Problem:
 f(a(),x) -> f(b(),f(c(),x))
 f(a(),f(b(),x)) -> f(b(),f(a(),x))
 f(d(),f(c(),x)) -> f(d(),f(a(),x))
 f(a(),f(c(),x)) -> f(c(),f(a(),x))

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {5}
   transitions:
    f1(9,2) -> 10*
    f1(9,4) -> 10*
    f1(11,10) -> 5*
    f1(9,1) -> 10*
    f1(9,3) -> 10*
    b1() -> 11*
    c1() -> 9*
    f0(3,1) -> 5*
    f0(3,3) -> 5*
    f0(4,2) -> 5*
    f0(4,4) -> 5*
    f0(1,2) -> 5*
    f0(1,4) -> 5*
    f0(2,1) -> 5*
    f0(2,3) -> 5*
    f0(3,2) -> 5*
    f0(3,4) -> 5*
    f0(4,1) -> 5*
    f0(4,3) -> 5*
    f0(1,1) -> 5*
    f0(1,3) -> 5*
    f0(2,2) -> 5*
    f0(2,4) -> 5*
    a0() -> 1*
    b0() -> 2*
    c0() -> 3*
    d0() -> 4*
  problem:
   
  Qed