YES

Problem:
 f(f(x)) -> f(g(f(x),x))
 f(f(x)) -> f(h(f(x),f(x)))
 g(x,y) -> y
 h(x,x) -> g(x,0())

Proof:
 DP Processor:
  DPs:
   f#(f(x)) -> g#(f(x),x)
   f#(f(x)) -> f#(g(f(x),x))
   f#(f(x)) -> h#(f(x),f(x))
   f#(f(x)) -> f#(h(f(x),f(x)))
   h#(x,x) -> g#(x,0())
  TRS:
   f(f(x)) -> f(g(f(x),x))
   f(f(x)) -> f(h(f(x),f(x)))
   g(x,y) -> y
   h(x,x) -> g(x,0())
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    f(f(x)) -> f(g(f(x),x))
    f(f(x)) -> f(h(f(x),f(x)))
    g(x,y) -> y
    h(x,x) -> g(x,0())
   interpretation:
    [h#](x0, x1) = x0 + x1 + 2,
    
    [g#](x0, x1) = x1 + 0,
    
    [f#](x0) = 1x0 + 0,
    
    [0] = 0,
    
    [h](x0, x1) = 2,
    
    [g](x0, x1) = 1x1 + 1,
    
    [f](x0) = 2x0 + 4
   orientation:
    f#(f(x)) = 3x + 5 >= x + 0 = g#(f(x),x)
    
    f#(f(x)) = 3x + 5 >= 2x + 2 = f#(g(f(x),x))
    
    f#(f(x)) = 3x + 5 >= 2x + 4 = h#(f(x),f(x))
    
    f#(f(x)) = 3x + 5 >= 3 = f#(h(f(x),f(x)))
    
    h#(x,x) = x + 2 >= 0 = g#(x,0())
    
    f(f(x)) = 4x + 6 >= 3x + 4 = f(g(f(x),x))
    
    f(f(x)) = 4x + 6 >= 4 = f(h(f(x),f(x)))
    
    g(x,y) = 1y + 1 >= y = y
    
    h(x,x) = 2 >= 1 = g(x,0())
   problem:
    DPs:
     
    TRS:
     f(f(x)) -> f(g(f(x),x))
     f(f(x)) -> f(h(f(x),f(x)))
     g(x,y) -> y
     h(x,x) -> g(x,0())
   Qed