Problem:
 app(nil(),YS) -> YS
 app(cons(X),YS) -> cons(X)
 from(X) -> cons(X)
 zWadr(nil(),YS) -> nil()
 zWadr(XS,nil()) -> nil()
 zWadr(cons(X),cons(Y)) -> cons(app(Y,cons(X)))
 prefix(L) -> cons(nil())

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {6,5,4,3}
   transitions:
    cons1(5) -> 6*
    cons1(2) -> 4,3
    cons1(1) -> 4,3
    cons1(8) -> 5*
    nil1() -> 5*
    app1(2,3) -> 8*
    app1(1,3) -> 8*
    app0(1,2) -> 3*
    app0(2,1) -> 3*
    app0(1,1) -> 3*
    app0(2,2) -> 3*
    nil0() -> 1*
    cons0(2) -> 2*
    cons0(1) -> 2*
    from0(2) -> 4*
    from0(1) -> 4*
    zWadr0(1,2) -> 5*
    zWadr0(2,1) -> 5*
    zWadr0(1,1) -> 5*
    zWadr0(2,2) -> 5*
    prefix0(2) -> 6*
    prefix0(1) -> 6*
    1 -> 3*
    2 -> 3*
    3 -> 8*
  problem:
   
  Qed