0.00/0.05	YES
0.00/0.05	
0.00/0.05	DP problem for innermost termination.
0.00/0.05	P =
0.00/0.05	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.05	  f2#(I0, I1, I2) -> f2#(I3, I2 + 1, I2) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 <= I0 /\ -1 <= I2 - 1 /\ I1 <= I2 - 1]
0.00/0.05	  f1#(I4, I5, I6) -> f2#(I7, I5 + 1, I5) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I7 <= I4]
0.00/0.05	R = 
0.00/0.05	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.05	  f2(I0, I1, I2) -> f2(I3, I2 + 1, I2) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 <= I0 /\ -1 <= I2 - 1 /\ I1 <= I2 - 1]
0.00/0.05	  f1(I4, I5, I6) -> f2(I7, I5 + 1, I5) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I7 <= I4]
0.00/0.05	
0.00/0.05	The dependency graph for this problem is:
0.00/0.05	  0 -> 2
0.00/0.05	  1 -> 
0.00/0.05	  2 -> 
0.00/0.05	Where:
0.00/0.05	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.05	  1) f2#(I0, I1, I2) -> f2#(I3, I2 + 1, I2) [0 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 <= I0 /\ -1 <= I2 - 1 /\ I1 <= I2 - 1]
0.00/0.05	  2) f1#(I4, I5, I6) -> f2#(I7, I5 + 1, I5) [0 <= I7 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1 /\ I7 <= I4]
0.00/0.05	
0.00/0.05	We have the following SCCs.
0.00/0.05	  
0.00/3.02	EOF