0.00/0.23	MAYBE
0.00/0.23	
0.00/0.23	DP problem for innermost termination.
0.00/0.23	P =
0.00/0.23	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.23	  f2#(I0, I1) -> f2#(I0, I2) 
0.00/0.23	  f1#(I3, I4) -> f2#(I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.23	R = 
0.00/0.23	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.23	  f2(I0, I1) -> f2(I0, I2) 
0.00/0.23	  f1(I3, I4) -> f2(I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.23	
0.00/0.23	The dependency graph for this problem is:
0.00/0.23	  0 -> 2
0.00/0.23	  1 -> 1
0.00/0.23	  2 -> 1
0.00/0.23	Where:
0.00/0.23	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.23	  1) f2#(I0, I1) -> f2#(I0, I2) 
0.00/0.23	  2) f1#(I3, I4) -> f2#(I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.23	
0.00/0.23	We have the following SCCs.
0.00/0.23	  { 1 }
0.00/0.23	
0.00/0.23	DP problem for innermost termination.
0.00/0.23	P =
0.00/0.23	  f2#(I0, I1) -> f2#(I0, I2) 
0.00/0.23	R = 
0.00/0.23	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.23	  f2(I0, I1) -> f2(I0, I2) 
0.00/0.23	  f1(I3, I4) -> f2(I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.23	
0.00/3.22	EOF