# MoreLogic

Require Export "Prop".

# Existential Quantification

*existential quantification*. We can express it with the following definition:

Inductive ex (X:Type) (P : X→Prop) : 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.### 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 ∃ 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 : ∃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.

Here is another example of how to work with existentials.

Lemma exists_example_3 :

∃(n:nat), even n ∧ beautiful n.

Proof.

(* WORKED IN CLASS *)

∃8.

split.

unfold even. simpl. reflexivity.

apply b_sum with (n:=3) (m:=5).

apply b_3. apply b_5.

Qed.

∃(n:nat), even n ∧ beautiful n.

Proof.

(* WORKED IN CLASS *)

∃8.

split.

unfold even. simpl. reflexivity.

apply b_sum with (n:=3) (m:=5).

apply b_3. apply b_5.

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 : ∀(X:Type) (P : X → Prop),

(∀x, P x) → ¬ (∃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 →

∀(X:Type) (P : X → Prop),

¬ (∃x, ¬ P x) → (∀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 : ∀(X:Type) (P Q : X → Prop),

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

Proof.

(* FILL IN HERE *) Admitted.

☐

# Evidence-carrying booleans.

*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.

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

Here's how we can define a sumbool for equality on nats

*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.)Theorem eq_nat_dec : ∀n m : nat, {n = m} + {n ≠ m}.

Proof.

(* WORKED IN CLASS *)

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

Here's a simple example illustrating the advantages of the sumbool form.

*transparent*, meaning that its definition is available when Coq tries to do reductions, which is important for the computational interpretation.)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' : ∀(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' : ∀(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.

(override' (override' f k1 x2) k1 x1) k2 = (override' f k1 x1) k2.

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 *)

☐
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,

#### 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.
[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 *)

Inductive appears_in {X:Type} (a:X) : list X → Prop :=

| ai_here : ∀l, appears_in a (a::l)

| ai_later : ∀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 : ∀(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 : ∀(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 *)

☐
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.)

#### 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.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.

*)

☐
First a pair of useful lemmas (we already proved these for lists
of naturals, but not for arbitrary lists).

#### 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...Lemma app_length : ∀(X:Type) (l1 l2 : list X),

length (l1 ++ l2) = length l1 + length l2.

Proof.

(* FILL IN HERE *) Admitted.

Lemma appears_in_app_split : ∀(X:Type) (x:X) (l:list X),

appears_in x l →

∃l1, ∃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
the labels assigned to a list of items: if there are more items
than labels, at least two items must have the same label. This
proof is much easier if you use the excluded_middle hypothesis
to show that appears_in is decidable, i.e. ∀ x
l, (appears_in x l) ∨ ¬ (appears_in x l). However, it is also
possible to make the proof go through

*without*assuming that appears_in is decidable; if you can manage to do this, you will not need the excluded_middle hypothesis.Theorem pigeonhole_principle: ∀(X:Type) (l1 l2:list X),

excluded_middle →

(∀x, appears_in x l1 → appears_in x l2) →

length l2 < length l1 →

repeats l1.

Proof.

intros X l1. induction l1 as [|x l1'].

(* FILL IN HERE *) Admitted.

☐

(* FILL IN HERE *)

(* $Date: 2014-10-14 15:52:05 -0400 (Tue, 14 Oct 2014) $ *)