YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [p](x0) = x0,
   
   [d](x0) = x0,
   
   [s](x0) = x0,
   
   [f](x0) = 4x0,
   
   [0](x0) = x0
  orientation:
   f(0(x1)) = 4x1 >= x1 = s(0(x1))
   
   d(0(x1)) = x1 >= x1 = 0(x1)
   
   d(s(x1)) = x1 >= x1 = s(s(d(p(s(x1)))))
   
   f(s(x1)) = 4x1 >= 4x1 = d(f(p(s(x1))))
   
   p(s(x1)) = x1 >= x1 = x1
  problem:
   d(0(x1)) -> 0(x1)
   d(s(x1)) -> s(s(d(p(s(x1)))))
   f(s(x1)) -> d(f(p(s(x1))))
   p(s(x1)) -> x1
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
                
    [p](x0) = x0,
    
              [0  0 ]  
    [d](x0) = [-& 2 ]x0,
    
              [0  -&]  
    [s](x0) = [0  0 ]x0,
    
              [0  -&]  
    [f](x0) = [-& -&]x0,
    
              [0  -&]  
    [0](x0) = [1  2 ]x0
   orientation:
               [1 2]      [0  -&]          
    d(0(x1)) = [3 4]x1 >= [1  2 ]x1 = 0(x1)
    
               [0 0]      [0 0]                      
    d(s(x1)) = [2 2]x1 >= [2 2]x1 = s(s(d(p(s(x1)))))
    
               [0  -&]      [0  -&]                   
    f(s(x1)) = [-& -&]x1 >= [-& -&]x1 = d(f(p(s(x1))))
    
               [0  -&]             
    p(s(x1)) = [0  0 ]x1 >= x1 = x1
   problem:
    d(s(x1)) -> s(s(d(p(s(x1)))))
    f(s(x1)) -> d(f(p(s(x1))))
    p(s(x1)) -> x1
   Bounds Processor:
    bound: 0
    enrichment: match
    automaton:
     final states: {2,7,1}
     transitions:
      f0(4) -> 8*
      f50() -> 2*
      s0(5) -> 6*
      s0(2) -> 3*
      s0(6) -> 1*
      d0(4) -> 5*
      d0(8) -> 7*
      p0(3) -> 4*
      1 -> 5*
      2 -> 4*
      7 -> 8*
    problem:
     
    Qed