NO

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

Proof:
 Nonconfluence Processor:
  terms: a(b(b(b(x13)))) *<- a(a(a(x13))) ->* b(b(b(a(x13))))
  Qed