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=1
  
  interpretation:
   [s](x0) = 4x0 + 3,
   
   [N](x0, x1) = x0 + 2x1,
   
   [0] = 1,
   
   [max](x0) = x0,
   
   [L](x0) = x0
  orientation:
   max(L(x)) = x >= x = x
   
   max(N(L(0()),L(y))) = 2y + 1 >= y = y
   
   max(N(L(s(x)),L(s(y)))) = 4x + 8y + 9 >= 4x + 8y + 3 = s(max(N(L(x),L(y))))
   
   max(N(L(x),N(y,z))) = x + 2y + 4z >= x + 2y + 4z = max(N(L(x),L(max(N(y,z)))))
  problem:
   max(L(x)) -> x
   max(N(L(x),N(y,z))) -> max(N(L(x),L(max(N(y,z)))))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                  [1 0 0]     [1 0 0]  
    [N](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                  [0 0 0]     [0 0 0]  ,
    
                [1 1 0]  
    [max](x0) = [0 1 1]x0
                [0 1 1]  ,
    
              [1 0 0]     [0]
    [L](x0) = [0 1 1]x0 + [1]
              [1 0 0]     [0]
   orientation:
                [1 1 1]    [1]         
    max(L(x)) = [1 1 1]x + [1] >= x = x
                [1 1 1]    [1]         
    
                          [1 0 0]    [1 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 + [0 0 0]z >= [0 0 0]x + [0 0 0]y + [0 0 0]z = max(N(L(x),L(max(N(y,z)))))
                          [0 0 0]    [0 0 0]    [0 0 0]     [0 0 0]    [0 0 0]    [0 0 0]                               
   problem:
    max(N(L(x),N(y,z))) -> max(N(L(x),L(max(N(y,z)))))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                   [1 0 0]     [1 1 1]     [0]
     [N](x0, x1) = [0 1 0]x0 + [0 0 0]x1 + [0]
                   [0 0 0]     [0 0 0]     [1],
     
                 [1 0 0]  
     [max](x0) = [0 0 0]x0
                 [0 0 0]  ,
     
               [1 0 0]  
     [L](x0) = [0 0 0]x0
               [0 0 0]  
    orientation:
                           [1 0 0]    [1 1 0]    [1 1 1]    [1]    [1 0 0]    [1 0 0]    [1 1 1]                               
     max(N(L(x),N(y,z))) = [0 0 0]x + [0 0 0]y + [0 0 0]z + [0] >= [0 0 0]x + [0 0 0]y + [0 0 0]z = max(N(L(x),L(max(N(y,z)))))
                           [0 0 0]    [0 0 0]    [0 0 0]    [0]    [0 0 0]    [0 0 0]    [0 0 0]                               
    problem:
     
    Qed