YES

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

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