YES

Problem:
 f(0(x1)) -> s(0(x1))
 d(0(x1)) -> 0(x1)
 d(s(x1)) -> s(s(d(x1)))
 f(s(x1)) -> d(f(x1))

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [d](x0) = x0,
   
   [s](x0) = x0,
   
   [f](x0) = 8x0,
   
   [0](x0) = 2x0
  orientation:
   f(0(x1)) = 10x1 >= 2x1 = s(0(x1))
   
   d(0(x1)) = 2x1 >= 2x1 = 0(x1)
   
   d(s(x1)) = x1 >= x1 = s(s(d(x1)))
   
   f(s(x1)) = 8x1 >= 8x1 = d(f(x1))
  problem:
   d(0(x1)) -> 0(x1)
   d(s(x1)) -> s(s(d(x1)))
   f(s(x1)) -> d(f(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  0 ]  
    [d](x0) = [-& 1 ]x0,
    
              [0  -&]  
    [s](x0) = [0  0 ]x0,
    
              [1  0 ]  
    [f](x0) = [-& -&]x0,
    
              [0  -&]  
    [0](x0) = [3  0 ]x0
   orientation:
               [3 0]      [0  -&]          
    d(0(x1)) = [4 1]x1 >= [3  0 ]x1 = 0(x1)
    
               [0 0]      [0 0]                
    d(s(x1)) = [1 1]x1 >= [0 1]x1 = s(s(d(x1)))
    
               [1  0 ]      [1  0 ]             
    f(s(x1)) = [-& -&]x1 >= [-& -&]x1 = d(f(x1))
   problem:
    d(s(x1)) -> s(s(d(x1)))
    f(s(x1)) -> d(f(x1))
   Bounds Processor:
    bound: 0
    enrichment: match
    automaton:
     final states: {5,1}
     transitions:
      f40() -> 2*
      s0(4) -> 1*
      s0(3) -> 4*
      d0(2) -> 3*
      d0(6) -> 5*
      f0(2) -> 6*
      1 -> 3*
      5 -> 6*
    problem:
     
    Qed