.

unfold not. intros H. destruct H. Qed.

Theorem contradiction_implies_anything : ∀P Q : Prop,

(P ∧ ¬P) → Q.

Proof.

(* WORKED IN CLASS *)

intros P Q [HP HNA]. unfold not in HNA.

apply HNA in HP. destruct HP. Qed.

Theorem double_neg : ∀P : Prop,

P → ~~P.

Proof.

(* WORKED IN CLASS *)

intros P H. unfold not. intros G. apply G. apply H. Qed.

¬False.

Proof.

unfold not. intros H. destruct H. Qed.

Theorem contradiction_implies_anything : ∀P Q : Prop,

(P ∧ ¬P) → Q.

Proof.

(* WORKED IN CLASS *)

intros P Q [HP HNA]. unfold not in HNA.

apply HNA in HP. destruct HP. Qed.

Theorem double_neg : ∀P : Prop,

P → ~~P.

Proof.

(* WORKED IN CLASS *)

intros P H. unfold not. intros G. apply G. apply H. Qed.

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

Write an informal proof of double_neg:*Theorem*: P implies ~~P, for any proposition P.

(* FILL IN HERE *)

(* Do not modify the following line: *)

Definition manual_grade_for_double_neg_inf : option (nat*string) := None.

☐
(* Do not modify the following line: *)

Definition manual_grade_for_double_neg_inf : option (nat*string) := None.

Theorem contrapositive : ∀(P Q : Prop),

(P → Q) → (¬Q → ¬P).

Proof.

(* FILL IN HERE *) Admitted.

☐
(P → Q) → (¬Q → ¬P).

Proof.

(* FILL IN HERE *) Admitted.

Theorem not_both_true_and_false : ∀P : Prop,

¬(P ∧ ¬P).

Proof.

(* FILL IN HERE *) Admitted.

☐
¬(P ∧ ¬P).

Proof.

(* FILL IN HERE *) Admitted.

#### Exercise: 1 star, advanced (informal_not_PNP)

Write an informal proof (in English) of the proposition ∀ P : Prop, ~(P ∧ ¬P).
(* FILL IN HERE *)

(* Do not modify the following line: *)

Definition manual_grade_for_informal_not_PNP : option (nat*string) := None.

☐
(* Do not modify the following line: *)

Definition manual_grade_for_informal_not_PNP : option (nat*string) := None.

Theorem not_true_is_false : ∀b : bool,

b ≠ true → b = false.

Proof.

intros [] H.

- (* b = true *)

unfold not in H.

apply ex_falso_quodlibet.

apply H. reflexivity.

- (* b = false *)

reflexivity.

Qed.

b ≠ true → b = false.

Proof.

intros [] H.

- (* b = true *)

unfold not in H.

apply ex_falso_quodlibet.

apply H. reflexivity.

- (* b = false *)

reflexivity.

Qed.

Since reasoning with ex_falso_quodlibet is quite common, Coq
provides a built-in tactic, exfalso, for applying it.

Theorem not_true_is_false' : ∀b : bool,

b ≠ true → b = false.

Proof.

intros [] H.

- (* b = true *)

unfold not in H.

exfalso. (* <=== *)

apply H. reflexivity.

- (* b = false *) reflexivity.

Qed.

b ≠ true → b = false.

Proof.

intros [] H.

- (* b = true *)

unfold not in H.

exfalso. (* <=== *)

apply H. reflexivity.

- (* b = false *) reflexivity.

Qed.

## Truth

Lemma True_is_true : True.

Proof. apply I. Qed.

Proof. apply I. Qed.

Unlike False, which is used extensively, True is used quite
rarely, since it is trivial (and therefore uninteresting) to prove
as a goal, and it carries no useful information as a hypothesis. But it can be quite useful when defining complex Props using
conditionals or as a parameter to higher-order Props.
We will see examples of such uses of True later on.
The handy "if and only if" connective, which asserts that two
propositions have the same truth value, is just the conjunction of
two implications.

## Logical Equivalence

Module MyIff.

Definition iff (P Q : Prop) := (P → Q) ∧ (Q → P).

Notation "P ↔ Q" := (iff P Q)

(at level 95, no associativity)

: type_scope.

End MyIff.

Theorem iff_sym : ∀P Q : Prop,

(P ↔ Q) → (Q ↔ P).

Proof.

(* WORKED IN CLASS *)

intros P Q [HAB HBA].

split.

- (* -> *) apply HBA.

- (* <- *) apply HAB. Qed.

Lemma not_true_iff_false : ∀b,

b ≠ true ↔ b = false.

Proof.

(* WORKED IN CLASS *)

intros b. split.

- (* -> *) apply not_true_is_false.

- (* <- *)

intros H. rewrite H. intros H'. discriminate H'.

Qed.

Definition iff (P Q : Prop) := (P → Q) ∧ (Q → P).

Notation "P ↔ Q" := (iff P Q)

(at level 95, no associativity)

: type_scope.

End MyIff.

Theorem iff_sym : ∀P Q : Prop,

(P ↔ Q) → (Q ↔ P).

Proof.

(* WORKED IN CLASS *)

intros P Q [HAB HBA].

split.

- (* -> *) apply HBA.

- (* <- *) apply HAB. Qed.

Lemma not_true_iff_false : ∀b,

b ≠ true ↔ b = false.

Proof.

(* WORKED IN CLASS *)

intros b. split.

- (* -> *) apply not_true_is_false.

- (* <- *)

intros H. rewrite H. intros H'. discriminate H'.

Qed.

Theorem or_distributes_over_and : ∀P Q R : Prop,

P ∨ (Q ∧ R) ↔ (P ∨ Q) ∧ (P ∨ R).

Proof.

(* FILL IN HERE *) Admitted.

☐
P ∨ (Q ∧ R) ↔ (P ∨ Q) ∧ (P ∨ R).

Proof.

(* FILL IN HERE *) Admitted.

From Coq Require Import Setoids.Setoid.

Here is a simple example demonstrating how these tactics work with
iff. First, let's prove a couple of basic iff equivalences...

Lemma mult_0 : ∀n m, n * m = 0 ↔ n = 0 ∨ m = 0.

Lemma or_assoc :

∀P Q R : Prop, P ∨ (Q ∨ R) ↔ (P ∨ Q) ∨ R.

Proof.

split.

- apply mult_eq_0.

- apply or_example.

Qed.

split.

- apply mult_eq_0.

- apply or_example.

Qed.

Lemma or_assoc :

∀P Q R : Prop, P ∨ (Q ∨ R) ↔ (P ∨ Q) ∨ R.

Proof.

intros P Q R. split.

- intros [H | [H | H]].

+ left. left. apply H.

+ left. right. apply H.

+ right. apply H.

- intros [[H | H] | H].

+ left. apply H.

+ right. left. apply H.

+ right. right. apply H.

Qed.

intros P Q R. split.

- intros [H | [H | H]].

+ left. left. apply H.

+ left. right. apply H.

+ right. apply H.

- intros [[H | H] | H].

+ left. apply H.

+ right. left. apply H.

+ right. right. apply H.

Qed.

We can now use these facts with rewrite and reflexivity to
give smooth proofs of statements involving equivalences. Here is
a ternary version of the previous mult_0 result:

Lemma mult_0_3 :

∀n m p, n * m * p = 0 ↔ n = 0 ∨ m = 0 ∨ p = 0.

Proof.

intros n m p.

rewrite mult_0. rewrite mult_0. rewrite or_assoc.

reflexivity.

Qed.

∀n m p, n * m * p = 0 ↔ n = 0 ∨ m = 0 ∨ p = 0.

Proof.

intros n m p.

rewrite mult_0. rewrite mult_0. rewrite or_assoc.

reflexivity.

Qed.

The apply tactic can also be used with ↔. When given an
equivalence as its argument, apply tries to guess which side of
the equivalence to use.

Lemma apply_iff_example :

∀n m : nat, n * m = 0 → n = 0 ∨ m = 0.

Proof.

intros n m H. apply mult_0. apply H.

Qed.

∀n m : nat, n * m = 0 → n = 0 ∨ m = 0.

Proof.

intros n m H. apply mult_0. apply H.

Qed.

## Existential Quantification

*existential quantification*. To say that there is some x of type T such that some property P holds of x, we write ∃ x : T, P. As with ∀, the type annotation : T can be omitted if Coq is able to infer from the context what the type of x should be.

*witness*of the existential. This is done in two steps: First, we explicitly tell Coq which witness t we have in mind by invoking the tactic ∃ t. Then we prove that P holds after all occurrences of x are replaced by t.

Lemma four_is_even : ∃n : nat, 4 = n + n.

Proof.

∃2. reflexivity.

Qed.

Proof.

∃2. reflexivity.

Qed.

Conversely, if we have an existential hypothesis ∃ x, P in
the context, we can destruct it to obtain a witness x and a
hypothesis stating that P holds of x.

Theorem exists_example_2 : ∀n,

(∃m, n = 4 + m) →

(∃o, n = 2 + o).

Proof.

(* WORKED IN CLASS *)

intros n [m Hm]. (* note implicit destruct here *)

∃(2 + m).

apply Hm. Qed.

(∃m, n = 4 + m) →

(∃o, n = 2 + o).

Proof.

(* WORKED IN CLASS *)

intros n [m Hm]. (* note implicit destruct here *)

∃(2 + m).

apply Hm. Qed.

#### Exercise: 1 star, standard, recommended (dist_not_exists)

Prove that "P holds for all x" implies "there is no x for which P does not hold." (Hint: destruct H as [x E] works on existential assumptions!)
Theorem dist_not_exists : ∀(X:Type) (P : X → Prop),

(∀x, P x) → ¬(∃x, ¬P x).

Proof.

(* FILL IN HERE *) Admitted.

☐
(∀x, P x) → ¬(∃x, ¬P x).

Proof.

(* FILL IN HERE *) Admitted.

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

Prove that existential quantification distributes over disjunction.
Theorem dist_exists_or : ∀(X:Type) (P Q : X → Prop),

(∃x, P x ∨ Q x) ↔ (∃x, P x) ∨ (∃x, Q x).

Proof.

(* FILL IN HERE *) Admitted.

☐
(∃x, P x ∨ Q x) ↔ (∃x, P x) ∨ (∃x, Q x).

Proof.

(* FILL IN HERE *) Admitted.

# Programming with Propositions

- If l is the empty list, then x cannot occur on it, so the property "x appears in l" is simply false.

- Otherwise, l has the form x' :: l'. In this case, x occurs in l if either it is equal to x' or it occurs in l'.

Fixpoint In {A : Type} (x : A) (l : list A) : Prop :=

match l with

| [] ⇒ False

| x' :: l' ⇒ x' = x ∨ In x l'

end.

match l with

| [] ⇒ False

| x' :: l' ⇒ x' = x ∨ In x l'

end.

When In is applied to a concrete list, it expands into a
concrete sequence of nested disjunctions.

Example In_example_1 : In 4 [1; 2; 3; 4; 5].

Proof.

(* WORKED IN CLASS *)

simpl. right. right. right. left. reflexivity.

Qed.

Example In_example_2 :

∀n, In n [2; 4] →

∃n', n = 2 * n'.

Proof.

(* WORKED IN CLASS *)

simpl.

intros n [H | [H | []]].

- ∃1. rewrite <- H. reflexivity.

- ∃2. rewrite <- H. reflexivity.

Qed.

Proof.

(* WORKED IN CLASS *)

simpl. right. right. right. left. reflexivity.

Qed.

Example In_example_2 :

∀n, In n [2; 4] →

∃n', n = 2 * n'.

Proof.

(* WORKED IN CLASS *)

simpl.

intros n [H | [H | []]].

- ∃1. rewrite <- H. reflexivity.

- ∃2. rewrite <- H. reflexivity.

Qed.

(Notice the use of the empty pattern to discharge the last case
We can also prove more generic, higher-level lemmas about In.
Note, in the next, how In starts out applied to a variable and
only gets expanded when we do case analysis on this variable:

*en passant*.)
Lemma In_map :

∀(A B : Type) (f : A → B) (l : list A) (x : A),

In x l →

In (f x) (map f l).

Proof.

intros A B f l x.

induction l as [|x' l' IHl'].

- (* l = nil, contradiction *)

simpl. intros [].

- (* l = x' :: l' *)

simpl. intros [H | H].

+ rewrite H. left. reflexivity.

+ right. apply IHl'. apply H.

Qed.

∀(A B : Type) (f : A → B) (l : list A) (x : A),

In x l →

In (f x) (map f l).

Proof.

intros A B f l x.

induction l as [|x' l' IHl'].

- (* l = nil, contradiction *)

simpl. intros [].

- (* l = x' :: l' *)

simpl. intros [H | H].

+ rewrite H. left. reflexivity.

+ right. apply IHl'. apply H.

Qed.

This way of defining propositions recursively, though convenient
in some cases, also has some drawbacks. In particular, it is
subject to Coq's usual restrictions regarding the definition of
recursive functions, e.g., the requirement that they be "obviously
terminating." In the next chapter, we will see how to define
propositions

*inductively*, a different technique with its own set of strengths and limitations.#### Exercise: 2 stars, standard (In_map_iff)

Lemma In_map_iff :

∀(A B : Type) (f : A → B) (l : list A) (y : B),

In y (map f l) ↔

∃x, f x = y ∧ In x l.

Proof.

(* FILL IN HERE *) Admitted.

☐
∀(A B : Type) (f : A → B) (l : list A) (y : B),

In y (map f l) ↔

∃x, f x = y ∧ In x l.

Proof.

(* FILL IN HERE *) Admitted.

Lemma In_app_iff : ∀A l l' (a:A),

In a (l++l') ↔ In a l ∨ In a l'.

Proof.

(* FILL IN HERE *) Admitted.

☐
In a (l++l') ↔ In a l ∨ In a l'.

Proof.

(* FILL IN HERE *) Admitted.

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

Recall that functions returning propositions can be seen as*properties*of their arguments. For instance, if P has type nat → Prop, then P n states that property P holds of n.

*not*just restate the left-hand side of All_In.)

Fixpoint All {T : Type} (P : T → Prop) (l : list T) : Prop

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

Lemma All_In :

∀T (P : T → Prop) (l : list T),

(∀x, In x l → P x) ↔

All P l.

Proof.

(* FILL IN HERE *) Admitted.

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

Lemma All_In :

∀T (P : T → Prop) (l : list T),

(∀x, In x l → P x) ↔

All P l.

Proof.

(* FILL IN HERE *) Admitted.

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

Complete the definition of the combine_odd_even function below. It takes as arguments two properties of numbers, Podd and Peven, and it should return a property P such that P n is equivalent to Podd n when n is odd and equivalent to Peven n otherwise.
Definition combine_odd_even (Podd Peven : nat → Prop) : nat → Prop

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

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

To test your definition, prove the following facts:

Theorem combine_odd_even_intro :

∀(Podd Peven : nat → Prop) (n : nat),

(oddb n = true → Podd n) →

(oddb n = false → Peven n) →

combine_odd_even Podd Peven n.

Proof.

(* FILL IN HERE *) Admitted.

Theorem combine_odd_even_elim_odd :

∀(Podd Peven : nat → Prop) (n : nat),

combine_odd_even Podd Peven n →

oddb n = true →

Podd n.

Proof.

(* FILL IN HERE *) Admitted.

Theorem combine_odd_even_elim_even :

∀(Podd Peven : nat → Prop) (n : nat),

combine_odd_even Podd Peven n →

oddb n = false →

Peven n.

Proof.

(* FILL IN HERE *) Admitted.

☐
∀(Podd Peven : nat → Prop) (n : nat),

(oddb n = true → Podd n) →

(oddb n = false → Peven n) →

combine_odd_even Podd Peven n.

Proof.

(* FILL IN HERE *) Admitted.

Theorem combine_odd_even_elim_odd :

∀(Podd Peven : nat → Prop) (n : nat),

combine_odd_even Podd Peven n →

oddb n = true →

Podd n.

Proof.

(* FILL IN HERE *) Admitted.

Theorem combine_odd_even_elim_even :

∀(Podd Peven : nat → Prop) (n : nat),

combine_odd_even Podd Peven n →

oddb n = false →

Peven n.

Proof.

(* FILL IN HERE *) Admitted.

# Applying Theorems to Arguments

*proofs*as first-class objects.

Check plus_comm.

(* ===> forall n m : nat, n + m = m + n *)

(* ===> forall n m : nat, n + m = m + n *)

Coq prints the
The reason is that the identifier plus_comm actually refers to a
Intuitively, this makes sense because the statement of a theorem
tells us what we can use that theorem for, just as the type of a
computational object tells us what we can do with that object —
e.g., if we have a term of type nat → nat → nat, we can give
it two nats as arguments and get a nat back. Similarly, if we
have an object of type n = m → n + n = m + m and we provide it
an "argument" of type n = m, we can derive n + n = m + m.
Operationally, this analogy goes even further: by applying a
theorem, as if it were a function, to hypotheses with matching
types, we can specialize its result without having to resort to
intermediate assertions. For example, suppose we wanted to prove
the following result:

*statement*of the plus_comm theorem in the same way that it prints the*type*of any term that we ask it to Check. Why?*proof object*— a data structure that represents a logical derivation establishing of the truth of the statement ∀ n m : nat, n + m = m + n. The type of this object*is*the statement of the theorem that it is a proof of.
Lemma plus_comm3 :

∀x y z, x + (y + z) = (z + y) + x.

∀x y z, x + (y + z) = (z + y) + x.

It appears at first sight that we ought to be able to prove this
by rewriting with plus_comm twice to make the two sides match.
The problem, however, is that the second rewrite will undo the
effect of the first.

Proof.

(* WORKED IN CLASS *)

intros x y z.

rewrite plus_comm.

rewrite plus_comm.

(* We are back where we started... *)

Abort.

(* WORKED IN CLASS *)

intros x y z.

rewrite plus_comm.

rewrite plus_comm.

(* We are back where we started... *)

Abort.

One simple way of fixing this problem, using only tools that we
already know, is to use assert to derive a specialized version
of plus_comm that can be used to rewrite exactly where we
want.

Lemma plus_comm3_take2 :

∀x y z, x + (y + z) = (z + y) + x.

Proof.

intros x y z.

rewrite plus_comm.

assert (H : y + z = z + y).

{ rewrite plus_comm. reflexivity. }

rewrite H.

reflexivity.

Qed.

∀x y z, x + (y + z) = (z + y) + x.

Proof.

intros x y z.

rewrite plus_comm.

assert (H : y + z = z + y).

{ rewrite plus_comm. reflexivity. }

rewrite H.

reflexivity.

Qed.

A more elegant alternative is to apply plus_comm directly to the
arguments we want to instantiate it with, in much the same way as
we apply a polymorphic function to a type argument.

Lemma plus_comm3_take3 :

∀x y z, x + (y + z) = (z + y) + x.

Proof.

intros x y z.

rewrite plus_comm.

rewrite (plus_comm y z).

reflexivity.

Qed.

∀x y z, x + (y + z) = (z + y) + x.

Proof.

intros x y z.

rewrite plus_comm.

rewrite (plus_comm y z).

reflexivity.

Qed.

Let us show another example of using a theorem or lemma
like a function. The following theorem says: any list l
containing some element must be nonempty.

Lemma in_not_nil :

∀A (x : A) (l : list A), In x l → l ≠ [].

Proof.

intros A x l H. unfold not. intro Hl. destruct l.

- simpl in H. destruct H.

- discriminate Hl.

Qed.

∀A (x : A) (l : list A), In x l → l ≠ [].

Proof.

intros A x l H. unfold not. intro Hl. destruct l.

- simpl in H. destruct H.

- discriminate Hl.

Qed.

What makes this interesting is that one quantified variable
(x) does not appear in the conclusion (l ≠ []).
We can use this lemma to prove the special case where x
is 42. Naively, the tactic apply in_not_nil will fail because
it cannot infer the value of x. There are several ways to work
around that...

Lemma in_not_nil_42 :

∀l : list nat, In 42 l → l ≠ [].

Proof.

(* WORKED IN CLASS *)

intros l H.

Fail apply in_not_nil.

Abort.

(* apply ... with ... *)

Lemma in_not_nil_42_take2 :

∀l : list nat, In 42 l → l ≠ [].

Proof.

intros l H.

apply in_not_nil with (x := 42).

apply H.

Qed.

(* apply ... in ... *)

Lemma in_not_nil_42_take3 :

∀l : list nat, In 42 l → l ≠ [].

Proof.

intros l H.

apply in_not_nil in H.

apply H.

Qed.

(* Explicitly apply the lemma to the value for x. *)

Lemma in_not_nil_42_take4 :

∀l : list nat, In 42 l → l ≠ [].

Proof.

intros l H.

apply (in_not_nil nat 42).

apply H.

Qed.

(* Explicitly apply the lemma to a hypothesis. *)

Lemma in_not_nil_42_take5 :

∀l : list nat, In 42 l → l ≠ [].

Proof.

intros l H.

apply (in_not_nil _ _ _ H).

Qed.

∀l : list nat, In 42 l → l ≠ [].

Proof.

(* WORKED IN CLASS *)

intros l H.

Fail apply in_not_nil.

Abort.

(* apply ... with ... *)

Lemma in_not_nil_42_take2 :

∀l : list nat, In 42 l → l ≠ [].

Proof.

intros l H.

apply in_not_nil with (x := 42).

apply H.

Qed.

(* apply ... in ... *)

Lemma in_not_nil_42_take3 :

∀l : list nat, In 42 l → l ≠ [].

Proof.

intros l H.

apply in_not_nil in H.

apply H.

Qed.

(* Explicitly apply the lemma to the value for x. *)

Lemma in_not_nil_42_take4 :

∀l : list nat, In 42 l → l ≠ [].

Proof.

intros l H.

apply (in_not_nil nat 42).

apply H.

Qed.

(* Explicitly apply the lemma to a hypothesis. *)

Lemma in_not_nil_42_take5 :

∀l : list nat, In 42 l → l ≠ [].

Proof.

intros l H.

apply (in_not_nil _ _ _ H).

Qed.

You can "use theorems as functions" in this way with almost all
tactics that take a theorem name as an argument. Note also that
theorem application uses the same inference mechanisms as function
application; thus, it is possible, for example, to supply
wildcards as arguments to be inferred, or to declare some
hypotheses to a theorem as implicit by default. These features
are illustrated in the proof below. (The details of how this proof
works are not critical — the goal here is just to illustrate what
can be done.)

Example lemma_application_ex :

∀{n : nat} {ns : list nat},

In n (map (fun m ⇒ m * 0) ns) →

n = 0.

Proof.

intros n ns H.

destruct (proj1 _ _ (In_map_iff _ _ _ _ _) H)

as [m [Hm _]].

rewrite mult_0_r in Hm. rewrite <- Hm. reflexivity.

Qed.

∀{n : nat} {ns : list nat},

In n (map (fun m ⇒ m * 0) ns) →

n = 0.

Proof.

intros n ns H.

destruct (proj1 _ _ (In_map_iff _ _ _ _ _) H)

as [m [Hm _]].

rewrite mult_0_r in Hm. rewrite <- Hm. reflexivity.

Qed.

We will see many more examples in later chapters.

# Coq vs. Set Theory

*Calculus of Inductive Constructions*, differs in some important ways from other formal systems that are used by mathematicians to write down precise and rigorous proofs. For example, in the most popular foundation for paper-and-pencil mathematics, Zermelo-Fraenkel Set Theory (ZFC), a mathematical object can potentially be a member of many different sets; a term in Coq's logic, on the other hand, is a member of at most one type. This difference often leads to slightly different ways of capturing informal mathematical concepts, but these are, by and large, about equally natural and easy to work with. For example, instead of saying that a natural number n belongs to the set of even numbers, we would say in Coq that even n holds, where even : nat → Prop is a property describing even numbers.

## Functional Extensionality

*functions*are equal to each other:

Example function_equality_ex

(fun x ⇒ 3 + x) = (fun x ⇒ (pred 4) + x).

_{1}:(fun x ⇒ 3 + x) = (fun x ⇒ (pred 4) + x).

Proof. reflexivity. Qed.

In common mathematical practice, two functions f and g are
considered equal if they produce the same outputs:
Informally speaking, an "extensional property" is one that
pertains to an object's observable behavior. Thus, functional
extensionality simply means that a function's identity is
completely determined by what we can observe from it — i.e., in
Coq terms, the results we obtain after applying it.
Functional extensionality is not part of Coq's built-in logic.
This means that some "reasonable" propositions are not provable.

(∀x, f x = g x) → f = g

This is known as the principle of *functional extensionality*.
Example function_equality_ex

(fun x ⇒ plus x 1) = (fun x ⇒ plus 1 x).

Proof.

(* Stuck *)

Abort.

_{2}:(fun x ⇒ plus x 1) = (fun x ⇒ plus 1 x).

Proof.

(* Stuck *)

Abort.

However, we can add functional extensionality to Coq's core using
the Axiom command.

Axiom functional_extensionality : ∀{X Y: Type}

{f g : X → Y},

(∀(x:X), f x = g x) → f = g.

{f g : X → Y},

(∀(x:X), f x = g x) → f = g.

Using Axiom has the same effect as stating a theorem and
skipping its proof using Admitted, but it alerts the reader that
this isn't just something we're going to come back and fill in
later!
We can now invoke functional extensionality in proofs:

Example function_equality_ex

(fun x ⇒ plus x 1) = (fun x ⇒ plus 1 x).

Proof.

apply functional_extensionality. intros x.

apply plus_comm.

Qed.

_{2}:(fun x ⇒ plus x 1) = (fun x ⇒ plus 1 x).

Proof.

apply functional_extensionality. intros x.

apply plus_comm.

Qed.

Naturally, we must be careful when adding new axioms into Coq's
logic, as they may render it
Unfortunately, there is no simple way of telling whether an axiom
is safe to add: hard work by highly-trained trained experts is
generally required to establish the consistency of any particular
combination of axioms.
Fortunately, it is known that adding functional extensionality, in
particular,
To check whether a particular proof relies on any additional
axioms, use the Print Assumptions command.

*inconsistent*— that is, they may make it possible to prove every proposition, including False, 2+2=5, etc.!*is*consistent.
Print Assumptions function_equality_ex

(* ===>

Axioms:

functional_extensionality :

forall (X Y : Type) (f g : X -> Y),

(forall x : X, f x = g x) -> f = g *)

_{2}.(* ===>

Axioms:

functional_extensionality :

forall (X Y : Type) (f g : X -> Y),

(forall x : X, f x = g x) -> f = g *)

#### Exercise: 4 stars, standard (tr_rev_correct)

One problem with the definition of the list-reversing function rev that we have is that it performs a call to app on each step; running app takes time asymptotically linear in the size of the list, which means that rev has quadratic running time. We can improve this with the following definition:
Fixpoint rev_append {X} (l

match l

| [] ⇒ l

| x :: l

end.

Definition tr_rev {X} (l : list X) : list X :=

rev_append l [].

_{1}l_{2}: list X) : list X :=match l

_{1}with| [] ⇒ l

_{2}| x :: l

_{1}' ⇒ rev_append l_{1}' (x :: l_{2})end.

Definition tr_rev {X} (l : list X) : list X :=

rev_append l [].

This version is said to be

*tail-recursive*, because the recursive call to the function is the last operation that needs to be performed (i.e., we don't have to execute ++ after the recursive call); a decent compiler will generate very efficient code in this case. Prove that the two definitions are indeed equivalent.
Lemma tr_rev_correct : ∀X, @tr_rev X = @rev X.

(* FILL IN HERE *) Admitted.

☐
(* FILL IN HERE *) Admitted.

## Propositions and Booleans

*booleans*(of type bool), and with

*propositions*(of type Prop).

Example even_42_bool : evenb 42 = true.

Proof. reflexivity. Qed.

... or that there exists some k such that n = double k.

Example even_42_prop : ∃k, 42 = double k.

Proof. ∃21. reflexivity. Qed.

Of course, it would be pretty strange if these two
characterizations of evenness did not describe the same set of
natural numbers! Fortunately, we can prove that they do...
We first need two helper lemmas.

Theorem evenb_double : ∀k, evenb (double k) = true.

Proof.

intros k. induction k as [|k' IHk'].

- reflexivity.

- simpl. apply IHk'.

Qed.

intros k. induction k as [|k' IHk'].

- reflexivity.

- simpl. apply IHk'.

Qed.

Theorem evenb_double_conv : ∀n,

∃k, n = if evenb n then double k

else S (double k).

Theorem even_bool_prop : ∀n,

evenb n = true ↔ ∃k, n = double k.

∃k, n = if evenb n then double k

else S (double k).

Proof.

(* Hint: Use the evenb_S lemma from Induction.v. *)

(* FILL IN HERE *) Admitted.

☐
(* Hint: Use the evenb_S lemma from Induction.v. *)

(* FILL IN HERE *) Admitted.

Theorem even_bool_prop : ∀n,

evenb n = true ↔ ∃k, n = double k.

Proof.

intros n. split.

- intros H. destruct (evenb_double_conv n) as [k Hk].

rewrite Hk. rewrite H. ∃k. reflexivity.

- intros [k Hk]. rewrite Hk. apply evenb_double.

Qed.

intros n. split.

- intros H. destruct (evenb_double_conv n) as [k Hk].

rewrite Hk. rewrite H. ∃k. reflexivity.

- intros [k Hk]. rewrite Hk. apply evenb_double.

Qed.

In view of this theorem, we say that the boolean computation
evenb n is reflected in the truth of the proposition ∃ k,
n = double k.
Similarly, to state that two numbers n and m are equal, we can
say either

- (1) that n =? m returns true, or
- (2) that n = m.

Theorem eqb_eq : ∀n

n

_{1}n_{2}: nat,n

_{1}=? n_{2}= true ↔ n_{1}= n_{2}.
Proof.

intros n

- apply eqb_true.

- intros H. rewrite H. rewrite <- eqb_refl. reflexivity.

Qed.

intros n

_{1}n_{2}. split.- apply eqb_true.

- intros H. rewrite H. rewrite <- eqb_refl. reflexivity.

Qed.

However, even when the boolean and propositional formulations of a
claim are equivalent from a purely logical perspective, they may
not be equivalent
In the case of even numbers above, when proving the
backwards direction of even_bool_prop (i.e., evenb_double,
going from the propositional to the boolean claim), we used a
simple induction on k. On the other hand, the converse (the
evenb_double_conv exercise) required a clever generalization,
since we can't directly prove
(evenb n = true) → (∃ k, n = double k).
For these examples, the propositional claims are more useful than
their boolean counterparts, but this is not always the case. For
instance, we cannot test whether a general proposition is true or
not in a function definition; as a consequence, the following code
fragment is rejected:

*operationally*.
Fail Definition is_even_prime n :=

if n = 2 then true

else false.

if n = 2 then true

else false.

Coq complains that n = 2 has type Prop, while it expects
an element of bool (or some other inductive type with two
elements). The reason for this error message has to do with the
Although general non-computable properties cannot be phrased as
boolean computations, it is worth noting that even many
Conversely, an important side benefit of stating facts using
booleans is enabling some proof automation through computation
with Coq terms, a technique known as

*computational*nature of Coq's core language, which is designed so that every function that it can express is computable and total. One reason for this is to allow the extraction of executable programs from Coq developments. As a consequence, Prop in Coq does*not*have a universal case analysis operation telling whether any given proposition is true or false, since such an operation would allow us to write non-computable functions.*computable*properties are easier to express using Prop than bool, since recursive function definitions are subject to significant restrictions in Coq. For instance, the next chapter shows how to define the property that a regular expression matches a given string using Prop. Doing the same with bool would amount to writing a regular expression matcher, which would be more complicated, harder to understand, and harder to reason about.*proof by reflection*. Consider the following statement:
Example even_1000 : ∃k, 1000 = double k.

The most direct proof of this fact is to give the value of k
explicitly.

Proof. ∃500. reflexivity. Qed.

On the other hand, the proof of the corresponding boolean
statement is even simpler:

Example even_1000' : evenb 1000 = true.

Proof. reflexivity. Qed.

Proof. reflexivity. Qed.

What is interesting is that, since the two notions are equivalent,
we can use the boolean formulation to prove the other one without
mentioning the value 500 explicitly:

Example even_1000'' : ∃k, 1000 = double k.

Proof. apply even_bool_prop. reflexivity. Qed.

Proof. apply even_bool_prop. reflexivity. Qed.

Although we haven't gained much in terms of proof-script
size in this case, larger proofs can often be made considerably
simpler by the use of reflection. As an extreme example, the Coq
proof of the famous
Another notable difference is that the negation of a "boolean
fact" is straightforward to state and prove: simply flip the
expected boolean result.

*4-color theorem*uses reflection to reduce the analysis of hundreds of different cases to a boolean computation.
Example not_even_1001 : evenb 1001 = false.

Proof.

(* WORKED IN CLASS *)

reflexivity.

Qed.

Proof.

(* WORKED IN CLASS *)

reflexivity.

Qed.

In contrast, propositional negation may be more difficult
to grasp.

Example not_even_1001' : ~(∃k, 1001 = double k).

Proof.

(* WORKED IN CLASS *)

rewrite <- even_bool_prop.

unfold not.

simpl.

intro H.

discriminate H.

Qed.

Proof.

(* WORKED IN CLASS *)

rewrite <- even_bool_prop.

unfold not.

simpl.

intro H.

discriminate H.

Qed.

Equality provides a complementary example: knowing that
n =? m = true is generally of little direct help in the middle
of a proof involving n and m; however, if we convert the
statement to the equivalent form n = m, we can rewrite with it.

Lemma plus_eqb_example : ∀n m p : nat,

n =? m = true → n + p =? m + p = true.

Proof.

(* WORKED IN CLASS *)

intros n m p H.

rewrite eqb_eq in H.

rewrite H.

rewrite eqb_eq.

reflexivity.

Qed.

n =? m = true → n + p =? m + p = true.

Proof.

(* WORKED IN CLASS *)

intros n m p H.

rewrite eqb_eq in H.

rewrite H.

rewrite eqb_eq.

reflexivity.

Qed.

We won't cover reflection in much detail, but it serves as a good
example showing the complementary strengths of booleans and
general propositions.

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

The following lemmas relate the propositional connectives studied in this chapter to the corresponding boolean operations.
Lemma andb_true_iff : ∀b

b

Proof.

(* FILL IN HERE *) Admitted.

Lemma orb_true_iff : ∀b

b

Proof.

(* FILL IN HERE *) Admitted.

☐
_{1}b_{2}:bool,b

_{1}&& b_{2}= true ↔ b_{1}= true ∧ b_{2}= true.Proof.

(* FILL IN HERE *) Admitted.

Lemma orb_true_iff : ∀b

_{1}b_{2},b

_{1}|| b_{2}= true ↔ b_{1}= true ∨ b_{2}= true.Proof.

(* FILL IN HERE *) Admitted.

#### Exercise: 1 star, standard (eqb_neq)

The following theorem is an alternate "negative" formulation of eqb_eq that is more convenient in certain situations (we'll see examples in later chapters).
Theorem eqb_neq : ∀x y : nat,

x =? y = false ↔ x ≠ y.

Proof.

(* FILL IN HERE *) Admitted.

☐
x =? y = false ↔ x ≠ y.

Proof.

(* FILL IN HERE *) Admitted.

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

Given a boolean operator eqb for testing equality of elements of some type A, we can define a function eqb_list for testing equality of lists with elements in A. Complete the definition of the eqb_list function below. To make sure that your definition is correct, prove the lemma eqb_list_true_iff.
Fixpoint eqb_list {A : Type} (eqb : A → A → bool)

(l

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

Lemma eqb_list_true_iff :

∀A (eqb : A → A → bool),

(∀a

∀l

Proof.

(* FILL IN HERE *) Admitted.

☐
(l

_{1}l_{2}: list A) : bool(* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Lemma eqb_list_true_iff :

∀A (eqb : A → A → bool),

(∀a

_{1}a_{2}, eqb a_{1}a_{2}= true ↔ a_{1}= a_{2}) →∀l

_{1}l_{2}, eqb_list eqb l_{1}l_{2}= true ↔ l_{1}= l_{2}.Proof.

(* FILL IN HERE *) Admitted.

#### Exercise: 2 stars, standard, recommended (All_forallb)

Recall the function forallb, from the exercise forall_exists_challenge in chapter Tactics:
Fixpoint forallb {X : Type} (test : X → bool) (l : list X) : bool :=

match l with

| [] ⇒ true

| x :: l' ⇒ andb (test x) (forallb test l')

end.

match l with

| [] ⇒ true

| x :: l' ⇒ andb (test x) (forallb test l')

end.

Prove the theorem below, which relates forallb to the All
property of the above exercise.

Theorem forallb_true_iff : ∀X test (l : list X),

forallb test l = true ↔ All (fun x ⇒ test x = true) l.

Proof.

(* FILL IN HERE *) Admitted.

forallb test l = true ↔ All (fun x ⇒ test x = true) l.

Proof.

(* FILL IN HERE *) Admitted.

Are there any important properties of the function forallb which
are not captured by this specification?

(* FILL IN HERE *)

☐
## Classical vs. Constructive Logic

*proofs*! In other words, the following intuitive reasoning principle is not derivable in Coq:

Definition excluded_middle := ∀P : Prop,

P ∨ ¬P.

P ∨ ¬P.

To understand operationally why this is the case, recall
that, to prove a statement of the form P ∨ Q, we use the left
and right tactics, which effectively require knowing which side
of the disjunction holds. But the universally quantified P in
excluded_middle is an
However, if we happen to know that P is reflected in some
boolean term b, then knowing whether it holds or not is trivial:
we just have to check the value of b.

*arbitrary*proposition, which we know nothing about. We don't have enough information to choose which of left or right to apply, just as Coq doesn't have enough information to mechanically decide whether P holds or not inside a function.
Theorem restricted_excluded_middle : ∀P b,

(P ↔ b = true) → P ∨ ¬P.

Proof.

intros P [] H.

- left. rewrite H. reflexivity.

- right. rewrite H. intros contra. discriminate contra.

Qed.

(P ↔ b = true) → P ∨ ¬P.

Proof.

intros P [] H.

- left. rewrite H. reflexivity.

- right. rewrite H. intros contra. discriminate contra.

Qed.

In particular, the excluded middle is valid for equations n = m,
between natural numbers n and m.

Theorem restricted_excluded_middle_eq : ∀(n m : nat),

n = m ∨ n ≠ m.

Proof.

intros n m.

apply (restricted_excluded_middle (n = m) (n =? m)).

symmetry.

apply eqb_eq.

Qed.

n = m ∨ n ≠ m.

Proof.

intros n m.

apply (restricted_excluded_middle (n = m) (n =? m)).

symmetry.

apply eqb_eq.

Qed.

It may seem strange that the general excluded middle is not
available by default in Coq; after all, any given claim must be
either true or false. Nonetheless, there is an advantage in not
assuming the excluded middle: statements in Coq can make stronger
claims than the analogous statements in standard mathematics.
Notably, if there is a Coq proof of ∃ x, P x, it is
possible to explicitly exhibit a value of x for which we can
prove P x — in other words, every proof of existence is
necessarily
Logics like Coq's, which do not assume the excluded middle, are
referred to as
More conventional logical systems such as ZFC, in which the
excluded middle does hold for arbitrary propositions, are referred
to as
The following example illustrates why assuming the excluded middle
may lead to non-constructive proofs:
Do you see what happened here? We used the excluded middle to
consider separately the cases where sqrt 2 ^ sqrt 2 is rational
and where it is not, without knowing which one actually holds!
Because of that, we wind up knowing that such a and b exist
but we cannot determine what their actual values are (at least,
using this line of argument).
As useful as constructive logic is, it does have its limitations:
There are many statements that can easily be proven in classical
logic but that have much more complicated constructive proofs, and
there are some that are known to have no constructive proof at
all! Fortunately, like functional extensionality, the excluded
middle is known to be compatible with Coq's logic, allowing us to
add it safely as an axiom. However, we will not need to do so in
this book: the results that we cover can be developed entirely
within constructive logic at negligible extra cost.
It takes some practice to understand which proof techniques must
be avoided in constructive reasoning, but arguments by
contradiction, in particular, are infamous for leading to
non-constructive proofs. Here's a typical example: suppose that
we want to show that there exists x with some property P,
i.e., such that P x. We start by assuming that our conclusion
is false; that is, ¬ ∃ x, P x. From this premise, it is not
hard to derive ∀ x, ¬ P x. If we manage to show that this
intermediate fact results in a contradiction, we arrive at an
existence proof without ever exhibiting a value of x for which
P x holds!
The technical flaw here, from a constructive standpoint, is that
we claimed to prove ∃ x, P x using a proof of
¬ ¬ (∃ x, P x). Allowing ourselves to remove double
negations from arbitrary statements is equivalent to assuming the
excluded middle, as shown in one of the exercises below. Thus,
this line of reasoning cannot be encoded in Coq without assuming
additional axioms.

*constructive*.*constructive logics*.*classical*.*Claim*: There exist irrational numbers a and b such that a ^ b is rational.*Proof*: It is not difficult to show that sqrt 2 is irrational. If sqrt 2 ^ sqrt 2 is rational, it suffices to take a = b = sqrt 2 and we are done. Otherwise, sqrt 2 ^ sqrt 2 is irrational. In this case, we can take a = sqrt 2 ^ sqrt 2 and b = sqrt 2, since a ^ b = sqrt 2 ^ (sqrt 2 * sqrt 2) = sqrt 2 ^ 2 = 2. ☐#### Exercise: 3 stars, standard (excluded_middle_irrefutable)

Proving the consistency of Coq with the general excluded middle axiom requires complicated reasoning that cannot be carried out within Coq itself. However, the following theorem implies that it is always safe to assume a decidability axiom (i.e., an instance of excluded middle) for any*particular*Prop P. Why? Because we cannot prove the negation of such an axiom. If we could, we would have both ¬ (P ∨ ¬P) and ¬ ¬ (P ∨ ¬P) (since P implies ¬ ¬ P, by the exercise below), which would be a contradiction. But since we can't, it is safe to add P ∨ ¬P as an axiom.
Theorem excluded_middle_irrefutable: ∀(P:Prop),

¬¬(P ∨ ¬P).

Proof.

(* FILL IN HERE *) Admitted.

☐
¬¬(P ∨ ¬P).

Proof.

(* FILL IN HERE *) Admitted.

#### Exercise: 3 stars, advanced (not_exists_dist)

It is a theorem of classical logic that the following two assertions are equivalent:
¬(∃x, ¬P x)

∀x, P x

The dist_not_exists theorem above proves one side of this
equivalence. Interestingly, the other direction cannot be proved
in constructive logic. Your job is to show that it is implied by
the excluded middle.
∀x, P x

Theorem not_exists_dist :

excluded_middle →

∀(X:Type) (P : X → Prop),

¬(∃x, ¬P x) → (∀x, P x).

Proof.

(* FILL IN HERE *) Admitted.

☐
excluded_middle →

∀(X:Type) (P : X → Prop),

¬(∃x, ¬P x) → (∀x, P x).

Proof.

(* FILL IN HERE *) Admitted.

#### Exercise: 5 stars, standard, optional (classical_axioms)

For those who like a challenge, here is an exercise taken from the Coq'Art book by Bertot and Casteran (p. 123). Each of the following four statements, together with excluded_middle, can be considered as characterizing classical logic. We can't prove any of them in Coq, but we can consistently add any one of them as an axiom if we wish to work in classical logic.
Definition peirce := ∀P Q: Prop,

((P→Q)→P)→P.

Definition double_negation_elimination := ∀P:Prop,

~~P → P.

Definition de_morgan_not_and_not := ∀P Q:Prop,

~(~P ∧ ¬Q) → P∨Q.

Definition implies_to_or := ∀P Q:Prop,

(P→Q) → (¬P∨Q).

(* FILL IN HERE *)

☐
((P→Q)→P)→P.

Definition double_negation_elimination := ∀P:Prop,

~~P → P.

Definition de_morgan_not_and_not := ∀P Q:Prop,

~(~P ∧ ¬Q) → P∨Q.

Definition implies_to_or := ∀P Q:Prop,

(P→Q) → (¬P∨Q).

(* FILL IN HERE *)

(* Wed Jan 9 12:02:45 EST 2019 *)