YES

Problem:
 f(f(x)) -> f(x)
 g(0()) -> g(f(0()))

Proof:
 DP Processor:
  DPs:
   g#(0()) -> f#(0())
   g#(0()) -> g#(f(0()))
  TRS:
   f(f(x)) -> f(x)
   g(0()) -> g(f(0()))
  TDG Processor:
   DPs:
    g#(0()) -> f#(0())
    g#(0()) -> g#(f(0()))
   TRS:
    f(f(x)) -> f(x)
    g(0()) -> g(f(0()))
   graph:
    g#(0()) -> g#(f(0())) -> g#(0()) -> g#(f(0()))
    g#(0()) -> g#(f(0())) -> g#(0()) -> f#(0())
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     g#(0()) -> g#(f(0()))
    TRS:
     f(f(x)) -> f(x)
     g(0()) -> g(f(0()))
    Matrix Interpretation Processor:
     dimension: 1
     interpretation:
      [g#](x0) = x0,
      
      [g](x0) = 0,
      
      [0] = 1,
      
      [f](x0) = 0
     orientation:
      g#(0()) = 1 >= 0 = g#(f(0()))
      
      f(f(x)) = 0 >= 0 = f(x)
      
      g(0()) = 0 >= 0 = g(f(0()))
     problem:
      DPs:
       
      TRS:
       f(f(x)) -> f(x)
       g(0()) -> g(f(0()))
     Qed