YES

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

Proof:
 String Reversal Processor:
  a(a(b(x1))) -> c(b(a(x1)))
  a(c(x1)) -> c(a(x1))
  a(c(b(x1))) -> c(b(a(x1)))
  b(c(x1)) -> a(b(x1))
  b(c(a(x1))) -> a(b(c(x1)))
  KBO Processor:
   weight function:
    w0 = 1
    w(c) = w(a) = 1
    w(b) = 0
   precedence:
    b > a > c
   problem:
    
   Qed