0.00/0.13	YES
0.00/0.13	
0.00/0.13	DP problem for innermost termination.
0.00/0.13	P =
0.00/0.13	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.13	  f3#(I0, I1) -> f2#(I0 - 1, I2) [0 <= I0 - 1 /\ I0 - 2 * y1 = 0 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
0.00/0.13	  f2#(I3, I4) -> f3#(I3, I5) [I3 - 2 * I6 = 0 /\ 0 <= I3 - 1]
0.00/0.13	  f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
0.00/0.13	R = 
0.00/0.13	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.13	  f3(I0, I1) -> f2(I0 - 1, I2) [0 <= I0 - 1 /\ I0 - 2 * y1 = 0 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
0.00/0.13	  f2(I3, I4) -> f3(I3, I5) [I3 - 2 * I6 = 0 /\ 0 <= I3 - 1]
0.00/0.13	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
0.00/0.13	
0.00/0.13	The dependency graph for this problem is:
0.00/0.13	  0 -> 3
0.00/0.13	  1 -> 
0.00/0.13	  2 -> 1
0.00/0.13	  3 -> 2
0.00/0.13	Where:
0.00/0.13	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.13	  1) f3#(I0, I1) -> f2#(I0 - 1, I2) [0 <= I0 - 1 /\ I0 - 2 * y1 = 0 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
0.00/0.13	  2) f2#(I3, I4) -> f3#(I3, I5) [I3 - 2 * I6 = 0 /\ 0 <= I3 - 1]
0.00/0.13	  3) f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
0.00/0.13	
0.00/0.13	We have the following SCCs.
0.00/0.13	  
0.00/3.11	EOF