YES

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

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