An afternoon of DTT/theorem provers: Agda and Coq

Note from 23rd August, 2017
I found this draft blog post lying around, written in the spring of 2015 while I was working at Imperial College London as a Research Associate in the Mobility Reading Group with Nobuko Yoshida. This was the fruit of a discussion with Tiago Cogumbreiro where we were comparing Coq and Agda for theorem proving. I believe we had a few more “side-by-side” comparisons written down, but this is all I have here: simple code for showing that addition of two even numbers yields an even number. It might prove to be a useful reference for someone else (at the time, I had a hard time remembering Coq syntax as I didn’t use it very often, and Tiago kindly gave me some pointers) so I have posted it “as is” now, over two years later.


Agda Coq
data Nat : Set where
  zero : Nat
  succ : Nat -> Nat

data Even : Nat -> Set where
  even_base : Even zero
  even_step : forall {n} -> Even n -> Even (succ (succ n))
Inductive Nat :=
  | zero : Nat
  | succ : Nat -> Nat.

Inductive Even: Nat -> Prop :=
  | even_base : Even zero
  | even_step : forall n, Even n -> Even (succ (succ n)).
_+_ : Nat -> Nat -> Nat
zero + n = n
(succ n) + m = succ (n + m)
Fixpoint add (n:Nat) (m:Nat) :=
  match n with
    | zero => m
    | succ n' => succ (add n' m)
  end.
even_sum : forall {n m} -> Even n -> Even m -> Even (n + m)
even_sum even_base x      = x
even_sum (even_step x) y  = even_step (even_sum x y)
Lemma add_succ:
  forall n m, add (succ n) m  = succ (add n m).
Proof.
  auto.
Qed.

Lemma even_sum :
  forall n m, Even n -> Even m -> Even (add n m).
Proof.
  intros n m even_n even_m.
  induction even_n.
   (* case even_base *)
   - simpl.
     assumption.
   (* case even_step *)
   - repeat (rewrite add_succ).
     apply even_step.
     assumption.
Qed.
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