# LogicLogic in Coq

Require Export MoreProp.

Coq's built-in logic is very small: the only primitives are Inductive definitions, universal quantification (), and implication (), while all the other familiar logical connectives — conjunction, disjunction, negation, existential quantification, even equality — can be encoded using just these.
This chapter explains the encodings and shows how the tactics we've seen can be used to carry out standard forms of logical reasoning involving these connectives.

# Conjunction

Logical conjunction is a binary operator on propositions:

Inductive and (P Q : Prop) : Prop :=
conj : P Q (and P Q).

Intuition: To construct evidence for and P Q, we must provide evidence for P and evidence for Q. Conversely, evidence for and P Q must be of the form conj p q, where p is evidence for P and q is evidence for Q.
More familiar syntax:

Notation "P Q" := (and P Q) : type_scope.

Consider the "type" of the constructor conj:

Check conj.
(* ===>  forall P Q : Prop, P -> Q -> P /λ Q *)

The tactics we already know can be used prove things involving conjunctions.

Theorem and_example :
(beautiful 0) (beautiful 3).
Proof.
apply conj.
Case "left". apply b_0.
Case "right". apply b_3. Qed.

Just for convenience, we can use the tactic split as a shorthand for apply conj.

Theorem and_example' :
(ev 0) (ev 4).
Proof.
split.
Case "left". apply ev_0.
Case "right". apply ev_SS. apply ev_SS. apply ev_0. Qed.

Conversely, the inversion tactic can be used to take a conjunction hypothesis in the context, calculate what evidence must have been used to build it, and add variables representing this evidence to the proof context.

Theorem proj1 : P Q : Prop,
P Q P.
Proof.
intros P Q H.
inversion H as [HP HQ].
apply HP. Qed.

Theorem and_commut : P Q : Prop,
P Q Q P.
Proof.
(* WORK IN CLASS *) Admitted.

## Iff

The handy "if and only if" connective is just the conjunction of two implications.

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

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

Theorem iff_sym : P Q : Prop,
(P Q) (Q P).
Proof.
(* WORK IN CLASS *) Admitted.

# Disjunction

Inductive or (P Q : Prop) : Prop :=
| or_introl : P or P Q
| or_intror : Q or P Q.

Notation "P Q" := (or P Q) : type_scope.

Check or_introl.
(* ===>  forall P Q : Prop, P -> P λ/ Q *)

Check or_intror.
(* ===>  forall P Q : Prop, Q -> P λ/ Q *)

Intuitively, there are two ways of giving evidence for P Q:
• give evidence for P (and say that it is P you are giving evidence for — this is the function of the or_introl constructor), or
• give evidence for Q, tagged with the or_intror constructor.
Since P Q has two constructors, doing inversion on a hypothesis of type P Q yields two subgoals.

Theorem or_commut : P Q : Prop,
P Q Q P.
Proof.
intros P Q H.
inversion H as [HP | HQ].
Case "left". apply or_intror. apply HP.
Case "right". apply or_introl. apply HQ. Qed.

From here on, we'll use the shorthand tactics left and right in place of apply or_introl and apply or_intror.

Theorem or_commut' : P Q : Prop,
P Q Q P.
Proof.
intros P Q H.
inversion H as [HP | HQ].
Case "left". right. apply HP.
Case "right". left. apply HQ. Qed.

To prove the following proposition, which tactics will we need besides intros and apply?
P Q : PropP  Q   P  Q.
(1) split, inversion, left and right
(2) inversion, left and right
(3) inversion, and one of left and right
(4) only left
(5) only right
(6) only inversion
(7) none of the above
P Q : PropP  (P  Q).
(1) split, inversion, left and right
(2) inversion, left and right
(3) inversion, and one of left and right
(4) only left
(5) only right
(6) only inversion
(7) none of the above
P Q : PropP  Q   P.
(1) split, inversion, left and right
(2) inversion, left and right
(3) inversion, and one of left and right
(4) only left
(5) only right
(6) only inversion
(7) none of the above
P Q : PropP  Q  Q  P.
(1) split, inversion, left and right
(2) inversion, left and right
(3) inversion, and one of left and right
(4) only left
(5) only right
(6) only inversion
(7) none of the above

# Falsehood

Logical falsehood can be represented in Coq as an inductively defined proposition with no constructors.

Inductive False : Prop := .

Intuition: False is a proposition for which there is no way to give evidence.

Theorem False_implies_nonsense :
False 2 + 2 = 5.
Proof.
intros contra.
inversion contra. Qed.

How does this work? The inversion tactic breaks contra into each of its possible cases, and yields a subgoal for each case. As contra is evidence for False, it has no possible cases, hence, there are no possible subgoals and the proof is done.
Conversely, the only way to prove False is if there is already something nonsensical or contradictory in the context:

Theorem nonsense_implies_False :
2 + 2 = 5 False.
Proof.
intros contra.
inversion contra. Qed.

Theorem ex_falso_quodlibet : (P:Prop),
False P.
Proof.
(* WORK IN CLASS *) Admitted.

# Negation

Definition not (P:Prop) := P False.

Intuition: If we could prove P, then we could prove False (and hence we could prove anything at all).

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

Check not.
(* ===> Prop -> Prop *)

Theorem not_False :
¬ False.
Proof.
unfold not. intros H. inversion 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.

Theorem five_not_even :
¬ ev 5.
Proof.
(* WORK IN CLASS *) Admitted.

Note that some theorems that are true in classical logic are not provable in Coq's (constructive) logic. E.g., let's look at how this proof gets stuck...

Theorem classic_double_neg : P : Prop,
~~P P.
Proof.
(* WORK IN CLASS *) Admitted.

## Inequality

Saying x y is just the same as saying ~(x = y).

Notation "x ≠ y" := (¬ (x = y)) : type_scope.

A useful proof idiom: If you are trying to prove a goal that is nonsensical, apply the lemma ex_falso_quodlibet to change the goal to False. This makes it easier to use assumptions of the form ¬P that are available in the context.

Theorem not_false_then_true : b : bool,
bfalse b = true.
Proof.
intros b H. destruct b.
Case "b = true". reflexivity.
Case "b = false".
unfold not in H.
apply ex_falso_quodlibet.
apply H. reflexivity. Qed.

To prove the following proposition, which tactics will we need besides intros and apply?
Xa b : X, (a=b (ab False.
(1) inversion, unfold, left and right
(2) inversion and unfold
(3) only inversion
(4) one of left and right
(5) only unfold
(6) none of the above
To prove the following proposition, which tactics will we need besides intros and apply?
P Q : Prop,  P  Q  ~~(P  Q).
(1) inversion, unfold, left and right
(2) inversion and unfold
(3) only inversion
(4) one of left and right
(5) only unfold
(6) none of the above
To prove the following proposition, which tactics will we need besides intros and apply?
A BPropA  (A  ~~B).
(1) inversion, unfold, left and right
(2) inversion and unfold
(3) only inversion
(4) one of left and right
(5) only unfold
(6) none of the above
To prove the following proposition, which tactics will we need besides intros and apply?
P QProp,  P  Q  ~~P  ~~Q.
(1) inversion, unfold, left and right
(2) inversion and unfold
(3) only inversion
(4) one of left and right
(5) only unfold
(6) none of the above
To prove the following proposition, which tactics will we need besides intros and apply?
A : Prop, 1=0  (A  ¬A).
(1) inversion, unfold, left and right
(2) inversion and unfold
(3) only inversion
(4) one of left and right
(5) only unfold
(6) none of the above

# Existential Quantification

Another critical logical connective is existential quantification. We can express it with the following definition:

Inductive ex (X:Type) (P : XProp) : Prop :=
ex_intro : (witness:X), P witness ex X P.

That is, ex is a family of propositions indexed by a type X and a property P over X. In order to give evidence for the assertion "there exists an x for which the property P holds" we must actually name a witness — a specific value x — and then give evidence for P x, i.e., evidence that x has the property P.
Standard notations for existentials:

Notation "'exists' x , p" := (ex _ (fun xp))
(at level 200, x ident, right associativity) : type_scope.
Notation "'exists' x : X , p" := (ex _ (fun x:Xp))
(at level 200, x ident, right associativity) : type_scope.

The details of how these notations work are not too important: the critical point is that they allow us to write x, P or x:X, P, just as we do with the quantifier.
To prove an existential statement, use apply ex_intro.

Example exists_example_1 : n, n + (n × n) = 6.
Proof.
apply ex_intro with (witness:=2).
reflexivity. Qed.

Note that we have to explicitly give the witness.
Or, instead of writing apply ex_intro with (witness:=e) all the time, we can use the convenient shorthand e, which means the same thing.

Example exists_example_1' : n, n + (n × n) = 6.
Proof.
2.
reflexivity. Qed.

Conversely, if we have an existential hypothesis in the context, we can eliminate it with inversion. Note the use of the as... pattern to name the variable that Coq introduces to name the witness value and get evidence that the hypothesis holds for the witness. (If we don't explicitly choose one, Coq will just call it witness, which makes proofs confusing.)

Theorem exists_example_2 : n,
(m, n = 4 + m)
(o, n = 2 + o).
Proof.
intros n H.
inversion H as [m Hm].
(2 + m).
apply Hm. Qed.

# Equality

Even Coq's equality relation is not built in. It has (roughly) the following inductive definition.

Inductive eq {X:Type} : X X Prop :=
refl_equal : x, eq x x.
Standard infix notation:

Notation "x = y" := (eq x y)
(at level 70, no associativity)
: type_scope.

The definition of = is a bit subtle. The way to think about it is that, given a set X, it defines a family of propositions "x is equal to y," indexed by pairs of values (x and y) from X. There is just one way of constructing evidence for members of this family: applying the constructor refl_equal to a type X and a value x : X yields evidence that x is equal to x.
We can use refl_equal to construct evidence that, for example, 2 = 2. Can we also use it to construct evidence that 1 + 1 = 2? Yes: indeed, it is the very same piece of evidence! The reason is that Coq treats as "the same" any two terms that are convertible according to a simple set of computation rules. These rules, which are similar to those used by Eval compute, include evaluation of function application, inlining of definitions, and simplification of matches.

Lemma four: 2 + 2 = 1 + 3.
Proof.
apply refl_equal.
Qed.

The reflexivity tactic that we have used to prove equalities up to now is essentially just short-hand for apply refl_equal.

Which of the following is correct proof object for the proposition
xx + 3 = 4
?
(1) refl_equal 4
(2) ex_intro nat (fun z (z + 3 = 4)) 1 refl_equal
(3) ex_intro nat (z + 3 = 4) 1 (refl_equal 4)
(4) ex_intro nat (fun z (z + 3 = 4)) 1 (refl_equal 4)
(5) ex_intro nat (fun z (z + 3 = 4)) 1 (refl_equal 1)
(6) none of the above

# Evidence-carrying booleans.

So far we've seen two different forms of equality predicates: eq, which produces a Prop, and the type-specific forms, like beq_nat, that produce boolean values. The former are more convenient to reason about, but we've relied on the latter to let us use equality tests in computations. While it is straightforward to write lemmas (e.g. beq_nat_true and beq_nat_false) that connect the two forms, using these lemmas quickly gets tedious.
It turns out that we can get the benefits of both forms at once by using a construct called sumbool.

Inductive sumbool (A B : Prop) : Set :=
| left : A sumbool A B
| right : B sumbool A B.

Notation "{ A } + { B }" := (sumbool A B) : type_scope.

Think of sumbool as being like the boolean type, but instead of its values being just true and false, they carry evidence of truth or falsity. This means that when we destruct them, we are left with the relevant evidence as a hypothesis — just as with or. (In fact, the definition of sumbool is almost the same as for or. The only difference is that values of sumbool are declared to be in Set rather than in Prop; this is a technical distinction that allows us to compute with them.)
Here's how we can define a sumbool for equality on nats

Theorem eq_nat_dec : n m : nat, {n = m} + {nm}.
Proof.
intros n.
induction n as [|n'].
Case "n = 0".
intros m.
destruct m as [|m'].
SCase "m = 0".
left. reflexivity.
SCase "m = S m'".
right. intros contra. inversion contra.
Case "n = S n'".
intros m.
destruct m as [|m'].
SCase "m = 0".
right. intros contra. inversion contra.
SCase "m = S m'".
destruct IHn' with (m := m') as [eq | neq].
left. apply f_equal. apply eq.
right. intros Heq. inversion Heq as [Heq']. apply neq. apply Heq'.
Defined.

Read as a theorem, this says that equality on nats is decidable: that is, given two nat values, we can always produce either evidence that they are equal or evidence that they are not. Read computationally, eq_nat_dec takes two nat values and returns a sumbool constructed with left if they are equal and right if they are not; this result can be tested with a match or, better, with an if-then-else, just like a regular boolean. (Notice that we ended this proof with Defined rather than Qed. The only difference this makes is that the proof becomes transparent, meaning that its definition is available when Coq tries to do reductions, which is important for the computational interpretation.)
Here's a simple example illustrating the advantages of the sumbool form.

Definition override' {X: Type} (f: natX) (k:nat) (x:X) : natX:=
fun (k':nat) ⇒ if eq_nat_dec k k' then x else f k'.

Theorem override_same' : (X:Type) x1 k1 k2 (f : natX),
f k1 = x1
(override' f k1 x1) k2 = f k2.
Proof.
intros X x1 k1 k2 f. intros Hx1.
unfold override'.
destruct (eq_nat_dec k1 k2). (* observe what appears as a hypothesis *)
Case "k1 = k2".
rewrite e.
symmetry. apply Hx1.
Case "k1 ≠ k2".
reflexivity. Qed.

Compare this to the more laborious proof (in MoreCoq.v) for the version of override defined using beq_nat, where we had to use the auxiliary lemma beq_nat_true to convert a fact about booleans to a Prop.