YES

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

Proof:
 String Reversal Processor:
  s(s(a(a(x1)))) -> a(a(s(s(x1))))
  s(s(b(b(a(a(b(b(x1)))))))) -> a(a(s(s(b(b(a(a(x1))))))))
  b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1))))))))
  a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1))))))))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [a](x0) = x0 + 1,
    
    [s](x0) = 2x0 + 2,
    
    [b](x0) = x0 + 1
   orientation:
    s(s(a(a(x1)))) = 4x1 + 14 >= 4x1 + 8 = a(a(s(s(x1))))
    
    s(s(b(b(a(a(b(b(x1)))))))) = 4x1 + 30 >= 4x1 + 24 = a(a(s(s(b(b(a(a(x1))))))))
    
    b(b(b(b(a(a(b(b(x1)))))))) = x1 + 8 >= x1 + 8 = b(b(a(a(b(b(a(a(x1))))))))
    
    a(a(a(a(b(b(a(a(x1)))))))) = x1 + 8 >= x1 + 8 = a(a(b(b(a(a(b(b(x1))))))))
   problem:
    b(b(b(b(a(a(b(b(x1)))))))) -> b(b(a(a(b(b(a(a(x1))))))))
    a(a(a(a(b(b(a(a(x1)))))))) -> a(a(b(b(a(a(b(b(x1))))))))
   Bounds Processor:
    bound: 0
    enrichment: match
    automaton:
     final states: {10,1}
     transitions:
      a0(7) -> 8*
      a0(16) -> 17*
      a0(3) -> 4*
      a0(2) -> 3*
      a0(12) -> 13*
      a0(17) -> 10*
      a0(13) -> 14*
      a0(6) -> 7*
      f30() -> 2*
      b0(11) -> 12*
      b0(4) -> 5*
      b0(14) -> 15*
      b0(15) -> 16*
      b0(2) -> 11*
      b0(8) -> 9*
      b0(9) -> 1*
      b0(5) -> 6*
      1 -> 11,12
      10 -> 3,4
    problem:
     
    Qed