(* The following lemma also lives under the name KSI in other versions of this proof *)
Axiom
EXTE :
forall (D : Set) (x y : D -> proc), (forall d : D, x d = y d) -> x = y.
Section BPA.
Variable x y z : proc.

Axiom A5 : seq x (seq y z) = seq (seq x y) z.