YES

Problem:
 f(f(X)) -> c()
 c() -> d()
 h(X) -> c()

Proof:
 DP Processor:
  DPs:
   f#(f(X)) -> c#()
   h#(X) -> c#()
  TRS:
   f(f(X)) -> c()
   c() -> d()
   h(X) -> c()
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [h#](x0) = x0 + 8,
    
    [c#] = 0,
    
    [f#](x0) = 8,
    
    [h](x0) = x0 + 2,
    
    [d] = 0,
    
    [c] = 1,
    
    [f](x0) = x0 + 4
   orientation:
    f#(f(X)) = 8 >= 0 = c#()
    
    h#(X) = X + 8 >= 0 = c#()
    
    f(f(X)) = X + 4 >= 1 = c()
    
    c() = 1 >= 0 = d()
    
    h(X) = X + 2 >= 1 = c()
   problem:
    DPs:
     
    TRS:
     f(f(X)) -> c()
     c() -> d()
     h(X) -> c()
   Qed