0.00/0.46	MAYBE
0.00/0.46	
0.00/0.46	DP problem for innermost termination.
0.00/0.46	P =
0.00/0.46	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.46	  f3#(I0, I1) -> f3#(I0 + 1, I2) [I0 <= I0 + 1 - 1 /\ 5 <= I0 - 1]
0.00/0.46	  f2#(I3, I4) -> f3#(I3, I5) [I3 <= 9]
0.00/0.46	  f2#(I6, I7) -> f2#(I6 + 1, I8) [I6 <= I6 + 1 - 1 /\ I6 <= 9 /\ -1 <= I6 - 1]
0.00/0.46	  f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.46	R = 
0.00/0.46	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.46	  f3(I0, I1) -> f3(I0 + 1, I2) [I0 <= I0 + 1 - 1 /\ 5 <= I0 - 1]
0.00/0.46	  f2(I3, I4) -> f3(I3, I5) [I3 <= 9]
0.00/0.46	  f2(I6, I7) -> f2(I6 + 1, I8) [I6 <= I6 + 1 - 1 /\ I6 <= 9 /\ -1 <= I6 - 1]
0.00/0.46	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.46	
0.00/0.46	The dependency graph for this problem is:
0.00/0.46	  0 -> 4
0.00/0.46	  1 -> 1
0.00/0.46	  2 -> 1
0.00/0.46	  3 -> 2, 3
0.00/0.46	  4 -> 2, 3
0.00/0.46	Where:
0.00/0.46	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.46	  1) f3#(I0, I1) -> f3#(I0 + 1, I2) [I0 <= I0 + 1 - 1 /\ 5 <= I0 - 1]
0.00/0.46	  2) f2#(I3, I4) -> f3#(I3, I5) [I3 <= 9]
0.00/0.46	  3) f2#(I6, I7) -> f2#(I6 + 1, I8) [I6 <= I6 + 1 - 1 /\ I6 <= 9 /\ -1 <= I6 - 1]
0.00/0.46	  4) f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.46	
0.00/0.46	We have the following SCCs.
0.00/0.46	  { 3 }
0.00/0.46	  { 1 }
0.00/0.46	
0.00/0.46	DP problem for innermost termination.
0.00/0.46	P =
0.00/0.46	  f3#(I0, I1) -> f3#(I0 + 1, I2) [I0 <= I0 + 1 - 1 /\ 5 <= I0 - 1]
0.00/0.46	R = 
0.00/0.46	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.46	  f3(I0, I1) -> f3(I0 + 1, I2) [I0 <= I0 + 1 - 1 /\ 5 <= I0 - 1]
0.00/0.46	  f2(I3, I4) -> f3(I3, I5) [I3 <= 9]
0.00/0.46	  f2(I6, I7) -> f2(I6 + 1, I8) [I6 <= I6 + 1 - 1 /\ I6 <= 9 /\ -1 <= I6 - 1]
0.00/0.46	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.46	
0.00/3.44	EOF