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) = 1x0 + x1 + -8,
    
    [c](x0) = 2x0 + 0,
    
    [f](x0, x1) = x0 + -1x1,
    
    [s](x0) = 2x0 + 0
   orientation:
    f#(s(x),y) = 3x + y + 1 >= 2x + 0 = f#(x,s(x))
    
    f#(x,s(y)) = 1x + 2y + 0 >= x + 1y + -8 = f#(y,x)
    
    f#(c(x),y) = 3x + y + 1 >= 2x + 0 = f#(x,s(x))
    
    f(s(x),y) = 2x + -1y + 0 >= 1x + -1 = f(x,s(x))
    
    f(x,s(y)) = x + 1y + -1 >= -1x + y = f(y,x)
    
    f(c(x),y) = 2x + -1y + 0 >= 1x + -1 = 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