Problem:
 f(0()) -> s(0())
 f(s(0())) -> s(s(0()))
 f(s(0())) -> *(s(s(0())),f(0()))
 f(+(x,s(0()))) -> +(s(s(0())),f(x))
 f(+(x,y)) -> *(f(x),f(y))

Proof:
 Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {5}
   transitions:
    *1(5,9) -> 10,5
    *1(10,10) -> 10,5
    f1(7) -> 9*
    f1(2) -> 10*
    f1(4) -> 10*
    f1(1) -> 10*
    f1(3) -> 10*
    +1(5,10) -> 10,5
    s1(5) -> 10,5
    s1(7) -> 10,5
    01() -> 7*
    s2(12) -> 9*
    f0(2) -> 5*
    f0(4) -> 5*
    f0(1) -> 5*
    f0(3) -> 5*
    02() -> 12*
    00() -> 1*
    s0(2) -> 2*
    s0(4) -> 2*
    s0(1) -> 2*
    s0(3) -> 2*
    *0(3,1) -> 3*
    *0(3,3) -> 3*
    *0(4,2) -> 3*
    *0(4,4) -> 3*
    *0(1,2) -> 3*
    *0(1,4) -> 3*
    *0(2,1) -> 3*
    *0(2,3) -> 3*
    *0(3,2) -> 3*
    *0(3,4) -> 3*
    *0(4,1) -> 3*
    *0(4,3) -> 3*
    *0(1,1) -> 3*
    *0(1,3) -> 3*
    *0(2,2) -> 3*
    *0(2,4) -> 3*
    +0(3,1) -> 4*
    +0(3,3) -> 4*
    +0(4,2) -> 4*
    +0(4,4) -> 4*
    +0(1,2) -> 4*
    +0(1,4) -> 4*
    +0(2,1) -> 4*
    +0(2,3) -> 4*
    +0(3,2) -> 4*
    +0(3,4) -> 4*
    +0(4,1) -> 4*
    +0(4,3) -> 4*
    +0(1,1) -> 4*
    +0(1,3) -> 4*
    +0(2,2) -> 4*
    +0(2,4) -> 4*
  problem:
   
  Qed