YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(0()),g(x)) -> f#(x,g(x))
   g#(s(x)) -> g#(x)
  TRS:
   f(s(0()),g(x)) -> f(x,g(x))
   g(s(x)) -> g(x)
  TDG Processor:
   DPs:
    f#(s(0()),g(x)) -> f#(x,g(x))
    g#(s(x)) -> g#(x)
   TRS:
    f(s(0()),g(x)) -> f(x,g(x))
    g(s(x)) -> g(x)
   graph:
    g#(s(x)) -> g#(x) -> g#(s(x)) -> g#(x)
    f#(s(0()),g(x)) -> f#(x,g(x)) -> f#(s(0()),g(x)) -> f#(x,g(x))
   EDG Processor:
    DPs:
     f#(s(0()),g(x)) -> f#(x,g(x))
     g#(s(x)) -> g#(x)
    TRS:
     f(s(0()),g(x)) -> f(x,g(x))
     g(s(x)) -> g(x)
    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(s(0()),g(x)) -> f(x,g(x))
      g(s(x)) -> g(x)
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [g#](x0) = x0 + 1,
       
       [f](x0, x1) = 0,
       
       [g](x0) = 0,
       
       [s](x0) = x0 + 1,
       
       [0] = 1
      orientation:
       g#(s(x)) = x + 2 >= x + 1 = g#(x)
       
       f(s(0()),g(x)) = 0 >= 0 = f(x,g(x))
       
       g(s(x)) = 0 >= 0 = g(x)
      problem:
       DPs:
        
       TRS:
        f(s(0()),g(x)) -> f(x,g(x))
        g(s(x)) -> g(x)
      Qed