YES

Problem:
 f(f(x)) -> f(c(f(x)))
 f(f(x)) -> f(d(f(x)))
 g(c(x)) -> x
 g(d(x)) -> x
 g(c(0())) -> g(d(1()))
 g(c(1())) -> g(d(0()))

Proof:
 DP Processor:
  DPs:
   f#(f(x)) -> f#(c(f(x)))
   f#(f(x)) -> f#(d(f(x)))
   g#(c(0())) -> g#(d(1()))
   g#(c(1())) -> g#(d(0()))
  TRS:
   f(f(x)) -> f(c(f(x)))
   f(f(x)) -> f(d(f(x)))
   g(c(x)) -> x
   g(d(x)) -> x
   g(c(0())) -> g(d(1()))
   g(c(1())) -> g(d(0()))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [g#](x0) = x0 + 0,
    
    [f#](x0) = x0 + 0,
    
    [1] = 3,
    
    [0] = 3,
    
    [g](x0) = 5x0 + -16,
    
    [d](x0) = -4x0 + 0,
    
    [c](x0) = -2x0 + 0,
    
    [f](x0) = 1x0 + 2
   orientation:
    f#(f(x)) = 1x + 2 >= -1x + 0 = f#(c(f(x)))
    
    f#(f(x)) = 1x + 2 >= -3x + 0 = f#(d(f(x)))
    
    g#(c(0())) = 1 >= 0 = g#(d(1()))
    
    g#(c(1())) = 1 >= 0 = g#(d(0()))
    
    f(f(x)) = 2x + 3 >= x + 2 = f(c(f(x)))
    
    f(f(x)) = 2x + 3 >= -2x + 2 = f(d(f(x)))
    
    g(c(x)) = 3x + 5 >= x = x
    
    g(d(x)) = 1x + 5 >= x = x
    
    g(c(0())) = 6 >= 5 = g(d(1()))
    
    g(c(1())) = 6 >= 5 = g(d(0()))
   problem:
    DPs:
     
    TRS:
     f(f(x)) -> f(c(f(x)))
     f(f(x)) -> f(d(f(x)))
     g(c(x)) -> x
     g(d(x)) -> x
     g(c(0())) -> g(d(1()))
     g(c(1())) -> g(d(0()))
   Qed