YES

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

Proof:
 Uncurry Processor (mirror):
  a2(a(),x) -> f(f(x,a()),a1(a1(a())))
  f(a1(x4),x5) -> a2(x4,x5)
  f(a(),x5) -> a1(x5)
  Bounds Processor:
   bound: 2
   enrichment: match
   automaton:
    final states: {5}
    transitions:
     a2() -> 15*
     a{1,0}(5) -> 5*
     a{1,1}(10) -> 13,11
     a{1,1}(5) -> 5*
     a{1,1}(11) -> 12*
     a{2,1}(5,5) -> 5*
     f1(14,12) -> 13*
     f1(13,12) -> 5*
     f1(10,10) -> 14*
     f1(5,10) -> 13*
     a1() -> 10*
     a{2,2}(5,10) -> 13*
     a{2,2}(10,12) -> 13,5
     a{2,2}(11,15) -> 18*
     a{2,0}(5,5) -> 5*
     a{1,2}(15) -> 16*
     a{1,2}(10) -> 14*
     a{1,2}(16) -> 17*
     a0() -> 5*
     f2(18,17) -> 13,5
     f2(12,15) -> 18*
     f0(5,5) -> 5*
   problem:
    
   Qed