Proving False with negative inductive types in Coq

Question

The third chapter of CPDT briefly discusses why negative inductive types are forbidden in Coq. If we had

Inductive term : Set :=
| App : term -> term -> term
| Abs : (term -> term) -> term.

then we could easily define a function

Definition uhoh (t : term) : term :=
  match t with
    | Abs f => f t
    | _ => t
  end.

so that the term uhoh (Abs uhoh) would be non-terminating, with which "we would be able to prove every theorem".

I understand the non-termination part, but I don't get how we can prove anything with it. How would one prove False using term as defined above?

Answer 1

Reading your question made me realize that I didn't quite understand Adam's argument either. But inconsistency in this case results quite easily from Cantor's usual diagonal argument (a never-ending source of paradoxes and puzzles in logic). Consider the following assumptions:

Section Diag.

Variable T : Type.

Variable test : T -> bool.

Variables x y : T.

Hypothesis xT : test x = true.
Hypothesis yF : test y = false.

Variable g : (T -> T) -> T.
Variable g_inv : T -> (T -> T).

Hypothesis gK : forall f, g_inv (g f) = f.

Definition kaboom (t : T) : T :=
  if test (g_inv t t) then y else x.

Lemma kaboom1 : forall t, kaboom t <> g_inv t t.
Proof.
  intros t H.
  unfold kaboom in H.
  destruct (test (g_inv t t)) eqn:E; congruence.
Qed.

Lemma kaboom2 : False.
Proof.
  assert (H := @kaboom1 (g kaboom)).
  rewrite -> gK in H.
  congruence.
Qed.

End Diag.

This is a generic development that could be instantiated with the term type defined in CPDT: T would be term , x and y would be two elements of term that we can test discriminate between (eg App (Abs id) (Abs id) and Abs id ). The key point is the last assumption: we assume that we have an invertible function g : (T -> T) -> T which, in your example, would be Abs . Using that function, we play the usual diagonalization trick: we define a function kaboom that is by construction different from every function T -> T , including itself. The contradiction results from that.