YES

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

Proof:
 DP Processor:
  DPs:
   f#(f(x)) -> g#(f(x))
   g#(g(x)) -> f#(x)
  TRS:
   f(f(x)) -> g(f(x))
   g(g(x)) -> f(x)
  EDG Processor:
   DPs:
    f#(f(x)) -> g#(f(x))
    g#(g(x)) -> f#(x)
   TRS:
    f(f(x)) -> g(f(x))
    g(g(x)) -> f(x)
   graph:
    g#(g(x)) -> f#(x) -> f#(f(x)) -> g#(f(x))
    f#(f(x)) -> g#(f(x)) -> g#(g(x)) -> f#(x)
   Subterm Criterion Processor:
    simple projection:
     pi(f#) = 0
     pi(g#) = 0
    problem:
     DPs:
      f#(f(x)) -> g#(f(x))
     TRS:
      f(f(x)) -> g(f(x))
      g(g(x)) -> f(x)
    Matrix Interpretation Processor:
     dimension: 1
     interpretation:
      [g#](x0) = 0,
      
      [f#](x0) = x0,
      
      [g](x0) = 1,
      
      [f](x0) = 1
     orientation:
      f#(f(x)) = 1 >= 0 = g#(f(x))
      
      f(f(x)) = 1 >= 1 = g(f(x))
      
      g(g(x)) = 1 >= 1 = f(x)
     problem:
      DPs:
       
      TRS:
       f(f(x)) -> g(f(x))
       g(g(x)) -> f(x)
     Qed