0.00/0.28	MAYBE
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.28	  f2#(I0, I1) -> f2#(I2, I3) 
0.00/0.28	  f1#(I4, I5) -> f2#(I6, I7) [-1 <= I5 - 1 /\ 0 <= I4 - 1]
0.00/0.28	R = 
0.00/0.28	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.28	  f2(I0, I1) -> f2(I2, I3) 
0.00/0.28	  f1(I4, I5) -> f2(I6, I7) [-1 <= I5 - 1 /\ 0 <= I4 - 1]
0.00/0.28	
0.00/0.28	The dependency graph for this problem is:
0.00/0.28	  0 -> 2
0.00/0.28	  1 -> 1
0.00/0.28	  2 -> 1
0.00/0.28	Where:
0.00/0.28	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.28	  1) f2#(I0, I1) -> f2#(I2, I3) 
0.00/0.28	  2) f1#(I4, I5) -> f2#(I6, I7) [-1 <= I5 - 1 /\ 0 <= I4 - 1]
0.00/0.28	
0.00/0.28	We have the following SCCs.
0.00/0.28	  { 1 }
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  f2#(I0, I1) -> f2#(I2, I3) 
0.00/0.28	R = 
0.00/0.28	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.28	  f2(I0, I1) -> f2(I2, I3) 
0.00/0.28	  f1(I4, I5) -> f2(I6, I7) [-1 <= I5 - 1 /\ 0 <= I4 - 1]
0.00/0.28	
0.00/3.26	EOF