YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 1 1]  
   [g](x0) = [1 1 0]x0
             [0 0 1]  ,
   
         [0]
   [0] = [1]
         [0],
   
             [1 0 0]     [0]
   [s](x0) = [0 1 1]x0 + [0]
             [0 0 0]     [1],
   
             [1 1 1]     [0]
   [f](x0) = [1 1 1]x0 + [0]
             [0 0 0]     [1]
  orientation:
             [1 1 1]    [1]    [1 1 1]    [0]       
   g(s(x)) = [1 1 1]x + [0] >= [1 1 1]x + [0] = f(x)
             [0 0 0]    [1]    [0 0 0]    [1]       
   
            [1]    [0]         
   f(0()) = [1] >= [1] = s(0())
            [1]    [1]         
   
             [1 1 1]    [1]    [1 1 1]    [0]             
   f(s(x)) = [1 1 1]x + [1] >= [1 1 1]x + [1] = s(s(g(x)))
             [0 0 0]    [1]    [0 0 0]    [1]             
   
            [1]    [0]      
   g(0()) = [1] >= [1] = 0()
            [0]    [0]      
  problem:
   
  Qed