Logic: Coqにおける論理
In previous chapters, we have seen many examples of factual
claims (propositions) and ways of presenting evidence of their
truth (proofs). In particular, we have worked extensively with
equality propositions of the form e1 = e2, with
implications (P > Q), and with quantified propositions (forall
x, P). In this chapter, we will see how Coq can be used to carry
out other familiar forms of logical reasoning.
Before diving into details, let's talk a bit about the status of
mathematical statements in Coq. Recall that Coq is a 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:
Note that all syntactically wellformed propositions have type
Prop in Coq, regardless of whether they are true.
Simply being a proposition is one thing; being provable is
something else!
Indeed, propositions don't just have types: they are
firstclass objects that can be manipulated in the same ways as
the other entities in Coq's world.
So far, we've seen one primary place that propositions can appear:
in Theorem (and Lemma and Example) declarations.
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.
We can later use this name in any situation where a proposition is
expected  for example, as the claim in a Theorem declaration.
We can also write parameterized propositions  that is,
functions that take arguments of some type and return a
proposition.
For instance, the following function takes a number
and returns a proposition asserting that this number is equal to
three:
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 : A > B) :=
forall x y : A, f x = f y > x = y.
Lemma succ_inj : injective S.
Proof.
intros n m H. injection H as H1. apply H1.
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:
(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.)
論理積、連言(Conjunction))
To prove a conjunction, use the split tactic. It will generate
two subgoals, one for each part of the statement:
Proof.
split.
 reflexivity.
 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 : forall A B : Prop, A > B > A /\ B.
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.
 reflexivity.
 reflexivity.
Qed.
☐
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.
If the proof context contains a hypothesis H of the form
A /\ B, writing destruct H as [HA HB] will remove H from the
context and add two new hypotheses: HA, stating that A is
true, and HB, stating that B is true.
Lemma and_example2 :
forall n m : nat, n = 0 /\ m = 0 > n + m = 0.
Proof.
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' :
forall 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'' :
forall 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 :
forall n m : nat, n + m = 0 > n * m = 0.
Proof.
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:
☐
Finally, we sometimes need to rearrange the order of conjunctions
and/or the grouping of multiway conjunctions. The following
commutativity and associativity theorems are handy in such
cases.
Theorem and_commut : forall P Q : Prop,
P /\ Q > Q /\ P.
Proof.
intros P Q [HP HQ].
split.
 apply HQ.
 apply HP. Qed.
練習問題: ★★, standard (and_assoc)
Theorem and_assoc : forall P Q R : Prop,
P /\ (Q /\ R) > (P /\ Q) /\ R.
Proof.
intros P Q R [HP [HQ HR]].
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.
論理和、選言(Disjunction)
Lemma or_example :
forall n m : nat, n = 0 \/ m = 0 > n * m = 0.
Proof.
intros n m [Hn  Hm].

rewrite Hn. reflexivity.

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...
... and here is a slightly more interesting example requiring both
left and right:
Lemma zero_or_succ :
forall n : nat, n = 0 \/ n = S (pred n).
Proof.
intros [n].
 left. reflexivity.
 right. reflexivity.
Qed.
☐
Falsehood and Negation
Module MyNot.
Definition not (P:Prop) := P > False.
Notation "~ x" := (not x) : type_scope.
Check not.
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 : forall (P:Prop),
False > P.
Proof.
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.
Show that Coq's definition of negation implies the intuitive one
mentioned above:
Exercise: 2 stars, standard, optional (not_implies_our_not)
☐
Inequality is a frequent enough example 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:
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.)
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 : forall P Q : Prop,
(P /\ ~P) > Q.
Proof.
intros P Q [HP HNA]. unfold not in HNA.
apply HNA in HP. destruct HP. Qed.
Theorem double_neg : forall P : Prop,
P > ~~P.
Proof.
intros P H. unfold not. intros G. apply G. apply H. Qed.
Exercise: 2 stars, advanced (double_neg_inf)
☐
Write an informal proof (in English) of the proposition forall P
: Prop, ~(P /\ ~P).
Exercise: 1 star, advanced (informal_not_PNP)
☐
Similarly, since inequality involves a negation, it 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
x<>y.
Theorem not_true_is_false : forall b : bool,
b <> true > b = false.
Proof.
intros [] H.

unfold not in H.
apply ex_falso_quodlibet.
apply H. reflexivity.

reflexivity.
Qed.
Since reasoning with ex_falso_quodlibet is quite common, Coq
provides a builtin tactic, exfalso, for applying it.
Theorem not_true_is_false' : forall b : bool,
b <> true > b = false.
Proof.
intros [] H.

unfold not in H.
exfalso. apply H. reflexivity.
 reflexivity.
Qed.
Truth
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 higherorder Props.
We will see examples of such uses of True later on.
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 : forall P Q : Prop,
(P <> Q) > (Q <> P).
Proof.
intros P Q [HAB HBA].
split.
 apply HBA.
 apply HAB. Qed.
Lemma not_true_iff_false : forall b,
b <> true <> b = false.
Proof.
intros b. split.
 apply not_true_is_false.

intros H. rewrite H. intros H'. discriminate H'.
Qed.
Exercise: 1 star, standard, optional (iff_properties)
Theorem iff_refl : forall P : Prop,
P <> P.
Proof.
Admitted.
Theorem iff_trans : forall P Q R : Prop,
(P <> Q) > (Q <> R) > (P <> R).
Proof.
Admitted.
Theorem or_distributes_over_and : forall P Q R : Prop,
P \/ (Q /\ R) <> (P \/ Q) /\ (P \/ R).
Proof.
Admitted.
P \/ (Q /\ R) <> (P \/ Q) /\ (P \/ R).
Proof.
Admitted.
☐
Some of Coq's tactics treat iff statements specially, avoiding
the need for some lowlevel proofstate manipulation. In
particular, rewrite and reflexivity can be used with iff
statements, not just equalities. To enable this behavior, we need
to import a Coq library that supports it:
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 : forall n m, n * m = 0 <> n = 0 \/ m = 0.
Proof.
split.
 apply mult_eq_0.
 apply or_example.
Qed.
Lemma or_assoc :
forall 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.
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 :
forall 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 :
forall n m : nat, n * m = 0 > n = 0 \/ m = 0.
Proof.
intros n m H. apply mult_0. apply H.
Qed.
Existential Quantification
Conversely, if we have an existential hypothesis exists 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 : forall n,
(exists m, n = 4 + m) >
(exists o, n = 2 + o).
Proof.
intros n [m Hm]. exists (2 + m).
apply Hm. Qed.
Exercise: 1 star, standard, recommended (dist_not_exists)
Theorem dist_not_exists : forall (X:Type) (P : X > Prop),
(forall x, P x) > ~ (exists x, ~ P x).
Proof.
Admitted.
☐
Prove that existential quantification distributes over
disjunction.
Exercise: 2 stars, standard (dist_exists_or)
Theorem dist_exists_or : forall (X:Type) (P Q : X > Prop),
(exists x, P x \/ Q x) <> (exists x, P x) \/ (exists x, Q x).
Proof.
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.
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.
simpl. right. right. right. left. reflexivity.
Qed.
Example In_example_2 :
forall n, In n [2; 4] >
exists n', n = 2 * n'.
Proof.
simpl.
intros n [H  [H  []]].
 exists 1. rewrite < H. reflexivity.
 exists 2. rewrite < H. reflexivity.
Qed.
(Notice the use of the empty pattern to discharge the last case
en passant.)
We can also prove more generic, higherlevel 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:
Lemma In_map :
forall (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'].

simpl. intros [].

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 :
forall (A B : Type) (f : A > B) (l : list A) (y : B),
In y (map f l) <>
exists x, f x = y /\ In x l.
Proof.
Admitted.
forall (A B : Type) (f : A > B) (l : list A) (y : B),
In y (map f l) <>
exists x, f x = y /\ In x l.
Proof.
Admitted.
☐
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.
Drawing inspiration from In, write a recursive function All
stating that some property P holds of all elements of a list
l. To make sure your definition is correct, prove the All_In
lemma below. (Of course, your definition should not just
restate the lefthand side of All_In.)
Exercise: 3 stars, standard, recommended (All)
Fixpoint All {T : Type} (P : T > Prop) (l : list T) : Prop
. Admitted.
Lemma All_In :
forall T (P : T > Prop) (l : list T),
(forall x, In x l > P x) <>
All P l.
Proof.
Admitted.
☐
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.
Exercise: 3 stars, standard (combine_odd_even)
To test your definition, prove the following facts:
Theorem combine_odd_even_intro :
forall (Podd Peven : nat > Prop) (n : nat),
(oddb n = true > Podd n) >
(oddb n = false > Peven n) >
combine_odd_even Podd Peven n.
Proof.
Admitted.
Theorem combine_odd_even_elim_odd :
forall (Podd Peven : nat > Prop) (n : nat),
combine_odd_even Podd Peven n >
oddb n = true >
Podd n.
Proof.
Admitted.
Theorem combine_odd_even_elim_even :
forall (Podd Peven : nat > Prop) (n : nat),
combine_odd_even Podd Peven n >
oddb n = false >
Peven n.
Proof.
Admitted.
☐
Applying Theorems to Arguments
Coq prints the 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?
The reason is that the identifier plus_comm actually refers to a
proof object  a data structure that represents a logical
derivation establishing of the truth of the statement forall n m
: nat, n + m = m + n. The type of this object is the statement
of the theorem that it is a proof of.
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:
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.
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 :
forall 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 :
forall 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 :
forall 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 :
forall l : list nat, In 42 l > l <> [].
Proof.
intros l H.
Fail apply in_not_nil.
Abort.
Lemma in_not_nil_42_take2 :
forall l : list nat, In 42 l > l <> [].
Proof.
intros l H.
apply in_not_nil with (x := 42).
apply H.
Qed.
Lemma in_not_nil_42_take3 :
forall l : list nat, In 42 l > l <> [].
Proof.
intros l H.
apply in_not_nil in H.
apply H.
Qed.
Lemma in_not_nil_42_take4 :
forall l : list nat, In 42 l > l <> [].
Proof.
intros l H.
apply (in_not_nil nat 42).
apply H.
Qed.
Lemma in_not_nil_42_take5 :
forall 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 :
forall {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
Functional Extensionality
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 outputs:
(forall x, f x = g x) > f = g
This is known as the principle of functional extensionality.
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 builtin logic.
This means that some "reasonable" propositions are not provable.
However, we can add functional extensionality to Coq's core using
the Axiom command.
Axiom functional_extensionality : forall {X Y: Type}
{f g : X > Y},
(forall (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_ex2 :
(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 inconsistent  that is, they may
make it 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 highlytrained 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, is consistent.
To check whether a particular proof relies on any additional
axioms, use the Print Assumptions command.
Exercise: 4 stars, standard (tr_rev_correct)
Fixpoint rev_append {X} (l1 l2 : list X) : list X :=
match l1 with
 [] => l2
 x :: l1' => rev_append l1' (x :: l2)
end.
Definition tr_rev {X} (l : list X) : list X :=
rev_append l [].
This version is said to be tailrecursive, 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.
☐
Propositions and Booleans
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 : forall k, evenb (double k) = true.
Proof.
intros k. induction k as [k' IHk'].
 reflexivity.
 simpl. apply IHk'.
Qed.
Proof.
intros k. induction k as [k' IHk'].
 reflexivity.
 simpl. apply IHk'.
Qed.
Theorem evenb_double_conv : forall n,
exists k, n = if evenb n then double k
else S (double k).
Proof.
Admitted.
exists k, n = if evenb n then double k
else S (double k).
Proof.
Admitted.
☐
Theorem even_bool_prop : forall n,
evenb n = true <> exists k, n = double k.
Proof.
intros n. split.
 intros H. destruct (evenb_double_conv n) as [k Hk].
rewrite Hk. rewrite H. exists 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 exists k,
n = double k.
Similarly, to state that two numbers n and m are equal, we can
say either
Again, these two notions are equivalent.
Theorem eqb_eq : forall n1 n2 : nat,
n1 =? n2 = true <> n1 = n2.
Proof.
intros n1 n2. 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 operationally.
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) > (exists 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:
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
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 noncomputable functions.
Although general noncomputable properties cannot be phrased as
boolean computations, it is worth noting that even many
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.
Conversely, 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:
The most direct proof of this fact is to give the value of k
explicitly.
Proof. exists 500. reflexivity. Qed.
On the other hand, the proof of the corresponding boolean
statement is even simpler:
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:
Although we haven't gained much in terms of proofscript
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 4color theorem uses reflection to reduce
the analysis of hundreds of different cases to a boolean
computation.
Another notable difference is that the negation of a "boolean
fact" is straightforward to state and prove: simply flip the
expected boolean result.
In contrast, propositional negation may be more difficult
to grasp.
Example not_even_1001' : ~(exists k, 1001 = double k).
Proof.
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 : forall n m p : nat,
n =? m = true > n + p =? m + p = true.
Proof.
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.
The following lemmas relate the propositional connectives studied
in this chapter to the corresponding boolean operations.
Exercise: 2 stars, standard (logical_connectives)
Lemma andb_true_iff : forall b1 b2:bool,
b1 && b2 = true <> b1 = true /\ b2 = true.
Proof.
Admitted.
Lemma orb_true_iff : forall b1 b2,
b1  b2 = true <> b1 = true \/ b2 = true.
Proof.
Admitted.
☐
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).
Exercise: 1 star, standard (eqb_neq)
☐
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.
Exercise: 3 stars, standard (eqb_list)
Fixpoint eqb_list {A : Type} (eqb : A > A > bool)
(l1 l2 : list A) : bool
. Admitted.
Lemma eqb_list_true_iff :
forall A (eqb : A > A > bool),
(forall a1 a2, eqb a1 a2 = true <> a1 = a2) >
forall l1 l2, eqb_list eqb l1 l2 = true <> l1 = l2.
Proof.
Admitted.
☐
Recall the function forallb, from the exercise
forall_exists_challenge in chapter Tactics:
Exercise: 2 stars, standard, recommended (All_forallb)
Fixpoint forallb {X : Type} (test : X > bool) (l : list X) : bool :=
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 : forall X test (l : list X),
forallb test l = true <> All (fun x => test x = true) l.
Proof.
Admitted.
Are there any important properties of the function forallb which
are not captured by this specification?
Classical vs. Constructive Logic
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 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.
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.
Theorem restricted_excluded_middle : forall P b,
(P <> b = true) > P \/ ~ P.
Proof.
intros P [] H.
 left. rewrite H. reflexivity.
 right. rewrite H. intros contra. discriminate contra.
Qed.
Theorem restricted_excluded_middle_eq : forall (n m : nat),
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 exists 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 constructive.
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.
The following example illustrates why assuming the excluded middle
may lead to nonconstructive proofs:
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. ☐
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
nonconstructive 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, ~ exists x, P x. From this premise, it is not
hard to derive forall 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 exists x, P x using a proof of
~ ~ (exists 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.
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.
Exercise: 3 stars, standard (excluded_middle_irrefutable)
☐
It is a theorem of classical logic that the following two
assertions are equivalent:
~ (exists x, ~ P x)
forall 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.
Exercise: 3 stars, advanced (not_exists_dist)
Theorem not_exists_dist :
excluded_middle >
forall (X:Type) (P : X > Prop),
~ (exists x, ~ P x) > (forall x, P x).
Proof.
Admitted.
☐
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.
Prove that all five propositions (these four plus
excluded_middle) are equivalent.