YES

Problem:
 f(0()) -> 1()
 f(s(x)) -> g(f(x))
 g(x) -> +(x,s(x))
 f(s(x)) -> +(f(x),s(f(x)))

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
             [3 2]     [3]
   [g](x0) = [0 0]x0 + [0],
   
             [2 2]     [1]
   [f](x0) = [0 0]x0 + [0],
   
             [1 0]     [0]
   [s](x0) = [2 3]x0 + [3],
   
                 [1 0]     [2 0]     [3]
   [+](x0, x1) = [0 0]x0 + [0 0]x1 + [0],
   
         [2]
   [0] = [0],
   
         [0]
   [1] = [0]
  orientation:
            [5]    [0]      
   f(0()) = [0] >= [0] = 1()
   
             [6 6]    [7]    [6 6]    [6]          
   f(s(x)) = [0 0]x + [0] >= [0 0]x + [0] = g(f(x))
   
          [3 2]    [3]    [3 0]    [3]            
   g(x) = [0 0]x + [0] >= [0 0]x + [0] = +(x,s(x))
   
             [6 6]    [7]    [6 6]    [6]                  
   f(s(x)) = [0 0]x + [0] >= [0 0]x + [0] = +(f(x),s(f(x)))
  problem:
   g(x) -> +(x,s(x))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
              [3 3]     [1]
    [g](x0) = [2 1]x0 + [0],
    
              [2 2]  
    [s](x0) = [1 0]x0,
    
                         
    [+](x0, x1) = x0 + x1
   orientation:
           [3 3]    [1]    [3 2]             
    g(x) = [2 1]x + [0] >= [1 1]x = +(x,s(x))
   problem:
    
   Qed