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:
                 [1 0]     [1 0]  
   [+](x0, x1) = [0 0]x0 + [0 0]x1,
   
             [3 2]     [1]
   [g](x0) = [0 0]x0 + [0],
   
             [2 2]     [1]
   [s](x0) = [2 0]x0 + [3],
   
         [0]
   [1] = [0],
   
             [2 1]     [2]
   [f](x0) = [0 0]x0 + [0],
   
         [0]
   [0] = [2]
  orientation:
            [4]    [0]      
   f(0()) = [0] >= [0] = 1()
   
             [6 4]    [7]    [6 3]    [7]          
   f(s(x)) = [0 0]x + [0] >= [0 0]x + [0] = g(f(x))
   
          [3 2]    [1]    [3 2]    [1]            
   g(x) = [0 0]x + [0] >= [0 0]x + [0] = +(x,s(x))
   
             [6 4]    [7]    [6 3]    [7]                  
   f(s(x)) = [0 0]x + [0] >= [0 0]x + [0] = +(f(x),s(f(x)))
  problem:
   f(s(x)) -> g(f(x))
   g(x) -> +(x,s(x))
   f(s(x)) -> +(f(x),s(f(x)))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
                       [2 0]  
    [+](x0, x1) = x0 + [0 0]x1,
    
              [3 2]     [2]
    [g](x0) = [0 1]x0 + [0],
    
              [1 1]     [1]
    [s](x0) = [2 3]x0 + [3],
    
              [1 2]  
    [f](x0) = [1 0]x0
   orientation:
              [5 7]    [7]    [5 6]    [2]          
    f(s(x)) = [1 1]x + [1] >= [1 0]x + [0] = g(f(x))
    
           [3 2]    [2]    [3 2]    [2]            
    g(x) = [0 1]x + [0] >= [0 1]x + [0] = +(x,s(x))
    
              [5 7]    [7]    [5 6]    [2]                  
    f(s(x)) = [1 1]x + [1] >= [1 0]x + [0] = +(f(x),s(f(x)))
   problem:
    g(x) -> +(x,s(x))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [+](x0, x1) = 3x0 + 4x1 + 1,
     
     [g](x0) = 7x0 + 4,
     
     [s](x0) = x0
    orientation:
     g(x) = 7x + 4 >= 7x + 1 = +(x,s(x))
    problem:
     
    Qed