YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(x),y) -> f#(x,s(x))
   f#(x,s(y)) -> f#(y,x)
   f#(c(x),y) -> f#(x,s(x))
  TRS:
   f(s(x),y) -> f(x,s(x))
   f(x,s(y)) -> f(y,x)
   f(c(x),y) -> f(x,s(x))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0, x1) = x0 + x1 + -16,
    
    [c](x0) = 1x0 + 4,
    
    [f](x0, x1) = 0,
    
    [s](x0) = x0
   orientation:
    f#(s(x),y) = x + y + -16 >= x + -16 = f#(x,s(x))
    
    f#(x,s(y)) = x + y + -16 >= x + y + -16 = f#(y,x)
    
    f#(c(x),y) = 1x + y + 4 >= x + -16 = f#(x,s(x))
    
    f(s(x),y) = 0 >= 0 = f(x,s(x))
    
    f(x,s(y)) = 0 >= 0 = f(y,x)
    
    f(c(x),y) = 0 >= 0 = f(x,s(x))
   problem:
    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)
     f(c(x),y) -> f(x,s(x))
   Arctic Interpretation Processor:
    dimension: 1
    interpretation:
     [f#](x0, x1) = 1x0 + x1,
     
     [c](x0) = 1x0 + -11,
     
     [f](x0, x1) = 3x0 + 2x1,
     
     [s](x0) = 2x0
    orientation:
     f#(s(x),y) = 3x + y >= 2x = f#(x,s(x))
     
     f#(x,s(y)) = 1x + 2y >= x + 1y = f#(y,x)
     
     f(s(x),y) = 5x + 2y >= 4x = f(x,s(x))
     
     f(x,s(y)) = 3x + 4y >= 2x + 3y = f(y,x)
     
     f(c(x),y) = 4x + 2y + -8 >= 4x = f(x,s(x))
    problem:
     DPs:
      
     TRS:
      f(s(x),y) -> f(x,s(x))
      f(x,s(y)) -> f(y,x)
      f(c(x),y) -> f(x,s(x))
    Qed