YES

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

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