YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [c](x0) = x0,
   
   [b](x0) = 4x0,
   
   [a](x0) = 8x0
  orientation:
   a(x1) = 8x1 >= 8x1 = b(b(x1))
   
   a(b(b(x1))) = 16x1 >= 16x1 = b(b(c(c(c(a(x1))))))
   
   b(b(x1)) = 8x1 >= x1 = c(c(c(x1)))
   
   c(c(c(b(b(x1))))) = 8x1 >= 8x1 = a(x1)
  problem:
   a(x1) -> b(b(x1))
   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
   c(c(c(b(b(x1))))) -> a(x1)
  String Reversal Processor:
   a(x1) -> b(b(x1))
   b(b(a(x1))) -> a(c(c(c(b(b(x1))))))
   b(b(c(c(c(x1))))) -> a(x1)
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [c](x0) = x0 + 1,
     
     [b](x0) = 2x0,
     
     [a](x0) = 4x0 + 4
    orientation:
     a(x1) = 4x1 + 4 >= 4x1 = b(b(x1))
     
     b(b(a(x1))) = 16x1 + 16 >= 16x1 + 16 = a(c(c(c(b(b(x1))))))
     
     b(b(c(c(c(x1))))) = 4x1 + 12 >= 4x1 + 4 = a(x1)
    problem:
     b(b(a(x1))) -> a(c(c(c(b(b(x1))))))
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
                [0  0 ]  
      [c](x0) = [-& -&]x0,
      
                [0 0]  
      [b](x0) = [0 1]x0,
      
                [0 3]  
      [a](x0) = [1 0]x0
     orientation:
                    [2 3]      [1 2]                         
      b(b(a(x1))) = [3 4]x1 >= [2 3]x1 = a(c(c(c(b(b(x1))))))
     problem:
      
     Qed