0.00/0.49	MAYBE
0.00/0.49	
0.00/0.49	DP problem for innermost termination.
0.00/0.49	P =
0.00/0.49	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.49	  f2#(I0, I1) -> f2#(I0 - 10, I1 - 1) [0 <= I1 - 1 /\ 100 <= I0 - 1]
0.00/0.49	  f2#(I2, I3) -> f2#(I2 + 11, I3 + 1) [0 <= I3 - 1 /\ I2 <= 100]
0.00/0.49	  f1#(I4, I5) -> f2#(I6, 1) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1]
0.00/0.49	R = 
0.00/0.49	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.49	  f2(I0, I1) -> f2(I0 - 10, I1 - 1) [0 <= I1 - 1 /\ 100 <= I0 - 1]
0.00/0.49	  f2(I2, I3) -> f2(I2 + 11, I3 + 1) [0 <= I3 - 1 /\ I2 <= 100]
0.00/0.49	  f1(I4, I5) -> f2(I6, 1) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1]
0.00/0.49	
0.00/0.49	The dependency graph for this problem is:
0.00/0.49	  0 -> 3
0.00/0.49	  1 -> 1, 2
0.00/0.49	  2 -> 1, 2
0.00/0.49	  3 -> 1, 2
0.00/0.49	Where:
0.00/0.49	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.49	  1) f2#(I0, I1) -> f2#(I0 - 10, I1 - 1) [0 <= I1 - 1 /\ 100 <= I0 - 1]
0.00/0.49	  2) f2#(I2, I3) -> f2#(I2 + 11, I3 + 1) [0 <= I3 - 1 /\ I2 <= 100]
0.00/0.49	  3) f1#(I4, I5) -> f2#(I6, 1) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1]
0.00/0.49	
0.00/0.49	We have the following SCCs.
0.00/0.49	  { 1, 2 }
0.00/0.49	
0.00/0.49	DP problem for innermost termination.
0.00/0.49	P =
0.00/0.49	  f2#(I0, I1) -> f2#(I0 - 10, I1 - 1) [0 <= I1 - 1 /\ 100 <= I0 - 1]
0.00/0.49	  f2#(I2, I3) -> f2#(I2 + 11, I3 + 1) [0 <= I3 - 1 /\ I2 <= 100]
0.00/0.49	R = 
0.00/0.49	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.49	  f2(I0, I1) -> f2(I0 - 10, I1 - 1) [0 <= I1 - 1 /\ 100 <= I0 - 1]
0.00/0.49	  f2(I2, I3) -> f2(I2 + 11, I3 + 1) [0 <= I3 - 1 /\ I2 <= 100]
0.00/0.49	  f1(I4, I5) -> f2(I6, 1) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1]
0.00/0.49	
0.00/3.47	EOF