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)
  CDG 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)
   Restore Modifier:
    DPs:
     f#(f(x)) -> g#(f(x))
     g#(g(x)) -> f#(x)
    TRS:
     f(f(x)) -> g(f(x))
     g(g(x)) -> f(x)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 2/4
     DPs:
      g#(g(x)) -> f#(x)
      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) = x0,
       
       [f#](x0) = x0 + 1,
       
       [g](x0) = x0 + 1,
       
       [f](x0) = x0 + 1
      orientation:
       g#(g(x)) = x + 1 >= x + 1 = f#(x)
       
       f#(f(x)) = x + 2 >= x + 1 = g#(f(x))
       
       f(f(x)) = x + 2 >= x + 2 = g(f(x))
       
       g(g(x)) = x + 2 >= x + 1 = f(x)
      problem:
       DPs:
        g#(g(x)) -> f#(x)
       TRS:
        f(f(x)) -> g(f(x))
        g(g(x)) -> f(x)
      Matrix Interpretation Processor:
       dimension: 1
       interpretation:
        [g#](x0) = x0,
        
        [f#](x0) = 0,
        
        [g](x0) = 1,
        
        [f](x0) = 1
       orientation:
        g#(g(x)) = 1 >= 0 = 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