YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(x),y) -> g#(x,y)
   f#(s(x),y) -> f#(x,g(x,y))
  TRS:
   f(s(x),y) -> f(x,g(x,y))
   f(0(),y) -> y
   g(x,y) -> y
  LPO Processor:
   argument filtering:
    pi(s) = [0]
    pi(f) = 1
    pi(g) = 1
    pi(0) = []
    pi(f#) = 0
    pi(g#) = []
   precedence:
    s > g# ~ f# ~ 0 ~ g ~ f
   problem:
    DPs:
     
    TRS:
     f(s(x),y) -> f(x,g(x,y))
     f(0(),y) -> y
     g(x,y) -> y
   Qed