0.00/0.04	MAYBE
0.00/0.04	
0.00/0.04	DP problem for innermost termination.
0.00/0.04	P =
0.00/0.04	  f4# -> f3# [0 <= 0]
0.00/0.04	  f3# -> f1# [0 <= 0]
0.00/0.04	  f2# -> f1# [0 <= 0]
0.00/0.04	  f1# -> f2# [0 <= 0]
0.00/0.04	R = 
0.00/0.04	  f4 -> f3 [0 <= 0]
0.00/0.04	  f3 -> f1 [0 <= 0]
0.00/0.04	  f2 -> f1 [0 <= 0]
0.00/0.04	  f1 -> f2 [0 <= 0]
0.00/0.04	
0.00/0.04	The dependency graph for this problem is:
0.00/0.04	  0 -> 1
0.00/0.04	  1 -> 3
0.00/0.04	  2 -> 3
0.00/0.04	  3 -> 2
0.00/0.04	Where:
0.00/0.04	  0) f4# -> f3# [0 <= 0]
0.00/0.04	  1) f3# -> f1# [0 <= 0]
0.00/0.04	  2) f2# -> f1# [0 <= 0]
0.00/0.04	  3) f1# -> f2# [0 <= 0]
0.00/0.04	
0.00/0.04	We have the following SCCs.
0.00/0.04	  { 2, 3 }
0.00/0.04	
0.00/0.04	DP problem for innermost termination.
0.00/0.04	P =
0.00/0.04	  f2# -> f1# [0 <= 0]
0.00/0.04	  f1# -> f2# [0 <= 0]
0.00/0.04	R = 
0.00/0.04	  f4 -> f3 [0 <= 0]
0.00/0.04	  f3 -> f1 [0 <= 0]
0.00/0.04	  f2 -> f1 [0 <= 0]
0.00/0.04	  f1 -> f2 [0 <= 0]
0.00/0.04	
0.00/3.02	EOF