YES

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

Proof:
 DP Processor:
  DPs:
   f#(g(x),s(0()),y) -> f#(y,y,g(x))
   g#(s(x)) -> g#(x)
  TRS:
   f(g(x),s(0()),y) -> f(y,y,g(x))
   g(s(x)) -> s(g(x))
   g(0()) -> 0()
  EDG Processor:
   DPs:
    f#(g(x),s(0()),y) -> f#(y,y,g(x))
    g#(s(x)) -> g#(x)
   TRS:
    f(g(x),s(0()),y) -> f(y,y,g(x))
    g(s(x)) -> s(g(x))
    g(0()) -> 0()
   graph:
    g#(s(x)) -> g#(x) -> g#(s(x)) -> g#(x)
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 1/4
    DPs:
     g#(s(x)) -> g#(x)
    TRS:
     f(g(x),s(0()),y) -> f(y,y,g(x))
     g(s(x)) -> s(g(x))
     g(0()) -> 0()
    Matrix Interpretation Processor:
     dimension: 1
     interpretation:
      [g#](x0) = x0 + 1,
      
      [f](x0, x1, x2) = x0 + x2,
      
      [s](x0) = x0 + 1,
      
      [0] = 0,
      
      [g](x0) = x0
     orientation:
      g#(s(x)) = x + 2 >= x + 1 = g#(x)
      
      f(g(x),s(0()),y) = x + y >= x + y = f(y,y,g(x))
      
      g(s(x)) = x + 1 >= x + 1 = s(g(x))
      
      g(0()) = 0 >= 0 = 0()
     problem:
      DPs:
       
      TRS:
       f(g(x),s(0()),y) -> f(y,y,g(x))
       g(s(x)) -> s(g(x))
       g(0()) -> 0()
     Qed