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:
 Matrix Interpretation Processor: dim=4
  
  interpretation:
             [1 0 0 0]  
             [0 1 0 0]  
   [b](x0) = [0 0 0 0]x0
             [0 1 0 0]  ,
   
             [1 0 0 1]     [0]
             [0 0 1 0]     [0]
   [a](x0) = [0 1 1 0]x0 + [1]
             [0 0 0 0]     [0],
   
             [1 0 0 0]  
             [0 0 0 0]  
   [c](x0) = [0 0 0 0]x0
             [0 0 0 1]  
  orientation:
                                [1 1 1 1]     [1]    [1 1 0 0]     [0]                             
                                [0 1 1 0]     [2]    [0 0 0 0]     [1]                             
   a(a(b(b(b(b(a(a(x1)))))))) = [0 1 1 0]x1 + [3] >= [0 0 0 0]x1 + [2] = a(a(c(c(a(a(b(b(x1))))))))
                                [0 0 0 0]     [0]    [0 0 0 0]     [0]                             
   
                    [1 0 0 1]     [0]    [1 0 0 1]                         
                    [0 0 0 0]     [1]    [0 0 0 0]                         
   a(a(c(c(x1)))) = [0 0 0 0]x1 + [2] >= [0 0 0 0]x1 = c(c(c(c(a(a(x1))))))
                    [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 0 0]x1 >= [0 0 0 0]x1 = b(b(c(c(b(b(x1))))))
                          [0 0 0 1]      [0 0 0 0]                         
  problem:
   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:
     a0(3) -> 4*
     a0(2) -> 3*
     b1(68) -> 69*
     b1(22) -> 23*
     b1(33) -> 34*
     b1(29) -> 30*
     b1(32) -> 33*
     b1(53) -> 54*
     b1(26) -> 27*
     b1(63) -> 64*
     b1(21) -> 22*
     b1(25) -> 26*
     b1(67) -> 68*
     b1(64) -> 65*
     b1(50) -> 51*
     b1(54) -> 55*
     b1(49) -> 50*
     b1(28) -> 29*
     b0(13) -> 8*
     b0(12) -> 13*
     b0(2) -> 9*
     b0(9) -> 10*
     c0(11) -> 12*
     c0(4) -> 5*
     c0(6) -> 7*
     c0(10) -> 11*
     c0(7) -> 1*
     c0(5) -> 6*
     f30() -> 2*
     c1(52) -> 53*
     c1(66) -> 67*
     c1(24) -> 25*
     c1(31) -> 32*
     c1(51) -> 52*
     c1(30) -> 31*
     c1(23) -> 24*
     c1(65) -> 66*
     27 -> 7*
     69 -> 5*
     7 -> 63*
     4 -> 49*
     6 -> 28*
     1 -> 3,4
     55 -> 6*
     34 -> 1,4
     5 -> 21*
   problem:
    
   Qed