YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(x),s(y)) -> f#(x,y)
   g#(0(),x) -> f#(x,x)
   g#(0(),x) -> g#(f(x,x),x)
  TRS:
   f(x,0()) -> s(0())
   f(s(x),s(y)) -> s(f(x,y))
   g(0(),x) -> g(f(x,x),x)
  Usable Rule Processor:
   DPs:
    f#(s(x),s(y)) -> f#(x,y)
    g#(0(),x) -> f#(x,x)
    g#(0(),x) -> g#(f(x,x),x)
   TRS:
    f(x,0()) -> s(0())
    f(s(x),s(y)) -> s(f(x,y))
   Matrix Interpretation Processor: dim=3
    
    usable rules:
     f(x,0()) -> s(0())
     f(s(x),s(y)) -> s(f(x,y))
    interpretation:
     [g#](x0, x1) = [0 1 0]x0 + [1 0 1]x1,
     
     [f#](x0, x1) = [0 0 1]x1,
     
               [0 0 0]     [0]
     [s](x0) = [0 0 0]x0 + [0]
               [0 0 1]     [1],
     
                   [0 0 1]     [0 0 1]     [0]
     [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                   [0 0 1]     [0 0 0]     [1],
     
           [0]
     [0] = [1]
           [0]
    orientation:
     f#(s(x),s(y)) = [0 0 1]y + [1] >= [0 0 1]y = f#(x,y)
     
     g#(0(),x) = [1 0 1]x + [1] >= [0 0 1]x = f#(x,x)
     
     g#(0(),x) = [1 0 1]x + [1] >= [1 0 1]x = g#(f(x,x),x)
     
                [0 0 1]    [0]    [0]         
     f(x,0()) = [0 0 0]x + [0] >= [0] = s(0())
                [0 0 1]    [1]    [1]         
     
                    [0 0 1]    [0 0 1]    [2]    [0 0 0]    [0]            
     f(s(x),s(y)) = [0 0 0]x + [0 0 0]y + [0] >= [0 0 0]x + [0] = s(f(x,y))
                    [0 0 1]    [0 0 0]    [2]    [0 0 1]    [2]            
    problem:
     DPs:
      
     TRS:
      f(x,0()) -> s(0())
      f(s(x),s(y)) -> s(f(x,y))
    Qed