YES

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

Proof:
 String Reversal Processor:
  a(a(b(b(b(b(a(a(x1)))))))) -> b(b(a(a(c(c(a(a(x1))))))))
  c(c(a(a(x1)))) -> a(a(c(c(c(c(x1))))))
  c(c(c(c(c(c(x1)))))) -> b(b(c(c(b(b(x1))))))
  Matrix Interpretation Processor: dim=4
   
   interpretation:
              [1 0 0 0]  
              [0 0 0 0]  
    [c](x0) = [0 0 1 0]x0
              [0 0 0 0]  ,
    
              [1 0 0 0]  
              [0 1 1 0]  
    [b](x0) = [0 0 0 0]x0
              [0 0 0 0]  ,
    
              [1 0 0 1]     [0]
              [0 1 0 0]     [0]
    [a](x0) = [1 0 0 0]x0 + [1]
              [0 1 0 0]     [0]
   orientation:
                                 [2 2 0 2]     [1]    [1 1 0 1]     [0]                             
                                 [1 1 0 1]     [1]    [1 1 0 1]     [1]                             
    a(a(b(b(b(b(a(a(x1)))))))) = [1 1 0 1]x1 + [1] >= [0 0 0 0]x1 + [0] = b(b(a(a(c(c(a(a(x1))))))))
                                 [1 1 0 1]     [1]    [0 0 0 0]     [0]                             
    
                     [1 1 0 1]     [0]    [1 0 0 0]     [0]                       
                     [0 0 0 0]     [0]    [0 0 0 0]     [0]                       
    c(c(a(a(x1)))) = [1 0 0 1]x1 + [1] >= [1 0 0 0]x1 + [1] = a(a(c(c(c(c(x1))))))
                     [0 0 0 0]     [0]    [0 0 0 0]     [0]                       
    
                           [1 0 0 0]      [1 0 0 0]                         
                           [0 0 0 0]      [0 0 0 0]                         
    c(c(c(c(c(c(x1)))))) = [0 0 1 0]x1 >= [0 0 0 0]x1 = b(b(c(c(b(b(x1))))))
                           [0 0 0 0]      [0 0 0 0]                         
   problem:
    c(c(a(a(x1)))) -> a(a(c(c(c(c(x1))))))
    c(c(c(c(c(c(x1)))))) -> b(b(c(c(b(b(x1))))))
   String Reversal Processor:
    a(a(c(c(x1)))) -> c(c(c(c(a(a(x1))))))
    c(c(c(c(c(c(x1)))))) -> b(b(c(c(b(b(x1))))))
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {8,1}
      transitions:
       f30() -> 2*
       c0(10) -> 11*
       c0(5) -> 6*
       c0(7) -> 1*
       c0(4) -> 5*
       c0(11) -> 12*
       c0(6) -> 7*
       a0(2) -> 3*
       a0(3) -> 4*
       b0(12) -> 13*
       b0(2) -> 9*
       b0(9) -> 10*
       b0(13) -> 8*
       b1(40) -> 41*
       b1(35) -> 36*
       b1(15) -> 16*
       b1(67) -> 68*
       b1(47) -> 48*
       b1(42) -> 43*
       b1(64) -> 65*
       b1(39) -> 40*
       b1(19) -> 20*
       b1(14) -> 15*
       b1(46) -> 47*
       b1(36) -> 37*
       b1(68) -> 69*
       b1(63) -> 64*
       b1(43) -> 44*
       b1(18) -> 19*
       c1(65) -> 66*
       c1(45) -> 46*
       c1(37) -> 38*
       c1(17) -> 18*
       c1(44) -> 45*
       c1(66) -> 67*
       c1(16) -> 17*
       c1(38) -> 39*
       1 -> 3,4
       4 -> 63*
       5 -> 42*
       6 -> 14*
       7 -> 35*
       20 -> 1*
       41 -> 5*
       48 -> 7*
       69 -> 6*
     problem:
      
     Qed