YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> b(b(b(x1)))
  a(b(x1)) -> a(a(a(x1)))
  a(x1) -> x1
  b(x1) -> x1
  Bounds Processor:
   bound: 0
   enrichment: match
   automaton:
    final states: {2,5,1}
    transitions:
     f20() -> 2*
     b0(2) -> 3*
     b0(4) -> 1*
     b0(3) -> 4*
     a0(6) -> 7*
     a0(2) -> 6*
     a0(7) -> 5*
     7 -> 5*
     4 -> 1*
     2 -> 3,6
     6 -> 7*
     1 -> 3,4
     5 -> 6,7
     3 -> 4*
   problem:
    
   Qed