YES

Problem:
 f(g(f(a()),h(a(),f(a())))) -> f(h(g(f(a()),a()),g(f(a()),f(a()))))

Proof:
 DP Processor:
  DPs:
   f#(g(f(a()),h(a(),f(a())))) -> f#(h(g(f(a()),a()),g(f(a()),f(a()))))
  TRS:
   f(g(f(a()),h(a(),f(a())))) -> f(h(g(f(a()),a()),g(f(a()),f(a()))))
  Usable Rule Processor:
   DPs:
    f#(g(f(a()),h(a(),f(a())))) -> f#(h(g(f(a()),a()),g(f(a()),f(a()))))
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [f#](x0) = x0 + 0,
     
     [g](x0, x1) = 4,
     
     [h](x0, x1) = -4x0 + 0,
     
     [f](x0) = 3x0 + 1,
     
     [a] = 1
    orientation:
     f#(g(f(a()),h(a(),f(a())))) = 4 >= 0 = f#(h(g(f(a()),a()),g(f(a()),f(a()))))
    problem:
     DPs:
      
     TRS:
      
    Qed