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))))
  EDG 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))))
   graph:
    f#(g(x)) -> f#(x) -> f#(g(x)) -> f#(x)
    f#(g(x)) -> f#(x) -> f#(g(x)) -> f#(a(g(g(f(x))),g(f(x))))
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(g(x)) -> f#(x)
    TRS:
     f(g(x)) -> f(a(g(g(f(x))),g(f(x))))
    Subterm Criterion Processor:
     simple projection:
      pi(f#) = 0
     problem:
      DPs:
       
      TRS:
       f(g(x)) -> f(a(g(g(f(x))),g(f(x))))
     Qed