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
  TDG 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
   graph:
    f#(s(x),y) -> f#(x,g(x,y)) -> f#(s(x),y) -> f#(x,g(x,y))
    f#(s(x),y) -> f#(x,g(x,y)) -> f#(s(x),y) -> g#(x,y)
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     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
    Subterm Criterion Processor:
     simple projection:
      pi(f#) = 0
     problem:
      DPs:
       
      TRS:
       f(s(x),y) -> f(x,g(x,y))
       f(0(),y) -> y
       g(x,y) -> y
     Qed