YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(x)) -> p#(s(x))
   f#(s(x)) -> f#(p(s(x)))
   f#(s(x)) -> f#(f(p(s(x))))
  TRS:
   f(s(x)) -> s(f(f(p(s(x)))))
   f(0()) -> 0()
   p(s(x)) -> x
  Matrix Interpretation Processor: dim=4
   
   usable rules:
    f(s(x)) -> s(f(f(p(s(x)))))
    f(0()) -> 0()
    p(s(x)) -> x
   interpretation:
    [p#](x0) = [0],
    
    [f#](x0) = [0 0 0 1]x0,
    
          [0]
          [0]
    [0] = [0]
          [1],
    
              [1 0 0 0]  
              [1 0 0 0]  
    [p](x0) = [0 1 0 0]x0
              [0 0 1 0]  ,
    
              [0 0 1 1]  
              [0 0 0 1]  
    [f](x0) = [0 0 1 0]x0
              [0 0 0 1]  ,
    
              [1 1 0 0]     [0]
              [0 0 1 0]     [0]
    [s](x0) = [0 0 0 1]x0 + [0]
              [0 0 1 1]     [1]
   orientation:
    f#(s(x)) = [0 0 1 1]x + [1] >= [0] = p#(s(x))
    
    f#(s(x)) = [0 0 1 1]x + [1] >= [0 0 0 1]x = f#(p(s(x)))
    
    f#(s(x)) = [0 0 1 1]x + [1] >= [0 0 0 1]x = f#(f(p(s(x))))
    
              [0 0 1 2]    [1]    [0 0 1 2]    [0]                   
              [0 0 1 1]    [1]    [0 0 1 0]    [0]                   
    f(s(x)) = [0 0 0 1]x + [0] >= [0 0 0 1]x + [0] = s(f(f(p(s(x)))))
              [0 0 1 1]    [1]    [0 0 1 1]    [1]                   
    
             [1]    [0]      
             [1]    [0]      
    f(0()) = [0] >= [0] = 0()
             [1]    [1]      
    
              [1 1 0 0]          
              [1 1 0 0]          
    p(s(x)) = [0 0 1 0]x >= x = x
              [0 0 0 1]          
   problem:
    DPs:
     
    TRS:
     f(s(x)) -> s(f(f(p(s(x)))))
     f(0()) -> 0()
     p(s(x)) -> x
   Qed