Problem:
 f(x1) -> n(c(c(x1)))
 c(f(x1)) -> f(c(c(x1)))
 c(c(x1)) -> c(x1)
 n(s(x1)) -> f(s(s(x1)))
 n(f(x1)) -> f(n(x1))

Proof:
 Bounds Processor:
  bound: 3
  enrichment: match
  automaton:
   final states: {4,3,2}
   transitions:
    f1(19) -> 20*
    s1(17) -> 18*
    s1(18) -> 19*
    n1(7) -> 8*
    c1(5) -> 6*
    c1(6) -> 7*
    c2(22) -> 23*
    c2(31) -> 32*
    c2(21) -> 22*
    f0(1) -> 2*
    n2(23) -> 24*
    n0(1) -> 4*
    c3(33) -> 34*
    c0(1) -> 3*
    s0(1) -> 1*
    1 -> 17,5
    5 -> 31*
    8 -> 2*
    19 -> 21*
    20 -> 4*
    21 -> 33*
    24 -> 20*
    32 -> 7*
    34 -> 23*
  problem:
   
  Qed