YES

Problem:
 a(a(a(x1))) -> a(c(a(a(x1))))
 c(c(c(x1))) -> a(x1)
 a(x1) -> x1

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0 1]  
   [c](x0) = [0 1]x0,
   
             [1 0]  
   [a](x0) = [0 0]x0
  orientation:
                 [3 2]      [3 2]                   
   a(a(a(x1))) = [2 1]x1 >= [2 1]x1 = a(c(a(a(x1))))
   
                 [2 3]      [1 0]          
   c(c(c(x1))) = [2 3]x1 >= [0 0]x1 = a(x1)
   
           [1 0]             
   a(x1) = [0 0]x1 >= x1 = x1
  problem:
   a(a(a(x1))) -> a(c(a(a(x1))))
   a(x1) -> x1
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  -&]  
    [c](x0) = [-& -&]x0,
    
              [1  -&]  
    [a](x0) = [0  1 ]x0
   orientation:
                  [3  -&]      [3  -&]                   
    a(a(a(x1))) = [2  3 ]x1 >= [2  -&]x1 = a(c(a(a(x1))))
    
            [1  -&]             
    a(x1) = [0  1 ]x1 >= x1 = x1
   problem:
    a(a(a(x1))) -> a(c(a(a(x1))))
   Arctic Interpretation Processor:
    dimension: 3
    interpretation:
               [0  -& 0 ]  
     [c](x0) = [0  -& 0 ]x0
               [0  -& -&]  ,
     
               [0  0  0 ]  
     [a](x0) = [0  1  0 ]x0
               [-& 0  0 ]  
    orientation:
                   [1 2 1]      [0 1 0]                   
     a(a(a(x1))) = [2 3 2]x1 >= [1 2 1]x1 = a(c(a(a(x1))))
                   [1 2 1]      [0 1 0]                   
    problem:
     
    Qed