0.00/0.46	MAYBE
0.00/0.46	
0.00/0.46	DP problem for innermost termination.
0.00/0.46	P =
0.00/0.46	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.46	  f2#(I0, I1) -> f2#(I0 - 1, I2) [I0 <= -1 /\ I0 - 1 <= I0 - 1 /\ I0 <= 1 /\ I0 <= 0]
0.00/0.46	  f1#(I3, I4) -> f2#(-1 * I4, I5) [0 <= I3 - 1 /\ -1 <= I4 - 1 /\ -1 * I4 <= 0]
0.00/0.46	R = 
0.00/0.46	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.46	  f2(I0, I1) -> f2(I0 - 1, I2) [I0 <= -1 /\ I0 - 1 <= I0 - 1 /\ I0 <= 1 /\ I0 <= 0]
0.00/0.46	  f1(I3, I4) -> f2(-1 * I4, I5) [0 <= I3 - 1 /\ -1 <= I4 - 1 /\ -1 * I4 <= 0]
0.00/0.46	
0.00/0.46	The dependency graph for this problem is:
0.00/0.46	  0 -> 2
0.00/0.46	  1 -> 1
0.00/0.46	  2 -> 1
0.00/0.46	Where:
0.00/0.46	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.46	  1) f2#(I0, I1) -> f2#(I0 - 1, I2) [I0 <= -1 /\ I0 - 1 <= I0 - 1 /\ I0 <= 1 /\ I0 <= 0]
0.00/0.46	  2) f1#(I3, I4) -> f2#(-1 * I4, I5) [0 <= I3 - 1 /\ -1 <= I4 - 1 /\ -1 * I4 <= 0]
0.00/0.46	
0.00/0.46	We have the following SCCs.
0.00/0.46	  { 1 }
0.00/0.46	
0.00/0.46	DP problem for innermost termination.
0.00/0.46	P =
0.00/0.46	  f2#(I0, I1) -> f2#(I0 - 1, I2) [I0 <= -1 /\ I0 - 1 <= I0 - 1 /\ I0 <= 1 /\ I0 <= 0]
0.00/0.46	R = 
0.00/0.46	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.46	  f2(I0, I1) -> f2(I0 - 1, I2) [I0 <= -1 /\ I0 - 1 <= I0 - 1 /\ I0 <= 1 /\ I0 <= 0]
0.00/0.46	  f1(I3, I4) -> f2(-1 * I4, I5) [0 <= I3 - 1 /\ -1 <= I4 - 1 /\ -1 * I4 <= 0]
0.00/0.46	
0.00/3.44	EOF