YES

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

Proof:
 DP Processor:
  DPs:
   -#(s(x),s(y)) -> -#(x,y)
   f#(s(x),y) -> -#(y,s(x))
   f#(s(x),y) -> p#(-(y,s(x)))
   f#(s(x),y) -> -#(s(x),y)
   f#(s(x),y) -> p#(-(s(x),y))
   f#(s(x),y) -> f#(p(-(s(x),y)),p(-(y,s(x))))
   f#(x,s(y)) -> -#(s(y),x)
   f#(x,s(y)) -> p#(-(s(y),x))
   f#(x,s(y)) -> -#(x,s(y))
   f#(x,s(y)) -> p#(-(x,s(y)))
   f#(x,s(y)) -> f#(p(-(x,s(y))),p(-(s(y),x)))
  TRS:
   -(x,0()) -> x
   -(s(x),s(y)) -> -(x,y)
   p(s(x)) -> x
   f(s(x),y) -> f(p(-(s(x),y)),p(-(y,s(x))))
   f(x,s(y)) -> f(p(-(x,s(y))),p(-(s(y),x)))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0, x1) = 2x0 + 2x1 + 0,
    
    [p#](x0) = x0 + 0,
    
    [-#](x0, x1) = 1x0 + -8x1 + -16,
    
    [f](x0, x1) = x0 + x1 + 0,
    
    [p](x0) = -2x0 + 0,
    
    [s](x0) = 4x0 + 1,
    
    [-](x0, x1) = 1x0,
    
    [0] = 9
   orientation:
    -#(s(x),s(y)) = 5x + -4y + 2 >= 1x + -8y + -16 = -#(x,y)
    
    f#(s(x),y) = 6x + 2y + 3 >= -4x + 1y + -7 = -#(y,s(x))
    
    f#(s(x),y) = 6x + 2y + 3 >= 1y + 0 = p#(-(y,s(x)))
    
    f#(s(x),y) = 6x + 2y + 3 >= 5x + -8y + 2 = -#(s(x),y)
    
    f#(s(x),y) = 6x + 2y + 3 >= 5x + 2 = p#(-(s(x),y))
    
    f#(s(x),y) = 6x + 2y + 3 >= 5x + 1y + 2 = f#(p(-(s(x),y)),p(-(y,s(x))))
    
    f#(x,s(y)) = 2x + 6y + 3 >= -8x + 5y + 2 = -#(s(y),x)
    
    f#(x,s(y)) = 2x + 6y + 3 >= 5y + 2 = p#(-(s(y),x))
    
    f#(x,s(y)) = 2x + 6y + 3 >= 1x + -4y + -7 = -#(x,s(y))
    
    f#(x,s(y)) = 2x + 6y + 3 >= 1x + 0 = p#(-(x,s(y)))
    
    f#(x,s(y)) = 2x + 6y + 3 >= 1x + 5y + 2 = f#(p(-(x,s(y))),p(-(s(y),x)))
    
    -(x,0()) = 1x >= x = x
    
    -(s(x),s(y)) = 5x + 2 >= 1x = -(x,y)
    
    p(s(x)) = 2x + 0 >= x = x
    
    f(s(x),y) = 4x + y + 1 >= 3x + -1y + 0 = f(p(-(s(x),y)),p(-(y,s(x))))
    
    f(x,s(y)) = x + 4y + 1 >= -1x + 3y + 0 = f(p(-(x,s(y))),p(-(s(y),x)))
   problem:
    DPs:
     
    TRS:
     -(x,0()) -> x
     -(s(x),s(y)) -> -(x,y)
     p(s(x)) -> x
     f(s(x),y) -> f(p(-(s(x),y)),p(-(y,s(x))))
     f(x,s(y)) -> f(p(-(x,s(y))),p(-(s(y),x)))
   Qed