YES

Problem:
 f(f(y,z),f(x,f(a(),x))) -> f(f(f(a(),z),f(x,a())),f(a(),y))

Proof:
 Bounds Processor:
  bound: 0
  enrichment: match
  automaton:
   final states: {1}
   transitions:
    f50() -> 2*
    f0(7,16) -> 1*
    f0(3,3) -> 5*
    f0(3,5) -> 6*
    f0(5,3) -> 5*
    f0(2,3) -> 5*
    f0(3,2) -> 16*,6,4
    f0(3,6) -> 4*
    f0(3,16) -> 4*
    f0(16,5) -> 7*
    f0(6,5) -> 7*
    f0(7,4) -> 1*
    a0() -> 3*
  problem:
   
  Qed