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:
    
   Restore Modifier:
    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()))))
    SCC Processor:
     #sccs: 1
     #rules: 1
     #arcs: 1/1
     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()))))
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [f#](x0) = x0,
       
       [g](x0, x1) = x0 + x1,
       
       [h](x0, x1) = 0,
       
       [f](x0) = x0,
       
       [a] = 1
      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