(** * Logic: Logic in Coq *)
Require Export MoreProp.
(** Coq's built-in logic is very small: the only primitives are
[Inductive] definitions, universal quantification ([forall]), 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 *)
(** The logical conjunction of propositions [P] and [Q] can be
represented using an [Inductive] definition with one
constructor. *)
Inductive and (P Q : Prop) : Prop :=
conj : P -> Q -> (and P Q).
(** Note that, like the definition of [ev] in a previous
chapter, this definition is parameterized; however, in this case,
the parameters are themselves propositions, rather than numbers. *)
(** The intuition behind this definition is simple: to
construct evidence for [and P Q], we must provide evidence
for [P] and evidence for [Q]. More precisely:
- [conj p q] can be taken as evidence for [and P Q] if [p]
is evidence for [P] and [q] is evidence for [Q]; and
- this is the _only_ way to give evidence for [and P Q] --
that is, if someone gives us evidence for [and P Q], we
know it must have the form [conj p q], where [p] is
evidence for [P] and [q] is evidence for [Q].
Since we'll be using conjunction a lot, let's introduce a more
familiar-looking infix notation for it. *)
Notation "P /\ Q" := (and P Q) : type_scope.
(** (The [type_scope] annotation tells Coq that this notation
will be appearing in propositions, not values.) *)
(** Consider the "type" of the constructor [conj]: *)
Check conj.
(* ===> forall P Q : Prop, P -> Q -> P /\ Q *)
(** Notice that it takes 4 inputs -- namely the propositions [P]
and [Q] and evidence for [P] and [Q] -- and returns as output the
evidence of [P /\ Q]. *)
(** Besides the elegance of building everything up from a tiny
foundation, what's nice about defining conjunction this way is
that we can prove statements involving conjunction using the
tactics that we already know. For example, if the goal statement
is a conjuction, we can prove it by applying the single
constructor [conj], which (as can be seen from the type of [conj])
solves the current goal and leaves the two parts of the
conjunction as subgoals to be proved separately. *)
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 : forall P Q : Prop,
P /\ Q -> P.
Proof.
intros P Q H.
inversion H as [HP HQ].
apply HP. Qed.
(** **** Exercise: 1 star, optional (proj2) *)
Theorem proj2 : forall P Q : Prop,
P /\ Q -> Q.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
Theorem and_commut : forall P Q : Prop,
P /\ Q -> Q /\ P.
Proof.
(* WORKED IN CLASS *)
intros P Q H.
inversion H as [HP HQ].
split.
Case "left". apply HQ.
Case "right". apply HP. Qed.
(** **** Exercise: 2 stars (and_assoc) *)
(** In the following proof, notice how the _nested pattern_ in the
[inversion] breaks the hypothesis [H : P /\ (Q /\ R)] down into
[HP: P], [HQ : Q], and [HR : R]. Finish the proof from there: *)
Theorem and_assoc : forall P Q R : Prop,
P /\ (Q /\ R) -> (P /\ Q) /\ R.
Proof.
intros P Q R H.
inversion H as [HP [HQ HR]].
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars (even__ev) *)
(** Now we can prove the other direction of the equivalence of [even]
and [ev], which we left hanging in chapter [Prop]. Notice that the
left-hand conjunct here is the statement we are actually interested
in; the right-hand conjunct is needed in order to make the
induction hypothesis strong enough that we can carry out the
reasoning in the inductive step. (To see why this is needed, try
proving the left conjunct by itself and observe where things get
stuck.) *)
Theorem even__ev : forall n : nat,
(even n -> ev n) /\ (even (S n) -> ev (S n)).
Proof.
(* Hint: Use induction on [n]. *)
(* FILL IN HERE *) 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_implies : forall P Q : Prop,
(P <-> Q) -> P -> Q.
Proof.
intros P Q H.
inversion H as [HAB HBA]. apply HAB. Qed.
Theorem iff_sym : forall P Q : Prop,
(P <-> Q) -> (Q <-> P).
Proof.
(* WORKED IN CLASS *)
intros P Q H.
inversion H as [HAB HBA].
split.
Case "->". apply HBA.
Case "<-". apply HAB. Qed.
(** **** Exercise: 1 star, 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 : forall P : Prop,
P <-> P.
Proof.
(* FILL IN HERE *) Admitted.
Theorem iff_trans : forall P Q R : Prop,
(P <-> Q) -> (Q <-> R) -> (P <-> R).
Proof.
(* FILL IN HERE *) Admitted.
(** Hint: If you have an iff hypothesis in the context, you can use
[inversion] to break it into two separate implications. (Think
about why this works.) *)
(** [] *)
(** Some of Coq's tactics treat [iff] statements specially, thus
avoiding the need for some low-level manipulation when reasoning
with them. In particular, [rewrite] can be used with [iff]
statements, not just equalities. *)
(* ############################################################ *)
(** * Disjunction *)
(** Disjunction ("logical or") can also be defined as an
inductive proposition. *)
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.
(** Consider the "type" of the constructor [or_introl]: *)
Check or_introl.
(* ===> forall P Q : Prop, P -> P \/ Q *)
(** It takes 3 inputs, namely the propositions [P], [Q] and
evidence of [P], and returns, as output, the evidence of [P \/ Q].
Next, look at the type of [or_intror]: *)
Check or_intror.
(* ===> forall P Q : Prop, Q -> P \/ Q *)
(** It is like [or_introl] but it requires evidence of [Q]
instead of evidence of [P]. *)
(** 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 : forall 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' : forall 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.
Theorem or_distributes_over_and_1 : forall P Q R : Prop,
P \/ (Q /\ R) -> (P \/ Q) /\ (P \/ R).
Proof.
intros P Q R. intros H. inversion H as [HP | [HQ HR]].
Case "left". split.
SCase "left". left. apply HP.
SCase "right". left. apply HP.
Case "right". split.
SCase "left". right. apply HQ.
SCase "right". right. apply HR. Qed.
(** **** Exercise: 2 stars (or_distributes_over_and_2) *)
Theorem or_distributes_over_and_2 : forall P Q R : Prop,
(P \/ Q) /\ (P \/ R) -> P \/ (Q /\ R).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, optional (or_distributes_over_and) *)
Theorem or_distributes_over_and : forall P Q R : Prop,
P \/ (Q /\ R) <-> (P \/ Q) /\ (P \/ R).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ################################################### *)
(** ** Relating [/\] and [\/] with [andb] and [orb] (advanced) *)
(** We've already seen several places where analogous structures
can be found in Coq's computational ([Type]) and logical ([Prop])
worlds. Here is one more: the boolean operators [andb] and [orb]
are clearly analogs of the logical connectives [/\] and [\/].
This analogy can be made more precise by the following theorems,
which show how to translate knowledge about [andb] and [orb]'s
behaviors on certain inputs into propositional facts about those
inputs. *)
Theorem andb_prop : forall b c,
andb b c = true -> b = true /\ c = true.
Proof.
(* WORKED IN CLASS *)
intros b c H.
destruct b.
Case "b = true". destruct c.
SCase "c = true". apply conj. reflexivity. reflexivity.
SCase "c = false". inversion H.
Case "b = false". inversion H. Qed.
Theorem andb_true_intro : forall b c,
b = true /\ c = true -> andb b c = true.
Proof.
(* WORKED IN CLASS *)
intros b c H.
inversion H.
rewrite H0. rewrite H1. reflexivity. Qed.
(** **** Exercise: 2 stars, optional (bool_prop) *)
Theorem andb_false : forall b c,
andb b c = false -> b = false \/ c = false.
Proof.
(* FILL IN HERE *) Admitted.
Theorem orb_prop : forall b c,
orb b c = true -> b = true \/ c = true.
Proof.
(* FILL IN HERE *) Admitted.
Theorem orb_false_elim : forall b c,
orb b c = false -> b = false /\ c = false.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ################################################### *)
(** * 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. *)
(** Since [False] has no constructors, inverting an assumption
of type [False] always yields zero subgoals, allowing us to
immediately prove any goal. *)
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.
(** Actually, since the proof of [False_implies_nonsense]
doesn't actually have anything to do with the specific nonsensical
thing being proved; it can easily be generalized to work for an
arbitrary [P]: *)
Theorem ex_falso_quodlibet : forall (P:Prop),
False -> P.
Proof.
(* WORKED IN CLASS *)
intros P contra.
inversion contra. Qed.
(** The Latin _ex falso quodlibet_ means, literally, "from
falsehood follows whatever you please." This theorem is also
known as the _principle of explosion_. *)
(* #################################################### *)
(** ** Truth *)
(** Since we have defined falsehood in Coq, one might wonder whether
it is possible to define truth in the same way. We can. *)
(** **** Exercise: 2 stars, advanced (True) *)
(** Define [True] as another inductively defined proposition. (The
intution is that [True] should be a proposition for which it is
trivial to give evidence.) *)
(* FILL IN HERE *)
(** [] *)
(** However, unlike [False], which we'll use extensively, [True] is
used fairly rarely. By itself, it is trivial (and therefore
uninteresting) to prove as a goal, and it carries no useful
information as a hypothesis. But it can be useful when defining
complex [Prop]s using conditionals, or as a parameter to
higher-order [Prop]s. *)
(* #################################################### *)
(** * Negation *)
(** The logical complement of a proposition [P] is written [not
P] or, for shorthand, [~P]: *)
Definition not (P:Prop) := P -> False.
(** The intuition is that, if [P] is not true, then anything at
all (even [False]) follows from assuming [P]. *)
Notation "~ x" := (not x) : type_scope.
Check not.
(* ===> Prop -> Prop *)
(** It takes a little practice to get used to working with
negation in Coq. Even though you can see perfectly well why
something is true, it can be a little hard at first to get things
into the right configuration so that Coq can see it! Here are
proofs of a few familiar facts about negation to get you warmed
up. *)
Theorem not_False :
~ False.
Proof.
unfold not. intros H. inversion H. Qed.
Theorem contradiction_implies_anything : forall P Q : Prop,
(P /\ ~P) -> Q.
Proof.
(* WORKED IN CLASS *)
intros P Q H. inversion H as [HP HNA]. unfold not in HNA.
apply HNA in HP. inversion HP. Qed.
Theorem double_neg : forall 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].
_Proof_:
(* FILL IN HERE *)
[]
*)
(** **** Exercise: 2 stars (contrapositive) *)
Theorem contrapositive : forall P Q : Prop,
(P -> Q) -> (~Q -> ~P).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star (not_both_true_and_false) *)
Theorem not_both_true_and_false : forall P : Prop,
~ (P /\ ~P).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 1 star, advanced (informal_not_PNP) *)
(** Write an informal proof (in English) of the proposition [forall P
: Prop, ~(P /\ ~P)]. *)
(* FILL IN HERE *)
(** [] *)
Theorem five_not_even :
~ ev 5.
Proof.
(* WORKED IN CLASS *)
unfold not. intros Hev5. inversion Hev5 as [|n Hev3 Heqn].
inversion Hev3 as [|n' Hev1 Heqn']. inversion Hev1. Qed.
(** **** Exercise: 1 star (ev_not_ev_S) *)
(** Theorem [five_not_even] confirms the unsurprising fact that five
is not an even number. Prove this more interesting fact: *)
Theorem ev_not_ev_S : forall n,
ev n -> ~ ev (S n).
Proof.
unfold not. intros n H. induction H. (* not n! *)
(* FILL IN HERE *) 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 : forall P : Prop,
~~P -> P.
Proof.
(* WORKED IN CLASS *)
intros P H. unfold not in H.
(* But now what? There is no way to "invent" evidence for [~P]
from evidence for [P]. *)
Abort.
(** **** Exercise: 5 stars, advanced, optional (classical_axioms) *)
(** For those who like a challenge, here is an exercise
taken from the Coq'Art book (p. 123). The following five
statements are often considered as characterizations of
classical logic (as opposed to constructive logic, which is
what is "built in" to Coq). We can't prove them in Coq, but
we can consistently add any one of them as an unproven axiom
if we wish to work in classical logic. Prove that these five
propositions are equivalent. *)
Definition peirce := forall P Q: Prop,
((P->Q)->P)->P.
Definition classic := forall P:Prop,
~~P -> P.
Definition excluded_middle := forall P:Prop,
P \/ ~P.
Definition de_morgan_not_and_not := forall P Q:Prop,
~(~P /\ ~Q) -> P\/Q.
Definition implies_to_or := forall P Q:Prop,
(P->Q) -> (~P\/Q).
(* FILL IN HERE *)
(** [] *)
(* ########################################################## *)
(** ** Inequality *)
(** Saying [x <> y] is just the same as saying [~(x = y)]. *)
Notation "x <> y" := (~ (x = y)) : type_scope.
(** Since inequality involves a negation, it again requires
a little practice to be able to work with it fluently. Here
is one very useful trick. If you are trying to prove a goal
that is nonsensical (e.g., the goal state is [false = true]),
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 -- in particular,
assumptions of the form [x<>y]. *)
Theorem not_false_then_true : forall b : bool,
b <> false -> 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.
(** **** Exercise: 2 stars (false_beq_nat) *)
Theorem false_beq_nat : forall n m : nat,
n <> m ->
beq_nat n m = false.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars, optional (beq_nat_false) *)
Theorem beq_nat_false : forall n m,
beq_nat n m = false -> n <> m.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars, optional (ble_nat_false) *)
Theorem ble_nat_false : forall n m,
ble_nat n m = false -> ~(n <= m).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ############################################################ *)
(** * Existential Quantification *)
(** Another critical logical connective is _existential
quantification_. We can express it with the following
definition: *)
Inductive ex (X:Type) (P : X->Prop) : Prop :=
ex_intro : forall (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].
*)
(** Coq's [Notation] facility can be used to introduce more
familiar notation for writing existentially quantified
propositions, exactly parallel to the built-in syntax for
universally quantified propositions. Instead of writing [ex nat
ev] to express the proposition that there exists some number that
is even, for example, we can write [exists x:nat, ev x]. (It is
not necessary to understand exactly how the [Notation] definition
works.) *)
Notation "'exists' x , p" := (ex _ (fun x => p))
(at level 200, x ident, right associativity) : type_scope.
Notation "'exists' x : X , p" := (ex _ (fun x:X => p))
(at level 200, x ident, right associativity) : type_scope.
(** We can use the usual set of tactics for
manipulating existentials. For example, to prove an
existential, we can [apply] the constructor [ex_intro]. Since the
premise of [ex_intro] involves a variable ([witness]) that does
not appear in its conclusion, we need to explicitly give its value
when we use [apply]. *)
Example exists_example_1 : exists 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 [exists e], which means
the same thing. *)
Example exists_example_1' : exists n, n + (n * n) = 6.
Proof.
exists 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 : forall n,
(exists m, n = 4 + m) ->
(exists o, n = 2 + o).
Proof.
intros n H.
inversion H as [m Hm].
exists (2 + m).
apply Hm. Qed.
(** **** Exercise: 1 star, optional (english_exists) *)
(** In English, what does the proposition
ex nat (fun n => beautiful (S n))
]]
mean? *)
(* FILL IN HERE *)
(** **** Exercise: 1 star (dist_not_exists) *)
(** Prove that "[P] holds for all [x]" implies "there is no [x] for
which [P] does not hold." *)
Theorem dist_not_exists : forall (X:Type) (P : X -> Prop),
(forall x, P x) -> ~ (exists x, ~ P x).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 3 stars, optional (not_exists_dist) *)
(** (The other direction of this theorem requires the classical "law
of the excluded middle".) *)
Theorem not_exists_dist :
excluded_middle ->
forall (X:Type) (P : X -> Prop),
~ (exists x, ~ P x) -> (forall x, P x).
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** **** Exercise: 2 stars (dist_exists_or) *)
(** Prove that existential quantification distributes over
disjunction. *)
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.
(* FILL IN HERE *) Admitted.
(** [] *)
(* Print dist_exists_or. *)
(* ###################################################### *)
(** * Equality *)
(** Even Coq's equality relation is not built in. It has (roughly)
the following inductive definition. *)
(* (We enclose the definition in a module to avoid confusion with the
standard library equality, which we have used extensively
already.) *)
Module MyEquality.
Inductive eq {X:Type} : X -> X -> Prop :=
refl_equal : forall 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]. *)
(** **** Exercise: 2 stars (leibniz_equality) *)
(** The inductive definitions of equality corresponds to _Leibniz equality_:
what we mean when we say "[x] and [y] are equal" is that every
property on [P] that is true of [x] is also true of [y]. *)
Lemma leibniz_equality : forall (X : Type) (x y: X),
x = y -> forall P : X -> Prop, P x -> P y.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(** 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 [match]es.
*)
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]. *)
End MyEquality.
(* ###################################################### *)
(** * 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 [nat]s *)
Theorem eq_nat_dec : forall n m : nat, {n = m} + {n <> m}.
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 [nat]s 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: nat->X) (k:nat) (x:X) : nat->X:=
fun (k':nat) => if eq_nat_dec k k' then x else f k'.
Theorem override_same' : forall (X:Type) x1 k1 k2 (f : nat->X),
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. *)
(** **** Exercise: 1 star (override_shadow') *)
Theorem override_shadow' : forall (X:Type) x1 x2 k1 k2 (f : nat->X),
(override' (override' f k1 x2) k1 x1) k2 = (override' f k1 x1) k2.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ####################################################### *)
(** ** Inversion, Again (Advanced) *)
(** We've seen [inversion] used with both equality hypotheses and
hypotheses about inductively defined propositions. Now that we've
seen that these are actually the same thing, we're in a position
to take a closer look at how [inversion] behaves...
In general, the [inversion] tactic
- takes a hypothesis [H] whose type [P] is inductively defined,
and
- for each constructor [C] in [P]'s definition,
- generates a new subgoal in which we assume [H] was
built with [C],
- adds the arguments (premises) of [C] to the context of
the subgoal as extra hypotheses,
- matches the conclusion (result type) of [C] against the
current goal and calculates a set of equalities that must
hold in order for [C] to be applicable,
- adds these equalities to the context (and, for convenience,
rewrites them in the goal), and
- if the equalities are not satisfiable (e.g., they involve
things like [S n = O]), immediately solves the subgoal. *)
(** _Example_: If we invert a hypothesis built with [or], there are two
constructors, so two subgoals get generated. The
conclusion (result type) of the constructor ([P \/ Q]) doesn't
place any restrictions on the form of [P] or [Q], so we don't get
any extra equalities in the context of the subgoal.
_Example_: If we invert a hypothesis built with [and], there is
only one constructor, so only one subgoal gets generated. Again,
the conclusion (result type) of the constructor ([P /\ Q]) doesn't
place any restrictions on the form of [P] or [Q], so we don't get
any extra equalities in the context of the subgoal. The
constructor does have two arguments, though, and these can be seen
in the context in the subgoal.
_Example_: If we invert a hypothesis built with [eq], there is
again only one constructor, so only one subgoal gets generated.
Now, though, the form of the [refl_equal] constructor does give us
some extra information: it tells us that the two arguments to [eq]
must be the same! The [inversion] tactic adds this fact to the
context. *)
(** **** Exercise: 1 star, optional (dist_and_or_eq_implies_and) *)
Lemma dist_and_or_eq_implies_and : forall P Q R,
P /\ (Q \/ R) /\ Q = R -> P/\Q.
Proof.
(* FILL IN HERE *) Admitted.
(** [] *)
(* ####################################################### *)
(** * Additional Exercises *)
(** **** Exercise: 3 stars (all_forallb) *)
(** Inductively define a property [all] of lists, parameterized by a
type [X] and a property [P : X -> Prop], such that [all X P l]
asserts that [P] is true for every element of the list [l]. *)
Inductive all (X : Type) (P : X -> Prop) : list X -> Prop :=
(* FILL IN HERE *)
.
(** Recall the function [forallb], from the exercise
[forall_exists_challenge] in chapter [Poly]: *)
Fixpoint forallb {X : Type} (test : X -> bool) (l : list X) : bool :=
match l with
| [] => true
| x :: l' => andb (test x) (forallb test l')
end.
(** Using the property [all], write down a specification for [forallb],
and prove that it satisfies the specification. Try to make your
specification as precise as possible.
Are there any important properties of the function [forallb] which
are not captured by your specification? *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 4 stars, advanced (filter_challenge) *)
(** One of the main purposes of Coq is to prove that programs match
their specifications. To this end, let's prove that our
definition of [filter] matches a specification. Here is the
specification, written out informally in English.
Suppose we have a set [X], a function [test: X->bool], and a list
[l] of type [list X]. Suppose further that [l] is an "in-order
merge" of two lists, [l1] and [l2], such that every item in [l1]
satisfies [test] and no item in [l2] satisfies test. Then [filter
test l = l1].
A list [l] is an "in-order merge" of [l1] and [l2] if it contains
all the same elements as [l1] and [l2], in the same order as [l1]
and [l2], but possibly interleaved. For example,
[1,4,6,2,3]
is an in-order merge of
[1,6,2]
and
[4,3].
Your job is to translate this specification into a Coq theorem and
prove it. (Hint: You'll need to begin by defining what it means
for one list to be a merge of two others. Do this with an
inductive relation, not a [Fixpoint].) *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 5 stars, advanced, optional (filter_challenge_2) *)
(** A different way to formally characterize the behavior of [filter]
goes like this: Among all subsequences of [l] with the property
that [test] evaluates to [true] on all their members, [filter test
l] is the longest. Express this claim formally and prove it. *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 4 stars, advanced (no_repeats) *)
(** The following inductively defined proposition... *)
Inductive appears_in {X:Type} (a:X) : list X -> Prop :=
| ai_here : forall l, appears_in a (a::l)
| ai_later : forall b l, appears_in a l -> appears_in a (b::l).
(** ...gives us a precise way of saying that a value [a] appears at
least once as a member of a list [l].
Here's a pair of warm-ups about [appears_in].
*)
Lemma appears_in_app : forall (X:Type) (xs ys : list X) (x:X),
appears_in x (xs ++ ys) -> appears_in x xs \/ appears_in x ys.
Proof.
(* FILL IN HERE *) Admitted.
Lemma app_appears_in : forall (X:Type) (xs ys : list X) (x:X),
appears_in x xs \/ appears_in x ys -> appears_in x (xs ++ ys).
Proof.
(* FILL IN HERE *) Admitted.
(** Now use [appears_in] to define a proposition [disjoint X l1 l2],
which should be provable exactly when [l1] and [l2] are
lists (with elements of type X) that have no elements in common. *)
(* FILL IN HERE *)
(** Next, use [appears_in] to define an inductive proposition
[no_repeats X l], which should be provable exactly when [l] is a
list (with elements of type [X]) where every member is different
from every other. For example, [no_repeats nat [1,2,3,4]] and
[no_repeats bool []] should be provable, while [no_repeats nat
[1,2,1]] and [no_repeats bool [true,true]] should not be. *)
(* FILL IN HERE *)
(** Finally, state and prove one or more interesting theorems relating
[disjoint], [no_repeats] and [++] (list append). *)
(* FILL IN HERE *)
(** [] *)
(** **** Exercise: 3 stars (nostutter) *)
(** Formulating inductive definitions of predicates is an important
skill you'll need in this course. Try to solve this exercise
without any help at all (except from your study group partner, if
you have one).
We say that a list of numbers "stutters" if it repeats the same
number consecutively. The predicate "[nostutter mylist]" means
that [mylist] does not stutter. Formulate an inductive definition
for [nostutter]. (This is different from the [no_repeats]
predicate in the exercise above; the sequence [1,4,1] repeats but
does not stutter.) *)
Inductive nostutter: list nat -> Prop :=
(* FILL IN HERE *)
.
(** Make sure each of these tests succeeds, but you are free
to change the proof if the given one doesn't work for you.
Your definition might be different from mine and still correct,
in which case the examples might need a different proof.
The suggested proofs for the examples (in comments) use a number
of tactics we haven't talked about, to try to make them robust
with respect to different possible ways of defining [nostutter].
You should be able to just uncomment and use them as-is, but if
you prefer you can also prove each example with more basic
tactics. *)
Example test_nostutter_1: nostutter [3;1;4;1;5;6].
(* FILL IN HERE *) Admitted.
(*
Proof. repeat constructor; apply beq_nat_false; auto. Qed.
*)
Example test_nostutter_2: nostutter [].
(* FILL IN HERE *) Admitted.
(*
Proof. repeat constructor; apply beq_nat_false; auto. Qed.
*)
Example test_nostutter_3: nostutter [5].
(* FILL IN HERE *) Admitted.
(*
Proof. repeat constructor; apply beq_nat_false; auto. Qed.
*)
Example test_nostutter_4: not (nostutter [3;1;1;4]).
(* FILL IN HERE *) Admitted.
(*
Proof. intro.
repeat match goal with
h: nostutter _ |- _ => inversion h; clear h; subst
end.
contradiction H1; auto. Qed.
*)
(** [] *)
(** **** Exercise: 4 stars, advanced (pigeonhole principle) *)
(** The "pigeonhole principle" states a basic fact about counting:
if you distribute more than [n] items into [n] pigeonholes, some
pigeonhole must contain at least two items. As is often the case,
this apparently trivial fact about numbers requires non-trivial
machinery to prove, but we now have enough... *)
(** First a pair of useful lemmas (we already proved these for lists
of naturals, but not for arbitrary lists). *)
Lemma app_length : forall (X:Type) (l1 l2 : list X),
length (l1 ++ l2) = length l1 + length l2.
Proof.
(* FILL IN HERE *) Admitted.
Lemma appears_in_app_split : forall (X:Type) (x:X) (l:list X),
appears_in x l ->
exists l1, exists l2, l = l1 ++ (x::l2).
Proof.
(* FILL IN HERE *) Admitted.
(** Now define a predicate [repeats] (analogous to [no_repeats] in the
exercise above), such that [repeats X l] asserts that [l] contains
at least one repeated element (of type [X]). *)
Inductive repeats {X:Type} : list X -> Prop :=
(* FILL IN HERE *)
.
(** Now here's a way to formalize the pigeonhole principle. List [l2]
represents a list of pigeonhole labels, and list [l1] represents an
assignment of items to labels: if there are more items than labels,
at least two items must have the same label. You will almost
certainly need to use the [excluded_middle] hypothesis. *)
Theorem pigeonhole_principle: forall (X:Type) (l1 l2:list X),
excluded_middle ->
(forall x, appears_in x l1 -> appears_in x l2) ->
length l2 < length l1 ->
repeats l1.
Proof. intros X l1. induction l1.
(* FILL IN HERE *) Admitted.
(** [] *)
(* $Date: 2013-07-17 16:19:11 -0400 (Wed, 17 Jul 2013) $ *)