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
  Bounds Processor:
   bound: 0
   enrichment: match
   automaton:
    final states: {2,1}
    transitions:
     f40() -> 2*
     s0(5) -> 6*
     s0(2) -> 3*
     s0(6) -> 1*
     f0(4) -> 5*
     p0(3) -> 4*
     1 -> 5*
     2 -> 4*
   problem:
    
   Qed