0.00/0.07 MAYBE 0.00/0.07 0.00/0.07 DP problem for innermost termination. 0.00/0.07 P = 0.00/0.07 f5# -> f4# [0 <= 0] 0.00/0.07 f4# -> f3# [0 <= 0] 0.00/0.07 f3# -> f1# [0 <= 0] 0.00/0.07 f1# -> f3# [0 <= 0] 0.00/0.07 f2# -> f1# [0 <= 0] 0.00/0.07 f1# -> f2# [0 <= 0] 0.00/0.07 R = 0.00/0.07 f5 -> f4 [0 <= 0] 0.00/0.07 f4 -> f3 [0 <= 0] 0.00/0.07 f3 -> f1 [0 <= 0] 0.00/0.07 f1 -> f3 [0 <= 0] 0.00/0.07 f2 -> f1 [0 <= 0] 0.00/0.07 f1 -> f2 [0 <= 0] 0.00/0.07 0.00/0.07 The dependency graph for this problem is: 0.00/0.07 0 -> 1 0.00/0.07 1 -> 2 0.00/0.07 2 -> 3, 5 0.00/0.07 3 -> 2 0.00/0.07 4 -> 3, 5 0.00/0.07 5 -> 4 0.00/0.07 Where: 0.00/0.07 0) f5# -> f4# [0 <= 0] 0.00/0.07 1) f4# -> f3# [0 <= 0] 0.00/0.07 2) f3# -> f1# [0 <= 0] 0.00/0.07 3) f1# -> f3# [0 <= 0] 0.00/0.07 4) f2# -> f1# [0 <= 0] 0.00/0.07 5) f1# -> f2# [0 <= 0] 0.00/0.07 0.00/0.07 We have the following SCCs. 0.00/0.07 { 2, 3, 4, 5 } 0.00/0.07 0.00/0.07 DP problem for innermost termination. 0.00/0.07 P = 0.00/0.07 f3# -> f1# [0 <= 0] 0.00/0.07 f1# -> f3# [0 <= 0] 0.00/0.07 f2# -> f1# [0 <= 0] 0.00/0.07 f1# -> f2# [0 <= 0] 0.00/0.07 R = 0.00/0.07 f5 -> f4 [0 <= 0] 0.00/0.07 f4 -> f3 [0 <= 0] 0.00/0.07 f3 -> f1 [0 <= 0] 0.00/0.07 f1 -> f3 [0 <= 0] 0.00/0.07 f2 -> f1 [0 <= 0] 0.00/0.07 f1 -> f2 [0 <= 0] 0.00/0.07 0.00/3.05 EOF