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))
  Usable Rule 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:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [f#](x0, x1) = 1x0 + x1,
     
     [c](x0) = 6x0 + 3,
     
     [s](x0) = 6x0 + 1
    orientation:
     f#(s(x),y) = 7x + y + 2 >= 6x + 1 = f#(x,s(x))
     
     f#(x,s(y)) = 1x + 6y + 1 >= x + 1y = f#(y,x)
     
     f#(c(x),y) = 7x + y + 4 >= 6x + 1 = f#(x,s(x))
    problem:
     DPs:
      
     TRS:
      
    Qed