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(h(0()))) -> g(d(1()))
 g(c(1())) -> g(d(h(0())))
 g(h(x)) -> g(x)

Proof:
 DP Processor:
  DPs:
   f#(f(x)) -> f#(c(f(x)))
   f#(f(x)) -> f#(d(f(x)))
   g#(c(h(0()))) -> g#(d(1()))
   g#(c(1())) -> g#(d(h(0())))
   g#(h(x)) -> g#(x)
  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(h(0()))) -> g(d(1()))
   g(c(1())) -> g(d(h(0())))
   g(h(x)) -> g(x)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [g#](x0) = x0 + -16,
    
    [f#](x0) = 4x0 + 0,
    
    [1] = 4,
    
    [h](x0) = 2x0 + 1,
    
    [0] = 2,
    
    [g](x0) = 5x0 + 0,
    
    [d](x0) = -4x0 + 0,
    
    [c](x0) = -3x0 + 0,
    
    [f](x0) = 1x0 + 2
   orientation:
    f#(f(x)) = 5x + 6 >= 2x + 4 = f#(c(f(x)))
    
    f#(f(x)) = 5x + 6 >= 1x + 4 = f#(d(f(x)))
    
    g#(c(h(0()))) = 1 >= 0 = g#(d(1()))
    
    g#(c(1())) = 1 >= 0 = g#(d(h(0())))
    
    g#(h(x)) = 2x + 1 >= x + -16 = g#(x)
    
    f(f(x)) = 2x + 3 >= -1x + 2 = f(c(f(x)))
    
    f(f(x)) = 2x + 3 >= -2x + 2 = f(d(f(x)))
    
    g(c(x)) = 2x + 5 >= x = x
    
    g(d(x)) = 1x + 5 >= x = x
    
    g(c(h(0()))) = 6 >= 5 = g(d(1()))
    
    g(c(1())) = 6 >= 5 = g(d(h(0())))
    
    g(h(x)) = 7x + 6 >= 5x + 0 = g(x)
   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(h(0()))) -> g(d(1()))
     g(c(1())) -> g(d(h(0())))
     g(h(x)) -> g(x)
   Qed