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()))))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0) = -8x0 + 0,
    
    [g](x0, x1) = 4x0 + 9,
    
    [h](x0, x1) = 0,
    
    [f](x0) = -8x0 + 0,
    
    [a] = 0
   orientation:
    f#(g(f(a()),h(a(),f(a())))) = 1 >= 0 = f#(h(g(f(a()),a()),g(f(a()),f(a()))))
    
    f(g(f(a()),h(a(),f(a())))) = 1 >= 0 = f(h(g(f(a()),a()),g(f(a()),f(a()))))
   problem:
    DPs:
     
    TRS:
     f(g(f(a()),h(a(),f(a())))) -> f(h(g(f(a()),a()),g(f(a()),f(a()))))
   Qed