0.00/0.38	MAYBE
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.38	  f2#(I0, I1) -> f2#(0, I2) [0 <= I0 - 1 /\ I0 <= 5]
0.00/0.38	  f2#(I3, I4) -> f2#(I3 + 1, I5) [5 <= I3 - 1]
0.00/0.38	  f1#(I6, I7) -> f2#(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.38	R = 
0.00/0.38	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.38	  f2(I0, I1) -> f2(0, I2) [0 <= I0 - 1 /\ I0 <= 5]
0.00/0.38	  f2(I3, I4) -> f2(I3 + 1, I5) [5 <= I3 - 1]
0.00/0.38	  f1(I6, I7) -> f2(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.38	
0.00/0.38	The dependency graph for this problem is:
0.00/0.38	  0 -> 3
0.00/0.38	  1 -> 
0.00/0.38	  2 -> 2
0.00/0.38	  3 -> 1, 2
0.00/0.38	Where:
0.00/0.38	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.38	  1) f2#(I0, I1) -> f2#(0, I2) [0 <= I0 - 1 /\ I0 <= 5]
0.00/0.38	  2) f2#(I3, I4) -> f2#(I3 + 1, I5) [5 <= I3 - 1]
0.00/0.38	  3) f1#(I6, I7) -> f2#(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.38	
0.00/0.38	We have the following SCCs.
0.00/0.38	  { 2 }
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  f2#(I3, I4) -> f2#(I3 + 1, I5) [5 <= I3 - 1]
0.00/0.38	R = 
0.00/0.38	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.38	  f2(I0, I1) -> f2(0, I2) [0 <= I0 - 1 /\ I0 <= 5]
0.00/0.38	  f2(I3, I4) -> f2(I3 + 1, I5) [5 <= I3 - 1]
0.00/0.38	  f1(I6, I7) -> f2(I7, I8) [-1 <= I7 - 1 /\ 0 <= I6 - 1]
0.00/0.38	
0.00/3.36	EOF