NO

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

Proof:
 Nonconfluence Processor:
  terms: c(b(x1013)) *<- c(a(a(x1013))) ->* a(b(c(x1013)))
  Qed