# LogicLogic in Coq

Set Warnings "-notation-overridden,-parsing,-deprecated-hint-without-locality".
From LF Require Export Tactics.
So far, we have seen...
• propositions: factual claims
• equality propositions (e1 = e2)
• implications (P Q)
• quantified propositions ( x, P)
• proofs: ways of presenting evidence for the truth of a proposition
In this chapter we will introduce several more flavors of both propositions and proofs.
Like everything in Coq, propositions have a type:
Check (3 = 3) : Prop.

Check ( n m : nat, n + m = m + n) : Prop.
Note that all syntactically well-formed propositions have type Prop in Coq, regardless of whether they are true.
Simply being a proposition is one thing; being provable is a different thing!
Check 2 = 2 : Prop.

Check 3 = 2 : Prop.

Check n : nat, n = 2 : Prop.
So far, we've seen one primary place that propositions can appear: in Theorem (and Lemma and Example) declarations.
Theorem plus_2_2_is_4 :
2 + 2 = 4.
Proof. reflexivity. Qed.
Propositions are first-class entities in Coq. For example, we can name them:
Definition plus_claim : Prop := 2 + 2 = 4.
Check plus_claim : Prop.

Theorem plus_claim_is_true :
plus_claim.
Proof. reflexivity. Qed.
We can also write 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 : natProp.
In Coq, functions that return propositions are said to define properties of their arguments.
For instance, here's a (polymorphic) property defining the familiar notion of an injective function.
Definition injective {A B} (f : AB) :=
x y : A, f x = f yx = y.

Lemma succ_inj : injective S.
Proof.
intros n m H. injection H as H1. apply H1.
Qed.
The equality operator = is a (binary) function that returns a Prop.
The expression n = m is syntactic sugar for eq n m (defined in Coq's standard library using the Notation mechanism). Because eq can be used with elements of any type, it is also polymorphic:
Check @eq : A : Type, AAProp.

What is the type of the following expression?
pred (S O) = O (1) Prop
(2) natProp
(3) n:nat, Prop
(4) natnat
(5) Not typeable
What is the type of the following expression?
n:nat, pred (S n) = n (1) Prop
(2) natProp
(3) n:nat, Prop
(4) natnat
(5) Not typeable
What is the type of the following expression?
n:nat, S (pred n) = n (1) Prop
(2) natProp
(3) natnat
(4) Not typeable
What is the type of the following expression?
n:nat, S (pred n) (1) Prop
(2) natProp
(3) natnat
(4) Not typeable
What is the type of the following expression?
fun n:natS (pred n) (1) Prop
(2) natProp
(3) natnat
(4) Not typeable
What is the type of the following expression?
fun n:natS (pred n) = n (1) Prop
(2) natProp
(3) natnat
(4) Not typeable
Which of the following is not a proposition?
(1) 3 + 2 = 4
(2) 3 + 2 = 5
(3) 3 + 2 =? 5
(4) ((3+2) =? 4) = false
(5) n, (((3+2) =? n) = true) n = 5
(6) All of these are propositions

# Logical Connectives

## Conjunction

The 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.
For any propositions A and B, if we assume that A is true and that B is true, we can conclude that A B is also true.
Lemma and_intro : A B : Prop, ABAB.
Proof.
intros A B HA HB. split.
- apply HA.
- apply HB.
Qed.

Example and_exercise :
n m : nat, n + m = 0 → n = 0 ∧ m = 0.
Proof.
(* WORK IN CLASS *) Admitted.
So much for proving conjunctive statements. To go in the other direction -- i.e., to use a conjunctive hypothesis to help prove something else -- we employ the destruct tactic.
When the current proof context contains a hypothesis H of the form A B, writing destruct H as [HA HB] will remove H from the context and replace it with two new hypotheses: HA, stating that A is true, and HB, stating that B is true.
Lemma and_example2 :
n m : nat, n = 0 ∧ m = 0 → n + m = 0.
Proof.
(* WORK IN CLASS *) Admitted.
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.
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.
For the present example, both ways work. But in other situations we may wind up with a conjunctive hypothesis in the middle of a proof...
Lemma and_example3 :
n m : nat, n + m = 0 → n × m = 0.
Proof.
(* WORK IN CLASS *) Admitted.
By the way, the infix notation is actually just syntactic sugar for and A B. That is, and is a Coq operator that takes two propositions as arguments and yields a proposition.
Check and : PropPropProp.

## Disjunction

Another important connective is the 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.)
To use a disjunctive hypothesis in a proof, we proceed by case analysis -- which, as with other data types like nat, can be done explicitly with destruct or implicitly with an intros pattern:
Lemma factor_is_O:
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.
We can see in this example that, when we perform case analysis on a disjunction A B, we must separately satisfy two proof obligations, each showing that the conclusion holds under a different assumption -- A in the first subgoal and B in the second. Note that the case analysis pattern [Hn | Hm] allows us to name the hypotheses that are generated for the subgoals.
Conversely, to show that a disjunction holds, it suffices to show that one of its sides holds. This can be 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_l : A B : Prop, AAB.
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.
(* WORK IN CLASS *) Admitted.

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

Lemma mult_is_O :
n m, n × m = 0 → n = 0 ∨ m = 0.
Proof.
(* FILL IN HERE *) Admitted.

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

Theorem or_commut : P Q : Prop,
PQQP.
Proof.
(* FILL IN HERE *) Admitted.

## Falsehood and Negation

So far, we have mostly been concerned with proving "positive" statements -- addition is commutative, appending lists is associative, etc. Of course, we may also be interested in negative results, demonstrating that some given proposition is not true. Such statements are expressed with the logical negation operator ¬.
To see how negation works, recall the principle of explosion from the Tactics chapter, which asserts that, if we assume a contradiction, then any other proposition can be derived.
Following this intuition, we could define ¬ P ("not P") as Q, P Q.
Coq actually makes a slightly different (but equivalent) choice, defining ¬ P as P False, where False is a specific contradictory proposition defined in the standard library.
Module NotPlayground.

Definition not (P:Prop) := PFalse.

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

Check not : PropProp.

End NotPlayground.
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),
FalseP.
Proof.
(* WORK IN CLASS *) Admitted.
Inequality is a frequent enough form of negated statement that there is a special notation for it, x y:
Notation "x <> y" := (~(x = y)).
We can use not to state that 0 and 1 are different elements of nat:
Theorem zero_not_one : 0 ≠ 1.
Proof.
unfold not.
intros contra.
discriminate contra.
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 make Coq 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.
(* WORK IN CLASS *) Admitted.

Theorem double_neg : P : Prop,
P → ~~P.
Proof.
(* WORK IN CLASS *) Admitted.
Since inequality involves a negation, it also requires a little practice to be able to work with it fluently. Here is one useful trick:
If you are trying to prove a goal that is nonsensical (e.g., the goal state is false = true), apply ex_falso_quodlibet to change the goal to False.
This makes it easier to use assumptions of the form ¬P that may be available in the context -- in particular, assumptions of the form xy.
Theorem not_true_is_false : b : bool,
btrueb = false.
Proof.
intros b H.
destruct b eqn:HE.
- (* b = true *)
unfold not in H.
apply ex_falso_quodlibet.
apply H. reflexivity.
- (* b = false *)
reflexivity.
Qed.

To prove the following proposition, which tactics will we need besides intros and apply?
X, a b : X, (a=b) ∧ (ab) → False. (1) destruct, unfold, left and right
(2) destruct and unfold
(3) only destruct
(4) left and/or right
(5) only unfold
(6) none of the above
Lemma quiz1: X, a b : X, (a=b) ∧ (ab) → False.
Proof.
intros X a b [H0 H1]. apply H1 in H0. apply H0.
Qed.

To prove the following proposition, which tactics will we need besides intros and apply?
P Q : Prop, PQ → ~~(PQ). (1) destruct, unfold, left and right
(2) destruct and unfold
(3) only destruct
(4) left and/or right
(5) only unfold
(6) none of the above
Lemma quiz2 : P Q : Prop, PQ → ~~(PQ).
Proof.
intros P Q H H1. apply H1 in H. apply H.
Qed.

To prove the following proposition, which tactics will we need besides intros and apply?
A B: Prop, A → (A ∨ ~~B). (1) destruct, unfold, left and right
(2) destruct and unfold
(3) only destruct
(4) left and/or right
(5) only unfold
(6) none of the above
Lemma quiz3 : A B: Prop, A → (A ∨ ~~B).
Proof.
intros P Q H. left. apply H.
Qed.

To prove the following proposition, which tactics will we need besides intros and apply?
P Q: Prop, PQ → ~~P ∨ ~~Q. (1) destruct, unfold, left and right
(2) destruct and unfold
(3) only destruct
(4) left and/or right
(5) only unfold
(6) none of the above
Lemma quiz4 : P Q: Prop, PQ → ~~P ∨ ~~Q.
Proof.
intros P Q [H0 | H0].
- (* left *)
left. intros H1. apply H1 in H0. apply H0.
- (* right *)
right. intros H1. apply H1 in H0. apply H0.
Qed.

To prove the following proposition, which tactics will we need besides intros and apply?
A : Prop, 1=0 → (A ∨ ¬A). (1) discriminate, unfold, left and right
(2) discriminate and unfold
(3) only discriminate
(4) left and/or right
(5) only unfold
(6) none of the above
Lemma quiz5 : A : Prop, 1=0 → (A ∨ ¬A).
Proof.
intros P H. discriminate H.
Qed.

## Truth

Besides False, Coq's standard library also defines True, a proposition that is trivially true. To prove it, we use the constant I : True, which is also defined in the standard library:
Lemma True_is_true : True.
Proof. apply I. Qed.
Unlike False, which is used extensively, True is used relatively rarely, since it is trivial (and therefore uninteresting) to prove as a goal, and conversely it provides no interesting information when used as a hypothesis.

## Logical Equivalence

The handy "if and only if" connective, which asserts that two propositions have the same truth value, is simply the conjunction of two implications.
Module IffPlayground.

Definition iff (P Q : Prop) := (PQ) ∧ (QP).

Notation "P <-> Q" := (iff P Q)
(at level 95, no associativity)
: type_scope.

End IffPlayground.
Theorem iff_sym : P Q : Prop,
(PQ) → (QP).
Proof.
(* WORK IN CLASS *) Admitted.

Lemma not_true_iff_false : b,
btrueb = false.
Proof.
(* WORK IN CLASS *) Admitted.
Example: using apply with .
Lemma apply_iff_example1:
P Q R : Prop, (PQ) → (QR) → (PR).
intros P Q R Hiff H HP. apply H. apply Hiff. apply HP.
Qed.

Lemma apply_iff_example2:
P Q R : Prop, (PQ) → (PR) → (QR).
intros P Q R Hiff H HQ. apply H. apply Hiff. apply HQ.
Qed.

#### Exercise: 1 star, standard, optional (iff_properties)

Using the above proof that is symmetric (iff_sym) as a guide, prove that it is also reflexive and transitive.
Theorem iff_refl : P : Prop,
PP.
Proof.
(* FILL IN HERE *) Admitted.

Theorem iff_trans : P Q R : Prop,
(PQ) → (QR) → (PR).
Proof.
(* FILL IN HERE *) Admitted.

## Setoids and Logical Equivalence

Some of Coq's tactics treat iff statements specially, avoiding the need for some low-level proof-state manipulation. In particular, rewrite and reflexivity can be used with iff statements, not just equalities. To enable this behavior, we have to import the Coq library that supports it:
From Coq Require Import Setoids.Setoid.
A "setoid" is a set equipped with an equivalence relation, such as = or .
Here is a simple example demonstrating how these tactics work with iff. First, let's prove a couple of basic iff equivalences.
Lemma mul_eq_0 : n m, n × m = 0 ↔ n = 0 ∨ m = 0.
Proof.
split.
- apply mult_is_O.
- apply factor_is_O.
Qed.

Theorem or_assoc :
P Q R : Prop, P ∨ (QR) ↔ (PQ) ∨ 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.

Lemma mul_eq_0_ternary :
n m p, n × m × p = 0 ↔ n = 0 ∨ m = 0 ∨ p = 0.
Proof.
intros n m p.
rewrite mul_eq_0. rewrite mul_eq_0. rewrite or_assoc.
reflexivity.
Qed.

## Existential Quantification

To prove a statement of the form x, P, we must show that P holds for some specific choice of value for x, known as the 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.
Definition Even x := n : nat, x = double n.

Lemma four_is_Even : Even 4.
Proof.
unfold Even. 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.
(* WORK IN CLASS *) Admitted.

#### Exercise: 3 stars, standard, optional (leb_plus_exists)

Theorem leb_plus_exists : n m, n <=? m = true x, m = n+x.
Proof.
(* FILL IN HERE *) Admitted.

Theorem plus_exists_leb : n m, ( x, m = n+x) → n <=? m = true.
Proof.
(* FILL IN HERE *) Admitted.

# Programming with Propositions

What does it mean to say that "an element x occurs in a list l"?
• If l is the empty list, then x cannot occur in 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 it is equal to x' or if it occurs in l'.
We can translate this directly into a straightforward recursive function taking an element and a list and returning a proposition (!):
Fixpoint In {A : Type} (x : A) (l : list A) : Prop :=
match l with
| [] ⇒ False
| x' :: l'x' = xIn 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.
(* WORK IN CLASS *) Admitted.

Example In_example_2 :
n, In n [2; 4] →
n', n = 2 × n'.
Proof.
(* WORK IN CLASS *) Admitted.
We can also prove more generic, higher-level lemmas about In.
Theorem In_map :
(A B : Type) (f : AB) (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.

# Applying Theorems to Arguments

Coq also treats proofs as first-class objects!
We have seen that we can use Check to ask Coq to print the type of an expression. We can also use it to ask what theorem a particular identifier refers to.
Check plus : natnatnat.
Check add_comm : n m : nat, n + m = m + n.
Coq checks the statement of the add_comm theorem (or prints it for us, if we leave off the part beginning with the colon) in the same way that it checks the type of any term (e.g., plus) that we ask it to Check.
Why?
The reason is that the identifier add_comm actually refers to a proof object, which 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 proposition that it is a proof of.
The type of an ordinary function tells us what we can do with it.
• 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, the statement of a theorem tells us what we can use that theorem for.
• if we have a term of type n = m n + n = m + m and we provide it two numbers n and m plus an "argument" of type n = m, we can derive n + n = m + m.
Coq actually allows us to apply a theorem as if it were a function.
This is often handy in proof scripts -- e.g., suppose we want too prove the following:
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 add_comm twice to make the two sides match. The problem, however, is that the second rewrite will undo the effect of the first.
Proof.
intros x y z.
(* We are back where we started... *)
Abort.
We can fix this by applying add_comm to the arguments we want it be to instantiated with. Then the rewrite can only happen in one place.
x y z, x + (y + z) = (z + y) + x.
Proof.
intros x y z.
reflexivity.
Qed.
Let's see another example of using a theorem like a function.
The following theorem says: any list l containing some element must be nonempty.
Theorem in_not_nil :
A (x : A) (l : list A), In x ll ≠ [].
Proof.
intros A x l H. unfold not. intro Hl.
rewrite Hl in H.
simpl in H.
apply H.
Qed.

We should be able to use this theorem to prove the special case where x is 42. However, naively, the tactic apply in_not_nil will fail because it cannot infer the value of x.
Lemma in_not_nil_42 :
l : list nat, In 42 ll ≠ [].
Proof.
intros l H.
Fail apply in_not_nil.
Abort.
There are several ways to work around this:
Use apply ... with ...
Lemma in_not_nil_42_take2 :
l : list nat, In 42 ll ≠ [].
Proof.
intros l H.
apply in_not_nil with (x := 42).
apply H.
Qed.
Use apply ... in ...
Lemma in_not_nil_42_take3 :
l : list nat, In 42 ll ≠ [].
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 ll ≠ [].
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 ll ≠ [].
Proof.
intros l H.
apply (in_not_nil _ _ _ H).
Qed.

Lemma quiz : a b : nat,
a = bb = 42 →
( (X : Type) (n m o : X),
n = mm = on = o) →
True.
Proof.
intros a b H1 H2 trans_eq.
Suppose we have
a, b : nat
H1 : a = b
H2 : b = 42
trans_eq : (X : Type) (n m o : X),
n = mm = on = o
What is the type of this proof object?
trans_eq nat a b 42 H1 H2
(1) a = b
(2) 42 = a
(3) a = 42
(4) Does not typecheck
Check trans_eq nat a b 42 H1 H2
: a = 42.
Suppose, again, that we have
a, b : nat
H1 : a = b
H2 : b = 42
trans_eq : (X : Type) (n m o : X),
n = mm = on = o
What is the type of this proof object?
trans_eq nat b 42 a H2
(1) b = a
(2) b = a 42 = a
(3) 42 = a b = a
(4) Does not typecheck
Check trans_eq nat b 42 a H2
: 42 = ab = a.
Suppose, again, that we have
a, b : nat
H1 : a = b
H2 : b = 42
trans_eq : (X : Type) (n m o : X),
n = mm = on = o
What is the type of this proof object?
trans_eq _ 42 a b
(1) a = b b = 42 a = 42
(2) 42 = a a = b 42 = b
(3) a = 42 42 = b a = b
(4) Does not typecheck
Check trans_eq _ 42 a b
: 42 = aa = b → 42 = b.
Suppose, again, that we have
a, b : nat
H1 : a = b
H2 : b = 42
trans_eq : (X : Type) (n m o : X),
n = mm = on = o
What is the type of this proof object?
trans_eq _ _ _ _ H2 H1
(1) b = a
(2) 42 = a
(3) a = 42
(4) Does not typecheck
Fail Check trans_eq _ _ _ _ H2 H1.

Abort.

# Coq vs. Set Theory

Coq's logical core, the Calculus of Inductive Constructions, is a "metalanguage for mathematics" in the same sense as familiar foundations for paper-and-pencil math, like Zermelo-Fraenkel Set Theory (ZFC).
Mostly, the differences are not too important. But there are a few points worth discussing...

## Functional Extensionality

Coq's logic is intentionally quite minimal. This means that there are occasionally some cases where translating standard mathematical reasoning into Coq can be cumbersome or even impossible, unless we enrich the core logic with additional axioms.
For example, we can write an equality proposition stating that two functions are equal to each other:
Example function_equality_ex1 :
(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 output on every input:
( x, f x = g x) → f = g This is known as the principle of functional extensionality.
However, functional extensionality is not part of Coq's built-in logic. This means that some apparently "obvious" propositions are not provable.
Example function_equality_ex2 :
(fun xplus x 1) = (fun xplus 1 x).
Proof.
(* Stuck *)
Abort.
However, if we like, we can add functional extensionality to Coq's core using the Axiom command.
Axiom functional_extensionality : {X Y: Type}
{f g : XY},
( (x:X), f x = g x) → f = g.
Defining something as an 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_ex2 :
(fun xplus x 1) = (fun xplus 1 x).
Proof.
apply functional_extensionality. intros x.
Qed.
Naturally, we must be careful when adding new axioms into Coq's logic, as this can render it inconsistent -- that is, it may become possible to prove every proposition, including False, 2+2=5, etc.!
Unfortunately, there is no simple way of telling whether an axiom is safe to add: hard work by highly trained mathematicians is often required to establish the consistency of any particular combination of axioms.
Fortunately, it is known that adding functional extensionality, in particular, is consistent.
To check whether a particular proof relies on any additional axioms, use the Print Assumptions command: Print Assumptions function_equality_ex2.
(* ===>
Axioms:
functional_extensionality :
forall (X Y : Type) (f g : X -> Y),
(forall x : X, f x = g x) -> f = g *)

(If you try this yourself, you may also see add_comm listed as an assumption, depending on whether the copy of Tactics.v in the local directory has the proof of add_comm filled in.)
Is this provable by just reflexivity, without functional_extensionality?
(fun xs 1 :: xs) = (fun xs  ++ xs)
(1) Yes
(2) No
Example cons_1_eq_ex : (fun xs ⇒ 1 :: xs) = (fun xs ⇒  ++ xs).
Proof. reflexivity. Qed.

## Propositions vs. Booleans

We've seen two different ways of expressing logical claims in Coq: with booleans (of type bool), and with propositions (of type Prop).
Here are the key differences between bool and Prop:
```                                           bool     Prop
====     ====
decidable?                       yes       no
useable with match?             yes       no
equalities rewritable?          no        yes
```

## Working with Decidable Properties

Since every function terminates on all inputs in Coq, a function of type nat bool is a decision procedure -- i.e., it yields true or false on all inputs.
• For example, even : nat bool is a decision procedure for the property "is even".
It follows that there are some properties of numbers that we cannot express as functions of type nat bool.
• For example, the property "is the code of a halting Turing machine" is undecidable, so there is no way to write it as a function of type nat bool.
On the other hand, natProp is the type of all properties of numbers that can be expressed in Coq's logic, including both decidable and undecidable ones.
• For example, "being the code of a halting Turing machine" is a perfectly legitimate mathematical property, and we can absolutely represent it as a Coq expression of type nat Prop.
Since Prop includes both decidable and undecidable properties, we have two choices when when we want to formalize a property that happens to be decidable: we can express it as a boolean computation or as a function into Prop.
For instance, to claim that a number n is even, we can say either...
... that even n evaluates to true...
Example even_42_bool : even 42 = true.
Proof. reflexivity. Qed.

... or that there exists some k such that n = double k.
Example even_42_prop : Even 42.
Proof. unfold Even. 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.
Lemma even_double : k, even (double k) = true.
Proof.
intros k. induction k as [|k' IHk'].
- reflexivity.
- simpl. apply IHk'.
Qed.

Lemma even_double_conv : n, k,
n = if even n then double k else S (double k).
Proof.
(* Hint: Use the even_S lemma from Induction.v. *)
(* FILL IN HERE *) Admitted.
Now the main theorem:
Theorem even_bool_prop : n,
even n = trueEven n.
Proof.
intros n. split.
- intros H. destruct (even_double_conv n) as [k Hk].
rewrite Hk. rewrite H. k. reflexivity.
- intros [k Hk]. rewrite Hk. apply even_double.
Qed.

In view of this theorem, we say that the boolean computation even 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.
Again, these two notions are equivalent.
Theorem eqb_eq : n1 n2 : nat,
n1 =? n2 = truen1 = n2.
Proof.
intros n1 n2. split.
- apply eqb_true.
- intros H. rewrite H. rewrite eqb_refl. reflexivity.
Qed.

Even when the boolean and propositional formulations of a claim are equivalent from a purely logical perspective, they are often not equivalent from the point of view of convenience for some specific purpose.
For example, we cannot test whether or not a Prop is true in a function definition; as a consequence, the following code fragment is rejected:
Fail
Definition is_even_prime n :=
if n = 2 then true
else false.
An important side benefit of stating facts using booleans is enabling some proof automation through computation with Coq terms, a technique known as proof by reflection.
Consider the following statement:
Example even_1000 : Even 1000.
The most direct way to prove this is to give the value of k explicitly.
Proof. unfold Even. 500. reflexivity. Qed.
The proof of the corresponding boolean statement is even simpler, because we don't have to invent the witness: Coq's computation mechanism does it for us!
Example even_1000' : even 1000 = true.
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'' : Even 1000.
Proof. apply even_bool_prop. reflexivity. Qed.
Another notable difference is that the negation of a "boolean fact" is straightforward to state and prove: simply flip the expected boolean result.
Example not_even_1001 : even 1001 = false.
Proof.
(* WORK IN CLASS *) Admitted.
In contrast, propositional negation can be more difficult to work with directly.
Example not_even_1001' : ~(Even 1001).
Proof.
(* WORK IN CLASS *) Admitted.
Equality provides a complementary example, where it is sometimes easier to work in the propositional world.
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 = truen + p =? m + p = true.
Proof.
(* WORK IN CLASS *) Admitted.
We won't discuss reflection any further for the moment, but it serves as a good example showing the different strengths of booleans and general propositions; being able to cross back and forth between the boolean and propositional worlds will often be convenient in later chapters.

## Classical vs. Constructive Logic

The following reasoning principle is not derivable in Coq (though, again, it can consistently be added):
Definition excluded_middle := P : Prop,
P ∨ ¬P.
Logics like Coq's, which do not assume the excluded middle, are referred to as constructive logics.
More conventional logical systems such as ZFC, in which the excluded middle does hold for arbitrary propositions, are referred to as classical.