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=3
   
   interpretation:
    [p#](x0) = [0 2 0]x0 + [2],
    
    [f#](x0) = [2 0 0]x0,
    
          [2]
    [0] = [0]
          [1],
    
              [0 1 0]     [0]
    [p](x0) = [0 0 2]x0 + [1]
              [0 0 2]     [2],
    
              [1 0 0]  
    [f](x0) = [2 0 1]x0
              [2 0 0]  ,
    
              [3 0 0]     [2]
    [s](x0) = [1 0 0]x0 + [0]
              [0 1 1]     [1]
   orientation:
    f#(s(x)) = [6 0 0]x + [4] >= [2 0 0]x + [2] = p#(s(x))
    
    f#(s(x)) = [6 0 0]x + [4] >= [2 0 0]x = f#(p(s(x)))
    
    f#(s(x)) = [6 0 0]x + [4] >= [2 0 0]x = f#(f(p(s(x))))
    
              [3 0 0]    [2]    [3 0 0]    [2]                   
    f(s(x)) = [6 1 1]x + [5] >= [1 0 0]x + [0] = s(f(f(p(s(x)))))
              [6 0 0]    [4]    [6 0 0]    [1]                   
    
             [2]    [2]      
    f(0()) = [5] >= [0] = 0()
             [4]    [1]      
    
              [1 0 0]    [0]         
    p(s(x)) = [0 2 2]x + [3] >= x = x
              [0 2 2]    [4]         
   problem:
    DPs:
     
    TRS:
     f(s(x)) -> s(f(f(p(s(x)))))
     f(0()) -> 0()
     p(s(x)) -> x
   Qed