YES

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

Proof:
 String Reversal Processor:
  s(a(x1)) -> a(s(x1))
  s(b(a(b(x1)))) -> a(s(b(a(x1))))
  b(b(a(b(x1)))) -> s(c(x1))
  s(c(x1)) -> b(a(b(a(x1))))
  a(a(b(a(x1)))) -> a(b(a(b(x1))))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [a](x0) = 2x0 + 2,
    
    [c](x0) = 8x0 + 15,
    
    [s](x0) = 2x0,
    
    [b](x0) = 2x0 + 2
   orientation:
    s(a(x1)) = 4x1 + 4 >= 4x1 + 2 = a(s(x1))
    
    s(b(a(b(x1)))) = 16x1 + 28 >= 16x1 + 26 = a(s(b(a(x1))))
    
    b(b(a(b(x1)))) = 16x1 + 30 >= 16x1 + 30 = s(c(x1))
    
    s(c(x1)) = 16x1 + 30 >= 16x1 + 30 = b(a(b(a(x1))))
    
    a(a(b(a(x1)))) = 16x1 + 30 >= 16x1 + 30 = a(b(a(b(x1))))
   problem:
    b(b(a(b(x1)))) -> s(c(x1))
    s(c(x1)) -> b(a(b(a(x1))))
    a(a(b(a(x1)))) -> a(b(a(b(x1))))
   Bounds Processor:
    bound: 1
    enrichment: match
    automaton:
     final states: {8,4,1}
     transitions:
      s0(3) -> 1*
      b0(5) -> 6*
      b0(7) -> 4*
      b0(10) -> 11*
      b0(2) -> 9*
      a1(17) -> 18*
      a1(19) -> 20*
      b1(20) -> 21*
      b1(18) -> 19*
      a0(2) -> 5*
      a0(9) -> 10*
      a0(6) -> 7*
      a0(11) -> 8*
      f40() -> 2*
      c0(2) -> 3*
      2 -> 17*
      21 -> 1*
      8 -> 18,5
      1 -> 9*
    problem:
     
    Qed