# PropPropositions and Evidence

Require Export Logic.

# From Boolean Functions to Propositions

*function*evenb that tests a number for evenness, yielding true if so. We can use this function to define the

*proposition*that some number n is even:

Definition even (n:nat) : Prop :=

evenb n = true.

That is, we can define "n is even" to mean "the function evenb
returns true when applied to n."
Note that here we have given a name
to a proposition using a Definition, just as we have
given names to expressions of other sorts. This isn't a fundamentally
new kind of proposition; it is still just an equality.
Another alternative is to define the concept of evenness
directly. Instead of going via the evenb function ("a number is
even if a certain computation yields true"), we can say what the
concept of evenness means by giving two different ways of
presenting

*evidence*that a number is even.# Inductively Defined Propositions

Inductive ev : nat → Prop :=

| ev_0 : ev O

| ev_SS : ∀n:nat, ev n → ev (S (S n)).

This definition says that there are two ways to give
evidence that a number m is even. First, 0 is even, and
ev_0 is evidence for this. Second, if m = S (S n) for some
n and we can give evidence e that n is even, then m is
also even, and ev_SS n e is the evidence.

#### Exercise: 1 star (double_even)

Theorem double_even : ∀n,

ev (double n).

Proof.

(* FILL IN HERE *) Admitted.

☐
We have seen that the proposition "n is even" can be
phrased in two different ways — indirectly, via a boolean testing
function evenb, or directly, by inductively describing what
constitutes evidence for evenness. These two ways of defining
evenness are about equally easy to state and work with. Which we
choose is basically a question of taste.
However, for many other properties of interest, the direct
inductive definition is preferable, since writing a testing
function may be awkward or even impossible.
One such property is beautiful. This is a perfectly sensible
definition of a set of numbers, but we cannot translate its
definition directly into a Coq Fixpoint (or into a recursive
function in any other common programming language). We might be
able to find a clever way of testing this property using a
Fixpoint (indeed, it is not too hard to find one in this case),
but in general this could require arbitrarily deep thinking. In
fact, if the property we are interested in is uncomputable, then
we cannot define it as a Fixpoint no matter how hard we try,
because Coq requires that all Fixpoints correspond to
terminating computations.
On the other hand, writing an inductive definition of what it
means to give evidence for the property beautiful is
straightforward.

### Discussion: Computational vs. Inductive Definitions

#### Exercise: 1 star (ev__even)

Here is a proof that the inductive definition of evenness implies the computational one.Theorem ev__even : ∀n,

ev n → even n.

Proof.

intros n E. induction E as [| n' E'].

Case "E = ev_0".

unfold even. reflexivity.

Case "E = ev_SS n' E'".

unfold even. apply IHE'.

Qed.

Could this proof also be carried out by induction on n instead
of E? If not, why not?

(* FILL IN HERE *)

☐
The induction principle for inductively defined propositions does
not follow quite the same form as that of inductively defined
sets. For now, you can take the intuitive view that induction on
evidence ev n is similar to induction on n, but restricts our
attention to only those numbers for which evidence ev n could be
generated. We'll look at the induction principle of ev in more
depth below, to explain what's really going on.
(* FILL IN HERE *)

☐

#### Exercise: 1 star (l_fails)

The following proof attempt will not succeed.
Theorem l : ∀n,

ev n.

Proof.

intros n. induction n.

Case "O". simpl. apply ev_0.

Case "S".

...

Intuitively, we expect the proof to fail because not every
number is even. However, what exactly causes the proof to fail?
ev n.

Proof.

intros n. induction n.

Case "O". simpl. apply ev_0.

Case "S".

...

☐

#### Exercise: 2 stars (ev_sum)

Here's another exercise requiring induction.Theorem ev_sum : ∀n m,

ev n → ev m → ev (n+m).

Proof.

(* FILL IN HERE *) Admitted.

☐

## Example

- Rule b_0: The number 0 is beautiful.
- Rule b_3: The number 3 is beautiful.
- Rule b_5: The number 5 is beautiful.
- Rule b_sum: If n and m are both beautiful, then so is their sum.

## Inference Rules

We will see many definitions like this one during the rest of the course, and for purposes of informal discussions, it is helpful to have a lightweight notation that makes them easy to read and write.*Inference rules*are one such notation:

(b_0) | |

beautiful 0 |

(b_3) | |

beautiful 3 |

(b_5) | |

beautiful 5 |

beautiful n beautiful m | (b_sum) |

beautiful (n+m) |

### Each of the textual rules above is reformatted here as an inference rule; the intended reading is that, if the

*premises*above the line all hold, then the

*conclusion*below the line follows. For example, the rule b_sum says that, if n and m are both beautiful numbers, then it follows that n+m is beautiful too. If a rule has no premises above the line, then its conclusion holds unconditionally.

*define*the property beautiful. That is, if we want to convince someone that some particular number is beautiful, our argument must be based on these rules. For a simple example, suppose we claim that the number 5 is beautiful. To support this claim, we just need to point out that rule b_5 says so. Or, if we want to claim that 8 is beautiful, we can support our claim by first observing that 3 and 5 are both beautiful (by rules b_3 and b_5) and then pointing out that their sum, 8, is therefore beautiful by rule b_sum. This argument can be expressed graphically with the following

*proof tree*:

----------- (b_3) ----------- (b_5)

beautiful 3 beautiful 5

------------------------------- (b_sum)

beautiful 8

beautiful 3 beautiful 5

------------------------------- (b_sum)

beautiful 8

----------- (b_5) ----------- (b_3)

beautiful 5 beautiful 3

------------------------------- (b_sum)

beautiful 8

beautiful 5 beautiful 3

------------------------------- (b_sum)

beautiful 8

#### Exercise: 1 star (varieties_of_beauty)

How many different ways are there to show that 8 is beautiful?(* FILL IN HERE *)

Inductive beautiful : nat → Prop :=

b_0 : beautiful 0

| b_3 : beautiful 3

| b_5 : beautiful 5

| b_sum : ∀n m, beautiful n → beautiful m → beautiful (n+m).

The first line declares that beautiful is a proposition — or,
more formally, a family of propositions "indexed by" natural
numbers. (That is, for each number n, the claim that "n is
beautiful" is a proposition.) Such a family of propositions is
often called a

The rules introduced this way have the same status as proven
theorems; that is, they are true axiomatically.
So we can use Coq's apply tactic with the rule names to prove
that particular numbers are beautiful.

*property*of numbers. Each of the remaining lines embodies one of the rules for beautiful numbers.Theorem three_is_beautiful: beautiful 3.

Proof.

(* This simply follows from the rule b_3. *)

apply b_3.

Qed.

Theorem eight_is_beautiful: beautiful 8.

Proof.

(* First we use the rule b_sum, telling Coq how to

instantiate n and m. *)

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

(* To solve the subgoals generated by b_sum, we must provide

evidence of beautiful 3 and beautiful 5. Fortunately we

have rules for both. *)

apply b_3.

apply b_5.

Qed.

Theorem beautiful_plus_eight: ∀n, beautiful n → beautiful (8+n).

Proof.

intros n B.

apply b_sum with (n:=8) (m:=n).

apply eight_is_beautiful.

apply B.

Qed.

Theorem b_times2: ∀n, beautiful n → beautiful (2×n).

Proof.

(* FILL IN HERE *) Admitted.

Proof.

(* FILL IN HERE *) Admitted.

Theorem b_timesm: ∀n m, beautiful n → beautiful (m×n).

Proof.

(* FILL IN HERE *) Admitted.

Proof.

(* FILL IN HERE *) Admitted.

☐

## Induction Over Evidence

*constructing*evidence that numbers are beautiful, we can also

*reason about*such evidence.

*only*ways to build evidence that numbers are beautiful.

- E is b_0 (and n is O),
- E is b_3 (and n is 3),
- E is b_5 (and n is 5), or
- E is b_sum n1 n2 E1 E2 (and n is n1+n2, where E1 is evidence that n1 is beautiful and E2 is evidence that n2 is beautiful).

### This permits us to

*analyze*any hypothesis of the form beautiful n to see how it was constructed, using the tactics we already know. In particular, we can use the induction tactic that we have already seen for reasoning about inductively defined

*data*to reason about inductively defined

*evidence*.

Inductive gorgeous : nat → Prop :=

g_0 : gorgeous 0

| g_plus3 : ∀n, gorgeous n → gorgeous (3+n)

| g_plus5 : ∀n, gorgeous n → gorgeous (5+n).

#### Exercise: 1 star (gorgeous_tree)

Write out the definition of gorgeous numbers using inference rule notation.☐

#### Exercise: 1 star (gorgeous_plus13)

Theorem gorgeous_plus13: ∀n,

gorgeous n → gorgeous (13+n).

Proof.

(* FILL IN HERE *) Admitted.

gorgeous n → gorgeous (13+n).

Proof.

(* FILL IN HERE *) Admitted.

☐

### It seems intuitively obvious that, although gorgeous and beautiful are presented using slightly different rules, they are actually the same property in the sense that they are true of the same numbers. Indeed, we can prove this.

Theorem gorgeous__beautiful : ∀n,

gorgeous n → beautiful n.

Proof.

intros n H.

induction H as [|n'|n'].

Case "g_0".

apply b_0.

Case "g_plus3".

apply b_sum. apply b_3.

apply IHgorgeous.

Case "g_plus5".

apply b_sum. apply b_5. apply IHgorgeous.

Qed.

Notice that the argument proceeds by induction on the
Let's see what happens if we try to prove this by induction on n
instead of induction on the evidence H.

*evidence*H!Theorem gorgeous__beautiful_FAILED : ∀n,

gorgeous n → beautiful n.

Proof.

intros. induction n as [| n'].

Case "n = 0". apply b_0.

Case "n = S n'". (* We are stuck! *)

Abort.

The problem here is that doing induction on n doesn't yield a
useful induction hypothesis. Knowing how the property we are
interested in behaves on the predecessor of n doesn't help us
prove that it holds for n. Instead, we would like to be able to
have induction hypotheses that mention other numbers, such as n -
3 and n - 5. This is given precisely by the shape of the
constructors for gorgeous.

#### Exercise: 2 stars (gorgeous_sum)

Theorem gorgeous_sum : ∀n m,

gorgeous n → gorgeous m → gorgeous (n + m).

Proof.

(* FILL IN HERE *) Admitted.

gorgeous n → gorgeous m → gorgeous (n + m).

Proof.

(* FILL IN HERE *) Admitted.

Theorem beautiful__gorgeous : ∀n, beautiful n → gorgeous n.

Proof.

(* FILL IN HERE *) Admitted.

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 3 stars, optional (g_times2)

Prove the g_times2 theorem below without using gorgeous__beautiful. You might find the following helper lemma useful.Lemma helper_g_times2 : ∀x y z, x + (z + y)= z + x + y.

Proof.

(* FILL IN HERE *) Admitted.

Theorem g_times2: ∀n, gorgeous n → gorgeous (2×n).

Proof.

intros n H. simpl.

induction H.

(* FILL IN HERE *) Admitted.

☐

## Inversion on Evidence

Theorem ev_minus2: ∀n,

ev n → ev (pred (pred n)).

Proof.

intros n E.

inversion E as [| n' E'].

Case "E = ev_0". simpl. apply ev_0.

Case "E = ev_SS n' E'". simpl. apply E'. Qed.

#### Exercise: 1 star, optional (ev_minus2_n)

What happens if we try to use destruct on n instead of inversion on E?(* FILL IN HERE *)

Theorem SSev__even : ∀n,

ev (S (S n)) → ev n.

Proof.

intros n E.

inversion E as [| n' E'].

apply E'. Qed.

## inversion revisited

#### Exercise: 1 star (inversion_practice)

Theorem SSSSev__even : ∀n,

ev (S (S (S (S n)))) → ev n.

Proof.

(* FILL IN HERE *) Admitted.

ev (S (S (S (S n)))) → ev n.

Proof.

(* FILL IN HERE *) Admitted.

The inversion tactic can also be used to derive goals by showing
the absurdity of a hypothesis.

Theorem even5_nonsense :

ev 5 → 2 + 2 = 9.

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 3 stars, advanced (ev_ev__ev)

Finding the appropriate thing to do induction on is a bit tricky here:Theorem ev_ev__ev : ∀n m,

ev (n+m) → ev n → ev m.

Proof.

(* FILL IN HERE *) Admitted.

☐

#### Exercise: 3 stars, optional (ev_plus_plus)

Here's an exercise that just requires applying existing lemmas. No induction or even case analysis is needed, but some of the rewriting may be tedious.Theorem ev_plus_plus : ∀n m p,

ev (n+m) → ev (n+p) → ev (m+p).

Proof.

(* FILL IN HERE *) Admitted.

☐

# Additional Exercises

#### Exercise: 4 stars (palindromes)

A palindrome is a sequence that reads the same backwards as forwards.- Define an inductive proposition pal on list X that
captures what it means to be a palindrome. (Hint: You'll need
three cases. Your definition should be based on the structure
of the list; just having a single constructor
c : ∀l, l = rev l → pal lmay seem obvious, but will not work very well.)
- Prove that
∀l, pal (l ++ rev l).
- Prove that
∀l, pal l → l = rev l.

(* FILL IN HERE *)

☐

#### Exercise: 5 stars, optional (palindrome_converse)

Using your definition of pal from the previous exercise, prove that
∀l, l = rev l → pal l.

(* FILL IN HERE *)

☐

#### Exercise: 4 stars, advanced (subsequence)

A list is a*subsequence*of another list if all of the elements in the first list occur in the same order in the second list, possibly with some extra elements in between. For example,
[1,2,3]

is a subsequence of each of the lists
[1,2,3]

[1,1,1,2,2,3]

[1,2,7,3]

[5,6,1,9,9,2,7,3,8]

but it is [1,1,1,2,2,3]

[1,2,7,3]

[5,6,1,9,9,2,7,3,8]

*not*a subsequence of any of the lists
[1,2]

[1,3]

[5,6,2,1,7,3,8]

[1,3]

[5,6,2,1,7,3,8]

- Define an inductive proposition subseq on list nat that
captures what it means to be a subsequence. (Hint: You'll need
three cases.)
- Prove that subsequence is reflexive, that is, any list is a
subsequence of itself.
- Prove that for any lists l1, l2, and l3, if l1 is a
subsequence of l2, then l1 is also a subsequence of l2 ++
l3.
- (Optional, harder) Prove that subsequence is transitive — that is, if l1 is a subsequence of l2 and l2 is a subsequence of l3, then l1 is a subsequence of l3. Hint: choose your induction carefully!

(* FILL IN HERE *)

☐
☐

#### Exercise: 2 stars, optional (R_provability)

Suppose we give Coq the following definition:
Inductive R : nat → list nat → Prop :=

| c1 : R 0 []

| c2 : ∀n l, R n l → R (S n) (n :: l)

| c3 : ∀n l, R (S n) l → R n l.

Which of the following propositions are provable?
| c1 : R 0 []

| c2 : ∀n l, R n l → R (S n) (n :: l)

| c3 : ∀n l, R (S n) l → R n l.

- R 2 [1,0]
- R 1 [1,2,1,0]
- R 6 [3,2,1,0]

# Relations

*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.

Proof.

(* WORKED IN CLASS *)

apply le_n. Qed.

Theorem test_le2 :

3 ≤ 6.

Proof.

(* WORKED IN CLASS *)

apply le_S. apply le_S. apply le_S. apply le_n. Qed.

Theorem test_le3 :

(2 ≤ 1) → 2 + 2 = 5.

Proof.

(* WORKED IN CLASS *)

intros H. inversion H. inversion H2. Qed.

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.(* FILL IN HERE *)

☐

#### Exercise: 2 stars (empty_relation)

Define an inductive binary relation empty_relation (on numbers) that never holds.(* FILL IN HERE *)

☐

#### 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.

Proof.

(* FILL IN HERE *) Admitted.

Theorem O_le_n : ∀n,

0 ≤ n.

Proof.

(* FILL IN HERE *) Admitted.

Theorem n_le_m__Sn_le_Sm : ∀n m,

n ≤ m → S n ≤ S m.

Proof.

(* FILL IN HERE *) Admitted.

Theorem Sn_le_Sm__n_le_m : ∀n m,

S n ≤ S m → n ≤ m.

Proof.

(* FILL IN HERE *) Admitted.

Theorem le_plus_l : ∀a b,

a ≤ a + b.

Proof.

(* FILL IN HERE *) Admitted.

Theorem plus_lt : ∀n1 n2 m,

n1 + n2 < m →

n1 < m ∧ n2 < m.

Proof.

unfold lt.

(* FILL IN HERE *) Admitted.

Theorem lt_S : ∀n m,

n < m →

n < S m.

Proof.

(* FILL IN HERE *) Admitted.

Theorem ble_nat_true : ∀n m,

ble_nat n m = true → n ≤ m.

Proof.

(* FILL IN HERE *) Admitted.

Theorem le_ble_nat : ∀n m,

n ≤ m →

ble_nat n m = true.

Proof.

(* 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.

Proof.

(* Hint: This theorem can be easily proved without using induction. *)

(* FILL IN HERE *) Admitted.

Theorem ble_nat_false : ∀n m,

ble_nat n m = false → ~(n ≤ m).

Proof.

(* FILL IN HERE *) Admitted.

ble_nat n m = false → ~(n ≤ m).

Proof.

(* FILL IN HERE *) Admitted.

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?(* FILL IN HERE *)

☐

End R.

# Programming with Propositions Revisited

*proposition*is a statement expressing a factual claim, like "two plus two equals four." In Coq, propositions are written as expressions of type Prop. .

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 :

plus_fact.

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., nat→Prop, 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."

Definition between (n m o: nat) : Prop :=

andb (ble_nat n o) (ble_nat o m) = true.

... and then partially apply them:

Definition teen : nat→Prop := between 13 19.

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

Definition true_for_zero (P:nat→Prop) : Prop :=

P 0.

### Here are two more examples of passing parameterized propositions as arguments to a function.

Definition true_for_all_numbers (P:nat→Prop) : 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:nat→Prop) : 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 :=

∀(P:nat→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.

Proof.

(* 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.

Proof.

(* 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.

Proof.

(* 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

(* FILL IN HERE *)

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: 2014-08-23 15:24:59 -0400 (Sat, 23 Aug 2014) $ *)