Problem:
 f(0()) -> cons(0())
 f(s(0())) -> f(p(s(0())))
 p(s(X)) -> X

Proof:
 Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {5,4}
   transitions:
    f1(12) -> 13*
    p1(11) -> 12*
    s1(10) -> 11*
    01() -> 8*
    cons1(8) -> 9*
    cons2(20) -> 21*
    f0(2) -> 4*
    f0(1) -> 4*
    f0(3) -> 4*
    02() -> 20*
    00() -> 1*
    cons0(2) -> 2*
    cons0(1) -> 2*
    cons0(3) -> 2*
    s0(2) -> 3*
    s0(1) -> 3*
    s0(3) -> 3*
    p0(2) -> 5*
    p0(1) -> 5*
    p0(3) -> 5*
    1 -> 5*
    2 -> 5*
    3 -> 5*
    8 -> 10*
    9 -> 4*
    10 -> 12*
    13 -> 4*
    21 -> 13,4
  problem:
   
  Qed