NO Problem: a(b(a(x1))) -> b(b(b(x1))) b(a(b(x1))) -> a(b(a(x1))) Proof: Unfolding Processor: loop length: 6 terms: a(b(a(a(b(b(x863)))))) b(b(b(a(b(b(x863)))))) b(b(a(b(a(b(x863)))))) b(b(a(a(b(a(x863)))))) b(b(a(b(b(b(x863)))))) b(a(b(a(b(b(x863)))))) context: [] substitution: x863 -> x863 Qed