YES

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

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