Fix for #71 exposed a bug in equality.
The test for equality is failing, but the failure seems to indicate that the proof was incorrect previously, but type checked due to #71.
Here's the equivalent proof in Coq, which doesn't type check:
Inductive Eq (A : Set) : A -> A -> Set :=
refl : forall a, Eq A a a.
Definition sym A x y (e : (Eq A x y)) : (Eq A y x) :=
Eq_rec A
(fun a => fun b => fun e => (Eq A y x))
(fun c => (refl A y))
x
y
e.
Definition trans A x y z (e1 : (Eq A x y)) (e2 : (Eq A y z)) : (Eq A x z) :=
Eq_rec A
(fun a => fun b => fun e => (Eq A b z))
(fun c =>
(Eq_rec A
(fun a => fun b => fun e => (Eq A b z))
(fun c =>
(refl A z))
e2))
e1.
Fix for #71 exposed a bug in equality. The test for equality is failing, but the failure seems to indicate that the proof was incorrect previously, but type checked due to #71. Here's the equivalent proof in Coq, which doesn't type check: