1.84/0.80	NO
1.84/0.80	
1.84/0.80	Problem:
1.84/0.80	 a(x1) -> x1
1.84/0.80	 a(b(x1)) -> b(a(a(c(a(a(x1))))))
1.84/0.80	 c(c(x1)) -> b(x1)
1.84/0.80	
1.84/0.80	Proof:
1.84/0.80	 String Reversal Processor:
1.84/0.80	  a(x1) -> x1
1.84/0.80	  b(a(x1)) -> a(a(c(a(a(b(x1))))))
1.84/0.80	  c(c(x1)) -> b(x1)
1.84/0.80	  Unfolding Processor:
1.84/0.80	   loop length: 7
1.84/0.80	   terms:
1.84/0.80	    b(a(a(x5241)))
1.84/0.80	    a(a(c(a(a(b(a(x5241)))))))
1.84/0.80	    a(a(c(a(b(a(x5241))))))
1.84/0.80	    a(a(c(b(a(x5241)))))
1.84/0.80	    a(a(c(a(a(c(a(a(b(x5241)))))))))
1.84/0.80	    a(a(c(a(c(a(a(b(x5241))))))))
1.84/0.80	    a(a(c(c(a(a(b(x5241)))))))
1.84/0.80	   context: a(a([]))
1.84/0.80	   substitution:
1.84/0.80	    x5241 -> b(x5241)
1.84/0.80	   Qed
1.84/0.80	EOF