YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(x),y) -> f#(x,y)
   f#(s(x),y) -> f#(x,f(x,y))
  TRS:
   f(s(x),y) -> f(x,f(x,y))
   f(0(),y) -> c(y,y)
  KBO Processor:
   argument filtering:
    pi(s) = [0]
    pi(f) = []
    pi(0) = []
    pi(c) = []
    pi(f#) = [0,1]
   weight function:
    w0 = 1
    w(f#) = w(c) = w(0) = w(f) = 1
    w(s) = 0
   precedence:
    0 ~ s > f > f# ~ c
   problem:
    DPs:
     
    TRS:
     f(s(x),y) -> f(x,f(x,y))
     f(0(),y) -> c(y,y)
   Qed