YES

Problem:
 0(x1) -> 1(x1)
 4(5(4(5(x1)))) -> 4(4(5(5(x1))))
 5(5(5(5(5(5(4(4(4(4(4(4(x1)))))))))))) -> 2(x1)

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [2](x0) = x0,
   
   [4](x0) = 4x0,
   
   [5](x0) = x0,
   
   [1](x0) = x0,
   
   [0](x0) = x0
  orientation:
   0(x1) = x1 >= x1 = 1(x1)
   
   4(5(4(5(x1)))) = 8x1 >= 8x1 = 4(4(5(5(x1))))
   
   5(5(5(5(5(5(4(4(4(4(4(4(x1)))))))))))) = 24x1 >= x1 = 2(x1)
  problem:
   0(x1) -> 1(x1)
   4(5(4(5(x1)))) -> 4(4(5(5(x1))))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  0 ]  
    [4](x0) = [-& 0 ]x0,
    
              [2 1]  
    [5](x0) = [1 3]x0,
    
              [0  0 ]  
    [1](x0) = [-& -&]x0,
    
              [1 2]  
    [0](x0) = [0 0]x0
   orientation:
            [1 2]      [0  0 ]          
    0(x1) = [0 0]x1 >= [-& -&]x1 = 1(x1)
    
                     [4 6]      [4 6]                   
    4(5(4(5(x1)))) = [4 6]x1 >= [4 6]x1 = 4(4(5(5(x1))))
   problem:
    4(5(4(5(x1)))) -> 4(4(5(5(x1))))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0 0]  
     [4](x0) = [0 0]x0,
     
               [1 2]  
     [5](x0) = [0 0]x0
    orientation:
                      [3 4]      [2 3]                   
     4(5(4(5(x1)))) = [3 4]x1 >= [2 3]x1 = 4(4(5(5(x1))))
    problem:
     
    Qed