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:
 DP Processor:
  DPs:
   max#(N(L(s(x)),L(s(y)))) -> max#(N(L(x),L(y)))
   max#(N(L(x),N(y,z))) -> max#(N(y,z))
   max#(N(L(x),N(y,z))) -> max#(N(L(x),L(max(N(y,z)))))
  TRS:
   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)))))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
    [max#](x0) = [2 0]x0,
    
                   [1]
    [s](x0) = x0 + [0],
    
                  [0 0]     [1 1]     [0]
    [N](x0, x1) = [0 1]x0 + [2 2]x1 + [2],
    
          [1]
    [0] = [3],
    
                [1 0]  
    [max](x0) = [2 0]x0,
    
              [1 1]     [0]
    [L](x0) = [0 0]x0 + [1]
   orientation:
    max#(N(L(s(x)),L(s(y)))) = [2 2]y + [4] >= [2 2]y + [2] = max#(N(L(x),L(y)))
    
    max#(N(L(x),N(y,z))) = [0 2]y + [6 6]z + [4] >= [2 2]z = max#(N(y,z))
    
    max#(N(L(x),N(y,z))) = [0 2]y + [6 6]z + [4] >= [6 6]z + [2] = max#(N(L(x),L(max(N(y,z)))))
    
                [1 1]          
    max(L(x)) = [2 2]x >= x = x
    
                          [1 1]    [1]         
    max(N(L(0()),L(y))) = [2 2]y + [2] >= y = y
    
                              [1 1]    [2]    [1 1]    [2]                       
    max(N(L(s(x)),L(s(y)))) = [2 2]y + [4] >= [2 2]y + [2] = s(max(N(L(x),L(y))))
    
                          [0 1]    [3 3]    [2]    [3 3]    [1]                              
    max(N(L(x),N(y,z))) = [0 2]y + [6 6]z + [4] >= [6 6]z + [2] = max(N(L(x),L(max(N(y,z)))))
   problem:
    DPs:
     
    TRS:
     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)))))
   Qed