YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0  2 ]  
   [0](x0) = [-& 1 ]x0,
   
               
   [p](x0) = x0,
   
             [2  -&]  
   [f](x0) = [0  0 ]x0,
   
             [0  0 ]  
   [s](x0) = [-& 0 ]x0
  orientation:
              [2 2]      [2 2]                      
   f(s(x1)) = [0 0]x1 >= [0 0]x1 = s(s(f(p(s(x1)))))
   
              [2 4]      [0  2 ]          
   f(0(x1)) = [0 2]x1 >= [-& 1 ]x1 = 0(x1)
   
              [0  0 ]             
   p(s(x1)) = [-& 0 ]x1 >= x1 = x1
  problem:
   f(s(x1)) -> s(s(f(p(s(x1)))))
   p(s(x1)) -> x1
  DP Processor:
   DPs:
    f#(s(x1)) -> p#(s(x1))
    f#(s(x1)) -> f#(p(s(x1)))
   TRS:
    f(s(x1)) -> s(s(f(p(s(x1)))))
    p(s(x1)) -> x1
   TDG Processor:
    DPs:
     f#(s(x1)) -> p#(s(x1))
     f#(s(x1)) -> f#(p(s(x1)))
    TRS:
     f(s(x1)) -> s(s(f(p(s(x1)))))
     p(s(x1)) -> x1
    graph:
     f#(s(x1)) -> f#(p(s(x1))) -> f#(s(x1)) -> f#(p(s(x1)))
     f#(s(x1)) -> f#(p(s(x1))) -> f#(s(x1)) -> p#(s(x1))
    CDG Processor:
     DPs:
      f#(s(x1)) -> p#(s(x1))
      f#(s(x1)) -> f#(p(s(x1)))
     TRS:
      f(s(x1)) -> s(s(f(p(s(x1)))))
      p(s(x1)) -> x1
     graph:
      
     Qed