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	  f2#(I0, I1) -> f2#(I1, I1 - 1) [I0 <= 0 /\ I1 <= I0 - 1 /\ I1 <= -1 /\ I1 <= 0]
0.00/0.45	  f1#(I2, I3) -> f2#(-1 * I3 - 1, -1 * I3 - 2) [0 <= I2 - 1 /\ -1 * I3 <= 0 /\ -1 * I3 <= 1 /\ -1 <= I3 - 1 /\ -1 * I3 <= 0]
0.00/0.45	R = 
0.00/0.45	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.45	  f2(I0, I1) -> f2(I1, I1 - 1) [I0 <= 0 /\ I1 <= I0 - 1 /\ I1 <= -1 /\ I1 <= 0]
0.00/0.45	  f1(I2, I3) -> f2(-1 * I3 - 1, -1 * I3 - 2) [0 <= I2 - 1 /\ -1 * I3 <= 0 /\ -1 * I3 <= 1 /\ -1 <= I3 - 1 /\ -1 * I3 <= 0]
0.00/0.45	
0.00/0.45	The dependency graph for this problem is:
0.00/0.45	  0 -> 2
0.00/0.45	  1 -> 1
0.00/0.45	  2 -> 1
0.00/0.45	Where:
0.00/0.45	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.45	  1) f2#(I0, I1) -> f2#(I1, I1 - 1) [I0 <= 0 /\ I1 <= I0 - 1 /\ I1 <= -1 /\ I1 <= 0]
0.00/0.45	  2) f1#(I2, I3) -> f2#(-1 * I3 - 1, -1 * I3 - 2) [0 <= I2 - 1 /\ -1 * I3 <= 0 /\ -1 * I3 <= 1 /\ -1 <= I3 - 1 /\ -1 * I3 <= 0]
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	  f2#(I0, I1) -> f2#(I1, I1 - 1) [I0 <= 0 /\ I1 <= I0 - 1 /\ I1 <= -1 /\ I1 <= 0]
0.00/0.45	R = 
0.00/0.45	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.45	  f2(I0, I1) -> f2(I1, I1 - 1) [I0 <= 0 /\ I1 <= I0 - 1 /\ I1 <= -1 /\ I1 <= 0]
0.00/0.45	  f1(I2, I3) -> f2(-1 * I3 - 1, -1 * I3 - 2) [0 <= I2 - 1 /\ -1 * I3 <= 0 /\ -1 * I3 <= 1 /\ -1 <= I3 - 1 /\ -1 * I3 <= 0]
0.00/0.45	
0.00/3.43	EOF