I'm used to writing human proofs in mathematics, but I'm very new to writing Agda. The following is a toy example of something that I can't figure out how to prove with Agda.
Informally, I want to write a function f which takes a natural number x and a pair of naturals. If the first element in the pair equals x, return the second element of the pair. Otherwise, return 0.
Here are my definitions for natural number equality:
data N : Set where
zero : N
s : N → N
data _≡_ {X : Set} : X → X → Set where
refl : (x : X) → (x ≡ x)
data _≢_ : N → N → Set where
< : {n : N} → (zero ≢ (s n))
> : {n : N} → ((s n) ≢ zero)
rec : {n m : N} → (n ≢ m) → ((s n) ≢ (s m))
data _=?_ (n m : N) : Set where
true : (n ≡ m) → (n =? m)
false : (n ≢ m) → (n =? m)
equal? : (n m : N) → (n =? m)
equal? zero zero = true (refl zero)
equal? zero (s _) = false <
equal? (s _) zero = false >
equal? (s n) (s m) with (equal? n m)
... | (true (refl a)) = (true (refl (s a)))
... | (false p) = (false (rec p))
and here is the function.
data Npair : Set where
pair : (n m : N) → Npair
f : N → Npair → N
f a (pair b c) with equal? a b
... | (true (refl _)) = c
... | (false _) = zero
I cannot prove
lemma : (x y : N) → (y ≡ (f x (pair x y)))
because when I try to introduce the refl constructor in the definition, it complains that
y != f x (pair x y) | equal? x x of type N
What do I have to change in order to prove this lemma?
In lemma, you need to pattern match on equal? x x, because f matches on that as well, and you can't reason about f's output until you do the same match. However, you get two cases for equal? x x:
lemma : (x y : N) → (y ≡ (f x (pair x y)))
lemma x y with equal? x x
... | true (refl _) = refl _
... | false _ = ?
Of this, the second case is not possible. To rule it out, you need to prove ∀ n → equal? n n ≡ true (refl _):
equal?-true : ∀ n → equal? n n ≡ true (refl _)
equal?-true zero = refl _
equal?-true (s n) with equal? n n | equal?-true n
... | true (refl _) | q = refl _
... | false x | ()
lemma : (x y : N) → (y ≡ (f x (pair x y)))
lemma x y with equal? x x | equal?-true x
... | true (refl _) | _ = refl _
... | false _ | ()
However, you don't need to do extra work if you define inequality just as the negation of equality, because then x ≢ x immediately implies ⊥.
data ⊥ : Set where
⊥-elim : ⊥ → {A : Set} → A
⊥-elim ()
_≢_ = λ {A : Set}(x y : A) → x ≡ y → ⊥
data _=?_ (n m : N) : Set where
true : (n ≡ m) → (n =? m)
false : (n ≢ m) → (n =? m)
equal? : ∀ n m → n =? m
equal? zero zero = true (refl zero)
equal? zero (s m) = false (λ ())
equal? (s n) zero = false (λ ())
equal? (s n) (s m) with equal? n m
... | true (refl _) = true (refl _)
... | false p = false λ {(refl _) → p (refl n)}
data Npair : Set where
pair : (n m : N) → Npair
f : N → Npair → N
f a (pair b c) with equal? a b
... | (true (refl _)) = c
... | (false _) = zero
lemma : (x y : N) → (y ≡ (f x (pair x y)))
lemma x y with equal? x x
... | true (refl .x) = refl y
... | false p = ⊥-elim (p (refl _))