One well known fact is the sum . Let’s try to prove this fact in Idris.

We start intuitively by defining our recursive sum function:

total sum : Nat -> Nat sum Z = Z sum (S n) = (S n) + sum n

Testing it a few times:

Idris> sum 3 6 : Nat Idris> sum 4 10 : Nat

Looks good.

Next, we will come up with out dependently typed function to prove the fact.

theorem_1_firsttry : (n : Nat) -> sum n = divNat (n * (n + 1)) 2 theorem_1_firsttry Z = ?a theorem_1_firsttry (S n) = ?b

The base case that we need to prove is of type `0 = divNat 0 2`

. Looks a bit tricky. Let’s try to use `divNatNZ`

along with a proof that 2 is not zero:

theorem_1_secondtry : (n : Nat) -> sum n = divNatNZ (n * (n + 1)) 2 (SIsNotZ {x = 1}) theorem_1_secondtry Z = ?a theorem_1_secondtry (S n) = ?b

Now the base case is just `Refl`

. Let’s put an inductive hypothesis as well:

theorem_1_secondtry : (n : Nat) -> sum n = divNatNZ (n * (n + 1)) 2 (SIsNotZ {x = 1}) theorem_1_secondtry Z = Refl theorem_1_secondtry (S n) = let IH = theorem_1_secondtry n in ?b

Idris tells us that we now need to prove:

b : S (plus n (sum n)) = ifThenElse (lte (plus (plus n 1) (mult n (S (plus n 1)))) 0) (Delay 0) (Delay (S (Prelude.Nat.divNatNZ, div' (S (plus (plus n 1) (mult n (S (plus n 1))))) 1 SIsNotZ (plus (plus n 1) (mult n (S (plus n 1)))) (minus (plus (plus n 1) (mult n (S (plus n 1)))) 1) 1)))

Woot.

Let’s take a slightly different route by doing a few algebraic tricks to get rid off division. Instead of proving that , we will prove .

total theorem_1 : (n : Nat) -> 2 * sum n = n * (n + 1) -- sum n = n * (n + 1) / 2 theorem_1 Z = Refl theorem_1 (S n) = ?b

Now we need to show that `b : S (plus (plus n (sum n)) (S (plus (plus n (sum n)) 0))) = S (plus (plus n 1) (mult n (S (plus n 1))))`

.

total theorem_1 : (n : Nat) -> 2 * sum n = n * (n + 1) -- sum n = n * (n + 1) / 2 theorem_1 Z = Refl theorem_1 (S n) = let IH = theorem_1 n in rewrite (multRightSuccPlus n (plus n 1)) in rewrite sym IH in rewrite (plusZeroRightNeutral (sum n)) in rewrite (plusZeroRightNeutral (plus n (sum n))) in rewrite (plusAssociative n (sum n) (sum n)) in rewrite (sym (plusSuccRightSucc (plus n (sum n)) (plus n (sum n)))) in rewrite plusCommutative (plus n 1) (plus (plus n (sum n)) (sum n)) in rewrite sym (plusSuccRightSucc n Z) in rewrite plusZeroRightNeutral n in rewrite (sym (plusSuccRightSucc (plus (plus n (sum n)) (sum n)) n)) in rewrite (sym (plusAssociative (n + sum n) (sum n) n)) in rewrite plusCommutative (sum n) n in Refl

Looks a bit big, but it works! With line 4 and 5 we get rid off multiplication and then all we need to do is some algebraic re-ordering of plus to show that both sides are equivalent.

Now that we proved it, you can use this fact in your favorite programming language š