YES

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

Proof:
 DP Processor:
  DPs:
   f#(g(x)) -> f#(x)
  TRS:
   f(g(x)) -> g(g(f(x)))
   f(g(x)) -> g(g(g(x)))
  Subterm Criterion Processor:
   simple projection:
    pi(f#) = 0
   problem:
    DPs:
     
    TRS:
     f(g(x)) -> g(g(f(x)))
     f(g(x)) -> g(g(g(x)))
   Qed