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