Problem:
 active(f(0())) -> mark(cons(0(),f(s(0()))))
 active(f(s(0()))) -> mark(f(p(s(0()))))
 active(p(s(0()))) -> mark(0())
 active(f(X)) -> f(active(X))
 active(cons(X1,X2)) -> cons(active(X1),X2)
 active(s(X)) -> s(active(X))
 active(p(X)) -> p(active(X))
 f(mark(X)) -> mark(f(X))
 cons(mark(X1),X2) -> mark(cons(X1,X2))
 s(mark(X)) -> mark(s(X))
 p(mark(X)) -> mark(p(X))
 proper(f(X)) -> f(proper(X))
 proper(0()) -> ok(0())
 proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
 proper(s(X)) -> s(proper(X))
 proper(p(X)) -> p(proper(X))
 f(ok(X)) -> ok(f(X))
 cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
 s(ok(X)) -> ok(s(X))
 p(ok(X)) -> ok(p(X))
 top(mark(X)) -> top(proper(X))
 top(ok(X)) -> top(active(X))

Proof:
 Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {10,9,8,7,6,5,4}
   transitions:
    top1(22) -> 10*
    active1(2) -> 22*
    active1(1) -> 22*
    active1(3) -> 22*
    proper1(2) -> 22*
    proper1(1) -> 22*
    proper1(3) -> 22*
    ok1(15) -> 15,7
    ok1(19) -> 22,9
    ok1(14) -> 14,6
    ok1(11) -> 11,5
    ok1(18) -> 18,8
    p1(2) -> 18*
    p1(1) -> 18*
    p1(3) -> 18*
    s1(2) -> 15*
    s1(1) -> 15*
    s1(3) -> 15*
    cons1(3,1) -> 14*
    cons1(3,3) -> 14*
    cons1(1,2) -> 14*
    cons1(2,1) -> 14*
    cons1(2,3) -> 14*
    cons1(3,2) -> 14*
    cons1(1,1) -> 14*
    cons1(1,3) -> 14*
    cons1(2,2) -> 14*
    f1(2) -> 11*
    f1(1) -> 11*
    f1(3) -> 11*
    01() -> 19*
    mark1(15) -> 15,7
    mark1(14) -> 14,6
    mark1(11) -> 11,5
    mark1(18) -> 18,8
    top2(23) -> 10*
    active0(2) -> 4*
    active0(1) -> 4*
    active0(3) -> 4*
    active2(19) -> 23*
    f0(2) -> 5*
    f0(1) -> 5*
    f0(3) -> 5*
    00() -> 1*
    mark0(2) -> 2*
    mark0(1) -> 2*
    mark0(3) -> 2*
    cons0(3,1) -> 6*
    cons0(3,3) -> 6*
    cons0(1,2) -> 6*
    cons0(2,1) -> 6*
    cons0(2,3) -> 6*
    cons0(3,2) -> 6*
    cons0(1,1) -> 6*
    cons0(1,3) -> 6*
    cons0(2,2) -> 6*
    s0(2) -> 7*
    s0(1) -> 7*
    s0(3) -> 7*
    p0(2) -> 8*
    p0(1) -> 8*
    p0(3) -> 8*
    proper0(2) -> 9*
    proper0(1) -> 9*
    proper0(3) -> 9*
    ok0(2) -> 3*
    ok0(1) -> 3*
    ok0(3) -> 3*
    top0(2) -> 10*
    top0(1) -> 10*
    top0(3) -> 10*
  problem:
   
  Qed