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))))
   Restore Modifier:
    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))))
    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))))
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [f#](x0) = x0 + 1,
       
       [a](x0, x1) = x0,
       
       [f](x0) = 0,
       
       [g](x0) = x0 + 1
      orientation:
       f#(g(x)) = x + 2 >= x + 1 = f#(x)
       
       f(g(x)) = 0 >= 0 = 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