YES

Problem:
 max(L(x)) -> x
 max(N(L(0()),L(y))) -> y
 max(N(L(s(x)),L(s(y)))) -> s(max(N(L(x),L(y))))
 max(N(L(x),N(y,z))) -> max(N(L(x),L(max(N(y,z)))))

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]  
   [s](x0) = [0 0 0]x0
             [0 0 0]  ,
   
                 
   [max](x0) = x0
                 ,
   
                 [1 0 0]       
   [N](x0, x1) = [0 0 0]x0 + x1
                 [0 1 0]       ,
   
               
   [L](x0) = x0
               ,
   
         [1]
   [0] = [1]
         [0]
  orientation:
                         
   max(L(x)) = x >= x = x
                         
   
                             [1]         
   max(N(L(0()),L(y))) = y + [0] >= y = y
                             [1]         
   
                             [1 0 0]    [1 0 0]     [1 0 0]    [1 0 0]                        
   max(N(L(s(x)),L(s(y)))) = [0 0 0]x + [0 0 0]y >= [0 0 0]x + [0 0 0]y = s(max(N(L(x),L(y))))
                             [0 0 0]    [0 0 0]     [0 0 0]    [0 0 0]                        
   
                         [1 0 0]    [1 0 0]         [1 0 0]    [1 0 0]                                   
   max(N(L(x),N(y,z))) = [0 0 0]x + [0 0 0]y + z >= [0 0 0]x + [0 0 0]y + z = max(N(L(x),L(max(N(y,z)))))
                         [0 1 0]    [0 1 0]         [0 1 0]    [0 1 0]                                   
  problem:
   max(L(x)) -> x
   max(N(L(s(x)),L(s(y)))) -> s(max(N(L(x),L(y))))
   max(N(L(x),N(y,z))) -> max(N(L(x),L(max(N(y,z)))))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [0]
    [s](x0) = x0 + [1]
                   [0],
    
                [1 0 0]     [0]
    [max](x0) = [1 0 1]x0 + [0]
                [0 0 1]     [1],
    
                  [1 0 1]     [1 1 0]     [0]
    [N](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [1]
                  [0 0 0]     [0 0 0]     [0],
    
              [1 0 0]  
    [L](x0) = [0 0 0]x0
              [0 1 1]  
   orientation:
                [1 0 0]    [0]         
    max(L(x)) = [1 1 1]x + [0] >= x = x
                [0 1 1]    [1]         
    
                              [1 1 1]    [1 0 0]    [1]    [1 1 1]    [1 0 0]    [0]                       
    max(N(L(s(x)),L(s(y)))) = [1 1 1]x + [1 0 0]y + [1] >= [1 1 1]x + [1 0 0]y + [1] = s(max(N(L(x),L(y))))
                              [0 0 0]    [0 0 0]    [1]    [0 0 0]    [0 0 0]    [1]                       
    
                          [1 1 1]    [1 0 1]    [1 1 0]    [1]    [1 1 1]    [1 0 1]    [1 1 0]    [0]                              
    max(N(L(x),N(y,z))) = [1 1 1]x + [1 0 1]y + [1 1 0]z + [1] >= [1 1 1]x + [1 0 1]y + [1 1 0]z + [0] = max(N(L(x),L(max(N(y,z)))))
                          [0 0 0]    [0 0 0]    [0 0 0]    [1]    [0 0 0]    [0 0 0]    [0 0 0]    [1]                              
   problem:
    max(L(x)) -> x
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                      [1]
     [max](x0) = x0 + [0]
                      [0],
     
                    [0]
     [L](x0) = x0 + [1]
                    [0]
    orientation:
                     [1]         
     max(L(x)) = x + [1] >= x = x
                     [0]         
    problem:
     
    Qed