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+(* An example showing that prop-extensionality is incompatible with
+   powerful extensions of the guard condition.
+   Unlike the example in guard2, it is not exploiting the fact that
+   the elimination of False always produces a subterm.
+
+   Example due to Cristobal Camarero on Coq-Club.
+   Adapted to nested types by Bruno Barras.
+ *)
+
+Axiom prop_ext: forall P Q, (P<->Q)->P=Q.
+
+Module Unboxed.
+
+Inductive True2:Prop:= I2: (False->True2)->True2.
+
+Theorem Heq: (False->True2)=True2.
+Proof.
+apply prop_ext. split.
+- intros. constructor. exact H.
+- intros. exact H.
+Qed.
+
+Fail Fixpoint con (x:True2) :False :=
+match x with
+I2 f => con (match Heq in _=T return T with eq_refl => f end)
+end.
+
+End Unboxed.
+
+(* This boxed example shows that it is not enough to just require
+   that the return type of the match on Heq is an inductive type
+ *)
+Module Boxed.
+
+Inductive box (T:Type) := Box (_:T).
+Definition unbox {T} (b:box T) : T := let (x) := b in x.
+
+Inductive True2:Prop:= I2: box(False->True2)->True2.
+
+Definition Heq: (False->True2) <-> True2 :=
+  conj (fun f => I2 (Box _ f)) (fun x _ => x).
+
+Fail Fixpoint con (x:True2) :False :=
+match x with
+I2 f => con (unbox(match prop_ext _ _ Heq in _=T return box T with eq_refl => f end))
+end.
+
+End Boxed.