0.00/0.50	MAYBE
0.00/0.50	
0.00/0.50	DP problem for innermost termination.
0.00/0.50	P =
0.00/0.50	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.50	  f2#(I0, I1) -> f2#(I0 + 1, I0 * I0 + I0 + I0 + 1) [-1 <= I0 - 1 /\ 9 <= I1 - 1]
0.00/0.50	  f1#(I2, I3) -> f2#(I3, I3 * I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1]
0.00/0.50	R = 
0.00/0.50	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.50	  f2(I0, I1) -> f2(I0 + 1, I0 * I0 + I0 + I0 + 1) [-1 <= I0 - 1 /\ 9 <= I1 - 1]
0.00/0.50	  f1(I2, I3) -> f2(I3, I3 * I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1]
0.00/0.50	
0.00/0.50	The dependency graph for this problem is:
0.00/0.50	  0 -> 2
0.00/0.50	  1 -> 1
0.00/0.50	  2 -> 1
0.00/0.50	Where:
0.00/0.50	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.50	  1) f2#(I0, I1) -> f2#(I0 + 1, I0 * I0 + I0 + I0 + 1) [-1 <= I0 - 1 /\ 9 <= I1 - 1]
0.00/0.50	  2) f1#(I2, I3) -> f2#(I3, I3 * I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1]
0.00/0.50	
0.00/0.50	We have the following SCCs.
0.00/0.50	  { 1 }
0.00/0.50	
0.00/0.50	DP problem for innermost termination.
0.00/0.50	P =
0.00/0.50	  f2#(I0, I1) -> f2#(I0 + 1, I0 * I0 + I0 + I0 + 1) [-1 <= I0 - 1 /\ 9 <= I1 - 1]
0.00/0.50	R = 
0.00/0.50	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.50	  f2(I0, I1) -> f2(I0 + 1, I0 * I0 + I0 + I0 + 1) [-1 <= I0 - 1 /\ 9 <= I1 - 1]
0.00/0.50	  f1(I2, I3) -> f2(I3, I3 * I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1]
0.00/0.50	
0.00/3.48	EOF