0.62/0.65	MAYBE
0.62/0.65	
0.62/0.65	DP problem for innermost termination.
0.62/0.65	P =
0.62/0.65	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.62/0.65	  f2#(I0, I1) -> f2#(29, I2) [25 = I0]
0.62/0.65	  f2#(I3, I4) -> f2#(20, I5) [30 <= I3 - 1]
0.62/0.65	  f2#(I6, I7) -> f2#(I6 - 1, I8) [I6 <= 30 /\ 25 <= I6 - 1]
0.62/0.65	  f2#(I9, I10) -> f2#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30]
0.62/0.65	  f1#(I12, I13) -> f2#(I13 + 20, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 <= I13 + 20 - 1]
0.62/0.65	R = 
0.62/0.65	  init(x1, x2) -> f1(rnd1, rnd2) 
0.62/0.65	  f2(I0, I1) -> f2(29, I2) [25 = I0]
0.62/0.65	  f2(I3, I4) -> f2(20, I5) [30 <= I3 - 1]
0.62/0.65	  f2(I6, I7) -> f2(I6 - 1, I8) [I6 <= 30 /\ 25 <= I6 - 1]
0.62/0.65	  f2(I9, I10) -> f2(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30]
0.62/0.65	  f1(I12, I13) -> f2(I13 + 20, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 <= I13 + 20 - 1]
0.62/0.65	
0.62/0.65	The dependency graph for this problem is:
0.62/0.65	  0 -> 5
0.62/0.65	  1 -> 3
0.62/0.65	  2 -> 4
0.62/0.65	  3 -> 1, 3
0.62/0.65	  4 -> 4
0.62/0.65	  5 -> 1, 2, 3, 4
0.62/0.65	Where:
0.62/0.65	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.62/0.65	  1) f2#(I0, I1) -> f2#(29, I2) [25 = I0]
0.62/0.65	  2) f2#(I3, I4) -> f2#(20, I5) [30 <= I3 - 1]
0.62/0.65	  3) f2#(I6, I7) -> f2#(I6 - 1, I8) [I6 <= 30 /\ 25 <= I6 - 1]
0.62/0.65	  4) f2#(I9, I10) -> f2#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30]
0.62/0.65	  5) f1#(I12, I13) -> f2#(I13 + 20, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 <= I13 + 20 - 1]
0.62/0.65	
0.62/0.65	We have the following SCCs.
0.62/0.65	  { 4 }
0.62/0.65	  { 1, 3 }
0.62/0.65	
0.62/0.65	DP problem for innermost termination.
0.62/0.65	P =
0.62/0.65	  f2#(I0, I1) -> f2#(29, I2) [25 = I0]
0.62/0.65	  f2#(I6, I7) -> f2#(I6 - 1, I8) [I6 <= 30 /\ 25 <= I6 - 1]
0.62/0.65	R = 
0.62/0.65	  init(x1, x2) -> f1(rnd1, rnd2) 
0.62/0.65	  f2(I0, I1) -> f2(29, I2) [25 = I0]
0.62/0.65	  f2(I3, I4) -> f2(20, I5) [30 <= I3 - 1]
0.62/0.65	  f2(I6, I7) -> f2(I6 - 1, I8) [I6 <= 30 /\ 25 <= I6 - 1]
0.62/0.65	  f2(I9, I10) -> f2(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30]
0.62/0.65	  f1(I12, I13) -> f2(I13 + 20, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 <= I13 + 20 - 1]
0.62/0.65	
0.62/3.63	EOF