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(x197)) *<- c(a(a(x197))) ->* a(b(c(x197))) Qed