YES

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

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