YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [c](x0) = 6x0,
   
   [b](x0) = 9x0,
   
   [a](x0) = 15x0
  orientation:
   a(a(x1)) = 30x1 >= 27x1 = b(b(b(x1)))
   
   b(b(x1)) = 18x1 >= 18x1 = c(c(c(x1)))
   
   c(c(c(c(x1)))) = 24x1 >= 24x1 = a(b(x1))
  problem:
   b(b(x1)) -> c(c(c(x1)))
   c(c(c(c(x1)))) -> a(b(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0 0]  
    [c](x0) = [0 0]x0,
    
              [0 0]  
    [b](x0) = [1 1]x0,
    
              [0  -&]  
    [a](x0) = [0  -&]x0
   orientation:
               [1 1]      [0 0]                
    b(b(x1)) = [2 2]x1 >= [0 0]x1 = c(c(c(x1)))
    
                     [0 0]      [0 0]             
    c(c(c(c(x1)))) = [0 0]x1 >= [0 0]x1 = a(b(x1))
   problem:
    c(c(c(c(x1)))) -> a(b(x1))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0 1]  
     [c](x0) = [0 1]x0,
     
               [0 0]  
     [b](x0) = [1 2]x0,
     
               [2  1 ]  
     [a](x0) = [0  -&]x0
    orientation:
                      [3 4]      [2 3]             
     c(c(c(c(x1)))) = [3 4]x1 >= [0 0]x1 = a(b(x1))
    problem:
     
    Qed