YES

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

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