1.48/0.70	NO
1.48/0.70	
1.48/0.70	Problem:
1.48/0.70	 a(a(a(a(x1)))) -> a(c(a(c(c(x1)))))
1.48/0.70	 c(c(c(x1))) -> a(a(a(x1)))
1.48/0.70	
1.48/0.70	Proof:
1.48/0.70	 String Reversal Processor:
1.48/0.70	  a(a(a(a(x1)))) -> c(c(a(c(a(x1)))))
1.48/0.70	  c(c(c(x1))) -> a(a(a(x1)))
1.48/0.70	  Unfolding Processor:
1.48/0.70	   loop length: 7
1.48/0.70	   terms:
1.48/0.70	    a(a(a(a(a(a(a(a(a(a(a(a(a(x2583)))))))))))))
1.48/0.70	    c(c(a(c(a(a(a(a(a(a(a(a(a(a(x2583))))))))))))))
1.48/0.70	    c(c(a(c(c(c(a(c(a(a(a(a(a(a(a(x2583)))))))))))))))
1.48/0.70	    c(c(a(a(a(a(a(c(a(a(a(a(a(a(a(x2583)))))))))))))))
1.48/0.70	    c(c(a(a(a(a(a(c(c(c(a(c(a(a(a(a(x2583))))))))))))))))
1.48/0.70	    c(c(a(a(a(a(a(a(a(a(a(c(a(a(a(a(x2583))))))))))))))))
1.48/0.70	    c(c(a(a(a(a(a(a(a(a(a(c(c(c(a(c(a(x2583)))))))))))))))))
1.48/0.70	   context: c(c([]))
1.48/0.70	   substitution:
1.48/0.70	    x2583 -> c(a(x2583))
1.48/0.70	   Qed
1.48/0.70	EOF