0.00/0.40	MAYBE
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  init#(x1, x2, x3) -> f2#(rnd1, rnd2, rnd3) 
0.00/0.40	  f2#(I0, I1, I2) -> f1#(I3, I1 + 1, I1) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ -1 <= I1 - 1 /\ I3 <= I0]
0.00/0.40	  f1#(I4, I5, I6) -> f1#(I7, I5 + 1, I6) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ I7 <= I4]
0.00/0.40	R = 
0.00/0.40	  init(x1, x2, x3) -> f2(rnd1, rnd2, rnd3) 
0.00/0.40	  f2(I0, I1, I2) -> f1(I3, I1 + 1, I1) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ -1 <= I1 - 1 /\ I3 <= I0]
0.00/0.40	  f1(I4, I5, I6) -> f1(I7, I5 + 1, I6) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ I7 <= I4]
0.00/0.40	
0.00/0.40	The dependency graph for this problem is:
0.00/0.40	  0 -> 1
0.00/0.40	  1 -> 2
0.00/0.40	  2 -> 2
0.00/0.40	Where:
0.00/0.40	  0) init#(x1, x2, x3) -> f2#(rnd1, rnd2, rnd3) 
0.00/0.40	  1) f2#(I0, I1, I2) -> f1#(I3, I1 + 1, I1) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ -1 <= I1 - 1 /\ I3 <= I0]
0.00/0.40	  2) f1#(I4, I5, I6) -> f1#(I7, I5 + 1, I6) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ I7 <= I4]
0.00/0.40	
0.00/0.40	We have the following SCCs.
0.00/0.40	  { 2 }
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  f1#(I4, I5, I6) -> f1#(I7, I5 + 1, I6) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ I7 <= I4]
0.00/0.40	R = 
0.00/0.40	  init(x1, x2, x3) -> f2(rnd1, rnd2, rnd3) 
0.00/0.40	  f2(I0, I1, I2) -> f1(I3, I1 + 1, I1) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ -1 <= I1 - 1 /\ I3 <= I0]
0.00/0.40	  f1(I4, I5, I6) -> f1(I7, I5 + 1, I6) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ I7 <= I4]
0.00/0.40	
0.00/3.38	EOF