0.00/0.38	MAYBE
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.38	  f2#(I0, I1) -> f2#(2, I0 + 1) [I0 <= I1 - 1 /\ 0 <= I0 - 1 /\ 2 <= I1 - 1]
0.00/0.38	  f1#(I2, I3) -> f2#(13, 17) 
0.00/0.38	R = 
0.00/0.38	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.38	  f2(I0, I1) -> f2(2, I0 + 1) [I0 <= I1 - 1 /\ 0 <= I0 - 1 /\ 2 <= I1 - 1]
0.00/0.38	  f1(I2, I3) -> f2(13, 17) 
0.00/0.38	
0.00/0.38	The dependency graph for this problem is:
0.00/0.38	  0 -> 2
0.00/0.38	  1 -> 1
0.00/0.38	  2 -> 1
0.00/0.38	Where:
0.00/0.38	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.38	  1) f2#(I0, I1) -> f2#(2, I0 + 1) [I0 <= I1 - 1 /\ 0 <= I0 - 1 /\ 2 <= I1 - 1]
0.00/0.38	  2) f1#(I2, I3) -> f2#(13, 17) 
0.00/0.38	
0.00/0.38	We have the following SCCs.
0.00/0.38	  { 1 }
0.00/0.38	
0.00/0.38	DP problem for innermost termination.
0.00/0.38	P =
0.00/0.38	  f2#(I0, I1) -> f2#(2, I0 + 1) [I0 <= I1 - 1 /\ 0 <= I0 - 1 /\ 2 <= I1 - 1]
0.00/0.38	R = 
0.00/0.38	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.38	  f2(I0, I1) -> f2(2, I0 + 1) [I0 <= I1 - 1 /\ 0 <= I0 - 1 /\ 2 <= I1 - 1]
0.00/0.38	  f1(I2, I3) -> f2(13, 17) 
0.00/0.38	
0.00/3.36	EOF