MorePropMore about Propositions and Evidence

Require Export "Prop".


A proposition parameterized by a number (such as ev or beautiful) can be thought of as a property — i.e., it defines a subset of nat, namely those numbers for which the proposition is provable. In the same way, a two-argument proposition can be thought of as a relation — i.e., it defines a set of pairs for which the proposition is provable.

One useful example is the "less than or equal to" relation on numbers.
The following definition should be fairly intuitive. It says that there are two ways to give evidence that one number is less than or equal to another: either observe that they are the same number, or give evidence that the first is less than or equal to the predecessor of the second.

Inductive le : nat nat Prop :=
  | le_n : n, le n n
  | le_S : n m, (le n m) (le n (S m)).

Notation "m <= n" := (le m n).

Proofs of facts about <= using the constructors le_n and le_S follow the same patterns as proofs about properties, like ev in chapter Prop. We can apply the constructors to prove <= goals (e.g., to show that 3<=3 or 3<=6), and we can use tactics like inversion to extract information from <= hypotheses in the context (e.g., to prove that (2 <= 1) 2+2=5.)
Here are some sanity checks on the definition. (Notice that, although these are the same kind of simple "unit tests" as we gave for the testing functions we wrote in the first few lectures, we must construct their proofs explicitly — simpl and reflexivity don't do the job, because the proofs aren't just a matter of simplifying computations.)

Theorem test_le1 :
  3 <= 3.
  apply le_n. Qed.

Theorem test_le2 :
  3 <= 6.
  apply le_S. apply le_S. apply le_S. apply le_n. Qed.

Theorem test_le3 :
  (2 <= 1) 2 + 2 = 5.
  intros H. inversion H. inversion H2. Qed.

The "strictly less than" relation n < m can now be defined in terms of le.

Definition lt (n m:nat) := le (S n) m.

Notation "m < n" := (lt m n).

Here are a few more simple relations on numbers:

Inductive square_of : nat nat Prop :=
  sq : n:nat, square_of n (n * n).

Inductive next_nat (n:nat) : nat Prop :=
  | nn : next_nat n (S n).

Inductive next_even (n:nat) : nat Prop :=
  | ne_1 : ev (S n) next_even n (S n)
  | ne_2 : ev (S (S n)) next_even n (S (S n)).

Exercise: 2 stars (total_relation)

Define an inductive binary relation total_relation that holds between every pair of natural numbers.


Exercise: 2 stars (empty_relation)

Define an inductive binary relation empty_relation (on numbers) that never holds.


Exercise: 2 stars, optional (le_exercises)

Here are a number of facts about the <= and < relations that we are going to need later in the course. The proofs make good practice exercises.

Lemma le_trans : m n o, m <= n n <= o m <= o.
  (* FILL IN HERE *) Admitted.

Theorem O_le_n : n,
  0 <= n.
  (* FILL IN HERE *) Admitted.

Theorem n_le_m__Sn_le_Sm : n m,
  n <= m S n <= S m.
  (* FILL IN HERE *) Admitted.

Theorem Sn_le_Sm__n_le_m : n m,
  S n <= S m n <= m.
  (* FILL IN HERE *) Admitted.

Theorem le_plus_l : a b,
  a <= a + b.
  (* FILL IN HERE *) Admitted.

Theorem plus_lt : n1 n2 m,
  n1 + n2 < m
  n1 < m n2 < m.
 unfold lt.
 (* FILL IN HERE *) Admitted.

Theorem lt_S : n m,
  n < m
  n < S m.
  (* FILL IN HERE *) Admitted.

Theorem ble_nat_true : n m,
  ble_nat n m = true n <= m.
  (* FILL IN HERE *) Admitted.

Theorem le_ble_nat : n m,
  n <= m
  ble_nat n m = true.
  (* Hint: This may be easiest to prove by induction on m. *)
  (* FILL IN HERE *) Admitted.

Theorem ble_nat_true_trans : n m o,
  ble_nat n m = true ble_nat m o = true ble_nat n o = true.
  (* Hint: This theorem can be easily proved without using induction. *)
  (* FILL IN HERE *) Admitted.

Exercise: 3 stars (R_provability)

Module R.
We can define three-place relations, four-place relations, etc., in just the same way as binary relations. For example, consider the following three-place relation on numbers:

Inductive R : nat nat nat Prop :=
   | c1 : R 0 0 0
   | c2 : m n o, R m n o R (S m) n (S o)
   | c3 : m n o, R m n o R m (S n) (S o)
   | c4 : m n o, R (S m) (S n) (S (S o)) R m n o
   | c5 : m n o, R m n o R n m o.

  • Which of the following propositions are provable?
    • R 1 1 2
    • R 2 2 6
  • If we dropped constructor c5 from the definition of R, would the set of provable propositions change? Briefly (1 sentence) explain your answer.
  • If we dropped constructor c4 from the definition of R, would the set of provable propositions change? Briefly (1 sentence) explain your answer.

Exercise: 3 stars, optional (R_fact)

Relation R actually encodes a familiar function. State and prove two theorems that formally connects the relation and the function. That is, if R m n o is true, what can we say about m, n, and o, and vice versa?


End R.

Programming with Propositions

A proposition is a statement expressing a factual claim, like "two plus two equals four." In Coq, propositions are written as expressions of type Prop. Although we haven't discussed this explicitly, we have already seen numerous examples.

Check (2 + 2 = 4).
(* ===> 2 + 2 = 4 : Prop *)

Check (ble_nat 3 2 = false).
(* ===> ble_nat 3 2 = false : Prop *)

Check (beautiful 8).
(* ===> beautiful 8 : Prop *)

Both provable and unprovable claims are perfectly good propositions. Simply being a proposition is one thing; being provable is something else!

Check (2 + 2 = 5).
(* ===> 2 + 2 = 5 : Prop *)

Check (beautiful 4).
(* ===> beautiful 4 : Prop *)

Both 2 + 2 = 4 and 2 + 2 = 5 are legal expressions of type Prop.
We've mainly seen one place that propositions can appear in Coq: in Theorem (and Lemma and Example) declarations.

Theorem plus_2_2_is_4 :
  2 + 2 = 4.
Proof. reflexivity. Qed.

But they can be used in many other ways. For example, we have also seen that we can give a name to a proposition using a Definition, just as we have given names to expressions of other sorts.

Definition plus_fact : Prop := 2 + 2 = 4.
Check plus_fact.
(* ===> plus_fact : Prop *)

We can later use this name in any situation where a proposition is expected — for example, as the claim in a Theorem declaration.

Theorem plus_fact_is_true :
Proof. reflexivity. Qed.

We've seen several ways of constructing propositions.
  • We can define a new proposition primitively using Inductive.
  • Given two expressions e1 and e2 of the same type, we can form the proposition e1 = e2, which states that their values are equal.
  • We can combine propositions using implication and quantification.
We have also seen parameterized propositions, such as even and beautiful.

Check (even 4).
(* ===> even 4 : Prop *)
Check (even 3).
(* ===> even 3 : Prop *)
Check even.
(* ===> even : nat -> Prop *)

The type of even, i.e., natProp, can be pronounced in three equivalent ways: (1) "even is a function from numbers to propositions," (2) "even is a family of propositions, indexed by a number n," or (3) "even is a property of numbers."
Propositions — including parameterized propositions — are first-class citizens in Coq. For example, we can define functions from numbers to propositions...

Definition between (n m o: nat) : Prop :=
  andb (ble_nat n o) (ble_nat o m) = true.

... and then partially apply them:

Definition teen : natProp := between 13 19.

We can even pass propositions — including parameterized propositions — as arguments to functions:

Definition true_for_zero (P:natProp) : Prop :=
  P 0.

Here are two more examples of passing parameterized propositions as arguments to a function.
The first function, true_for_all_numbers, takes a proposition P as argument and builds the proposition that P is true for all natural numbers.

Definition true_for_all_numbers (P:natProp) : Prop :=
  n, P n.

The second, preserved_by_S, takes P and builds the proposition that, if P is true for some natural number n', then it is also true by the successor of n' — i.e. that P is preserved by successor:

Definition preserved_by_S (P:natProp) : Prop :=
  n', P n' P (S n').

Finally, we can put these ingredients together to define a proposition stating that induction is valid for natural numbers:

Definition natural_number_induction_valid : Prop :=
    true_for_zero P
    preserved_by_S P
    true_for_all_numbers P.

Exercise: 3 stars (combine_odd_even)

Complete the definition of the combine_odd_even function below. It takes as arguments two properties of numbers Podd and Peven. As its result, it should return a new property P such that P n is equivalent to Podd n when n is odd, and equivalent to Peven n otherwise.

Definition combine_odd_even (Podd Peven : nat Prop) : nat Prop :=
  (* FILL IN HERE *) admit.

To test your definition, see whether you can prove the following facts:

Theorem combine_odd_even_intro :
  (Podd Peven : nat Prop) (n : nat),
    (oddb n = true Podd n)
    (oddb n = false Peven n)
    combine_odd_even Podd Peven n.
  (* FILL IN HERE *) Admitted.

Theorem combine_odd_even_elim_odd :
  (Podd Peven : nat Prop) (n : nat),
    combine_odd_even Podd Peven n
    oddb n = true
    Podd n.
  (* FILL IN HERE *) Admitted.

Theorem combine_odd_even_elim_even :
  (Podd Peven : nat Prop) (n : nat),
    combine_odd_even Podd Peven n
    oddb n = false
    Peven n.
  (* FILL IN HERE *) Admitted.

One more quick digression, for adventurous souls: if we can define parameterized propositions using Definition, then can we also define them using Fixpoint? Of course we can! However, this kind of "recursive parameterization" doesn't correspond to anything very familiar from everyday mathematics. The following exercise gives a slightly contrived example.

Exercise: 4 stars, optional (true_upto_n__true_everywhere)

Define a recursive function true_upto_n__true_everywhere that makes true_upto_n_example work.

Fixpoint true_upto_n__true_everywhere

Example true_upto_n_example :
    (true_upto_n__true_everywhere 3 (fun n => even n))
  = (even 3 -> even 2 -> even 1 -> forall m : nat, even m).
Proof. reflexivity.  Qed.

(* $Date: 2013-07-17 16:19:11 -0400 (Wed, 17 Jul 2013) $ *)