YES

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

Proof:
 DP Processor:
  DPs:
   f#(x,f(a(),a())) -> f#(x,a())
   f#(x,f(a(),a())) -> f#(f(a(),a()),a())
   f#(x,f(a(),a())) -> f#(f(f(a(),a()),a()),f(x,a()))
  TRS:
   f(x,f(a(),a())) -> f(f(f(a(),a()),a()),f(x,a()))
  Usable Rule Processor:
   DPs:
    f#(x,f(a(),a())) -> f#(x,a())
    f#(x,f(a(),a())) -> f#(f(a(),a()),a())
    f#(x,f(a(),a())) -> f#(f(f(a(),a()),a()),f(x,a()))
   TRS:
    
   Matrix Interpretation Processor: dim=4
    
    usable rules:
     
    interpretation:
     [f#](x0, x1) = [1 0 1 0]x0 + [0 1 1 1]x1,
     
                   [0 0 0 0]     [0 0 0 0]  
                   [0 0 1 0]     [0 0 0 0]  
     [f](x0, x1) = [0 0 0 0]x0 + [1 0 0 0]x1
                   [1 0 0 0]     [0 0 0 0]  ,
     
           [1]
           [0]
     [a] = [1]
           [0]
    orientation:
     f#(x,f(a(),a())) = [1 0 1 0]x + [3] >= [1 0 1 0]x + [1] = f#(x,a())
     
     f#(x,f(a(),a())) = [1 0 1 0]x + [3] >= [2] = f#(f(a(),a()),a())
     
     f#(x,f(a(),a())) = [1 0 1 0]x + [3] >= [1 0 1 0]x + [2] = f#(f(f(a(),a()),a()),f(x,a()))
    problem:
     DPs:
      
     TRS:
      
    Qed