3.18/1.10	NO
3.18/1.10	
3.18/1.10	Problem:
3.18/1.10	 a(x1) -> x1
3.18/1.10	 a(b(x1)) -> b(a(a(c(b(a(x1))))))
3.18/1.10	 b(x1) -> x1
3.18/1.10	 c(c(x1)) -> x1
3.18/1.10	
3.18/1.10	Proof:
3.18/1.10	 Unfolding Processor:
3.18/1.10	  loop length: 7
3.18/1.10	  terms:
3.18/1.10	   a(b(c(b(x16624))))
3.18/1.10	   b(a(a(c(b(a(c(b(x16624))))))))
3.18/1.10	   b(a(a(c(a(c(b(x16624)))))))
3.18/1.10	   b(a(a(c(c(b(x16624))))))
3.18/1.10	   b(a(a(b(x16624))))
3.18/1.10	   b(a(b(a(a(c(b(a(x16624))))))))
3.18/1.10	   b(a(b(a(c(b(a(x16624)))))))
3.18/1.10	  context: b([])
3.18/1.10	  substitution:
3.18/1.10	   x16624 -> a(x16624)
3.18/1.10	  Qed
3.18/1.10	EOF