0.00/0.44	MAYBE
0.00/0.44	
0.00/0.44	DP problem for innermost termination.
0.00/0.44	P =
0.00/0.44	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.44	  f2#(I0, I1, I2) -> f2#(I3, I4, I2) [-1 <= y1 - 1 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I3 <= I0 /\ 0 <= I0 - 1 /\ 0 <= I3 - 1 /\ I1 + y1 = I4]
0.00/0.44	  f1#(I5, I6, I7) -> f2#(I8, 0, I6) [0 <= I8 - 1 /\ 0 <= I5 - 1 /\ -1 <= I6 - 1 /\ I8 <= I5]
0.00/0.44	R = 
0.00/0.44	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.44	  f2(I0, I1, I2) -> f2(I3, I4, I2) [-1 <= y1 - 1 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I3 <= I0 /\ 0 <= I0 - 1 /\ 0 <= I3 - 1 /\ I1 + y1 = I4]
0.00/0.44	  f1(I5, I6, I7) -> f2(I8, 0, I6) [0 <= I8 - 1 /\ 0 <= I5 - 1 /\ -1 <= I6 - 1 /\ I8 <= I5]
0.00/0.44	
0.00/0.44	The dependency graph for this problem is:
0.00/0.44	  0 -> 2
0.00/0.44	  1 -> 1
0.00/0.44	  2 -> 1
0.00/0.44	Where:
0.00/0.44	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.44	  1) f2#(I0, I1, I2) -> f2#(I3, I4, I2) [-1 <= y1 - 1 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I3 <= I0 /\ 0 <= I0 - 1 /\ 0 <= I3 - 1 /\ I1 + y1 = I4]
0.00/0.44	  2) f1#(I5, I6, I7) -> f2#(I8, 0, I6) [0 <= I8 - 1 /\ 0 <= I5 - 1 /\ -1 <= I6 - 1 /\ I8 <= I5]
0.00/0.44	
0.00/0.44	We have the following SCCs.
0.00/0.44	  { 1 }
0.00/0.44	
0.00/0.44	DP problem for innermost termination.
0.00/0.44	P =
0.00/0.44	  f2#(I0, I1, I2) -> f2#(I3, I4, I2) [-1 <= y1 - 1 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I3 <= I0 /\ 0 <= I0 - 1 /\ 0 <= I3 - 1 /\ I1 + y1 = I4]
0.00/0.44	R = 
0.00/0.44	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.44	  f2(I0, I1, I2) -> f2(I3, I4, I2) [-1 <= y1 - 1 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I3 <= I0 /\ 0 <= I0 - 1 /\ 0 <= I3 - 1 /\ I1 + y1 = I4]
0.00/0.44	  f1(I5, I6, I7) -> f2(I8, 0, I6) [0 <= I8 - 1 /\ 0 <= I5 - 1 /\ -1 <= I6 - 1 /\ I8 <= I5]
0.00/0.44	
0.00/3.42	EOF