YES

Problem:
 g(s(x)) -> f(x)
 f(0()) -> s(0())
 f(s(x)) -> s(s(g(x)))
 g(0()) -> 0()

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