0.00/0.40	MAYBE
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.40	  f2#(I0, I1) -> f2#(I0 - 2, I2) [I0 <= -1 /\ I0 - 2 <= I0 - 1 /\ I0 <= 0 /\ I0 <= 2]
0.00/0.40	  f1#(I3, I4) -> f2#(-1 * I4, I5) [0 <= I3 - 1 /\ -1 <= I4 - 1 /\ -1 * I4 <= 0]
0.00/0.40	R = 
0.00/0.40	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.40	  f2(I0, I1) -> f2(I0 - 2, I2) [I0 <= -1 /\ I0 - 2 <= I0 - 1 /\ I0 <= 0 /\ I0 <= 2]
0.00/0.40	  f1(I3, I4) -> f2(-1 * I4, I5) [0 <= I3 - 1 /\ -1 <= I4 - 1 /\ -1 * I4 <= 0]
0.00/0.40	
0.00/0.40	The dependency graph for this problem is:
0.00/0.40	  0 -> 2
0.00/0.40	  1 -> 1
0.00/0.40	  2 -> 1
0.00/0.40	Where:
0.00/0.40	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.40	  1) f2#(I0, I1) -> f2#(I0 - 2, I2) [I0 <= -1 /\ I0 - 2 <= I0 - 1 /\ I0 <= 0 /\ I0 <= 2]
0.00/0.40	  2) f1#(I3, I4) -> f2#(-1 * I4, I5) [0 <= I3 - 1 /\ -1 <= I4 - 1 /\ -1 * I4 <= 0]
0.00/0.40	
0.00/0.40	We have the following SCCs.
0.00/0.40	  { 1 }
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  f2#(I0, I1) -> f2#(I0 - 2, I2) [I0 <= -1 /\ I0 - 2 <= I0 - 1 /\ I0 <= 0 /\ I0 <= 2]
0.00/0.40	R = 
0.00/0.40	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.40	  f2(I0, I1) -> f2(I0 - 2, I2) [I0 <= -1 /\ I0 - 2 <= I0 - 1 /\ I0 <= 0 /\ I0 <= 2]
0.00/0.40	  f1(I3, I4) -> f2(-1 * I4, I5) [0 <= I3 - 1 /\ -1 <= I4 - 1 /\ -1 * I4 <= 0]
0.00/0.40	
0.00/3.38	EOF