0.00/0.52	MAYBE
0.00/0.52	
0.00/0.52	DP problem for innermost termination.
0.00/0.52	P =
0.00/0.52	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.52	  f2#(I0, I1) -> f2#(I0 + 2, I1 + 1) [-1 <= I1 - 1 /\ -1 <= I0 - 1 /\ I1 <= I0]
0.00/0.52	  f2#(I2, I3) -> f2#(I2 + 4, I3) [-1 <= I2 - 1 /\ I2 <= I3 - 1]
0.00/0.52	  f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ 1 <= I5 - 1 /\ -1 <= I7 - 1]
0.00/0.52	R = 
0.00/0.52	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.52	  f2(I0, I1) -> f2(I0 + 2, I1 + 1) [-1 <= I1 - 1 /\ -1 <= I0 - 1 /\ I1 <= I0]
0.00/0.52	  f2(I2, I3) -> f2(I2 + 4, I3) [-1 <= I2 - 1 /\ I2 <= I3 - 1]
0.00/0.52	  f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ 1 <= I5 - 1 /\ -1 <= I7 - 1]
0.00/0.52	
0.00/0.52	The dependency graph for this problem is:
0.00/0.52	  0 -> 3
0.00/0.52	  1 -> 1
0.00/0.52	  2 -> 1, 2
0.00/0.52	  3 -> 1, 2
0.00/0.52	Where:
0.00/0.52	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.52	  1) f2#(I0, I1) -> f2#(I0 + 2, I1 + 1) [-1 <= I1 - 1 /\ -1 <= I0 - 1 /\ I1 <= I0]
0.00/0.52	  2) f2#(I2, I3) -> f2#(I2 + 4, I3) [-1 <= I2 - 1 /\ I2 <= I3 - 1]
0.00/0.52	  3) f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ 1 <= I5 - 1 /\ -1 <= I7 - 1]
0.00/0.52	
0.00/0.52	We have the following SCCs.
0.00/0.52	  { 2 }
0.00/0.52	  { 1 }
0.00/0.52	
0.00/0.52	DP problem for innermost termination.
0.00/0.52	P =
0.00/0.52	  f2#(I0, I1) -> f2#(I0 + 2, I1 + 1) [-1 <= I1 - 1 /\ -1 <= I0 - 1 /\ I1 <= I0]
0.00/0.52	R = 
0.00/0.52	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.52	  f2(I0, I1) -> f2(I0 + 2, I1 + 1) [-1 <= I1 - 1 /\ -1 <= I0 - 1 /\ I1 <= I0]
0.00/0.52	  f2(I2, I3) -> f2(I2 + 4, I3) [-1 <= I2 - 1 /\ I2 <= I3 - 1]
0.00/0.52	  f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ 1 <= I5 - 1 /\ -1 <= I7 - 1]
0.00/0.52	
0.00/3.49	EOF