0.00/0.45	MAYBE
0.00/0.45	
0.00/0.45	DP problem for innermost termination.
0.00/0.45	P =
0.00/0.45	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.45	  f3#(I0, I1) -> f3#(I0 + 1, I2) [I0 <= I0 + 1 - 1 /\ 0 <= I0 - 1]
0.00/0.45	  f2#(I3, I4) -> f2#(1, I5) [0 = I3]
0.00/0.45	  f2#(I6, I7) -> f3#(I6, I8) [I6 <= 9]
0.00/0.45	  f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.45	R = 
0.00/0.45	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.45	  f3(I0, I1) -> f3(I0 + 1, I2) [I0 <= I0 + 1 - 1 /\ 0 <= I0 - 1]
0.00/0.45	  f2(I3, I4) -> f2(1, I5) [0 = I3]
0.00/0.45	  f2(I6, I7) -> f3(I6, I8) [I6 <= 9]
0.00/0.45	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.45	
0.00/0.45	The dependency graph for this problem is:
0.00/0.45	  0 -> 4
0.00/0.45	  1 -> 1
0.00/0.45	  2 -> 3
0.00/0.45	  3 -> 1
0.00/0.45	  4 -> 2, 3
0.00/0.45	Where:
0.00/0.45	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.45	  1) f3#(I0, I1) -> f3#(I0 + 1, I2) [I0 <= I0 + 1 - 1 /\ 0 <= I0 - 1]
0.00/0.45	  2) f2#(I3, I4) -> f2#(1, I5) [0 = I3]
0.00/0.45	  3) f2#(I6, I7) -> f3#(I6, I8) [I6 <= 9]
0.00/0.45	  4) f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.45	
0.00/0.45	We have the following SCCs.
0.00/0.45	  { 1 }
0.00/0.45	
0.00/0.45	DP problem for innermost termination.
0.00/0.45	P =
0.00/0.45	  f3#(I0, I1) -> f3#(I0 + 1, I2) [I0 <= I0 + 1 - 1 /\ 0 <= I0 - 1]
0.00/0.45	R = 
0.00/0.45	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.45	  f3(I0, I1) -> f3(I0 + 1, I2) [I0 <= I0 + 1 - 1 /\ 0 <= I0 - 1]
0.00/0.45	  f2(I3, I4) -> f2(1, I5) [0 = I3]
0.00/0.45	  f2(I6, I7) -> f3(I6, I8) [I6 <= 9]
0.00/0.45	  f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1]
0.00/0.45	
0.00/3.43	EOF