0.00/0.18	YES
0.00/0.18	
0.00/0.18	DP problem for innermost termination.
0.00/0.18	P =
0.00/0.18	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.18	  f3#(I0, I1) -> f3#(I0 - 1, I1) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I1 - 1]
0.00/0.18	  f2#(I2, I3) -> f3#(I4, I3) [-1 <= y1 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ y1 - 100 * y2 <= 99 /\ 0 <= y1 - 100 * y2 /\ y1 - 100 * y2 = I4]
0.00/0.18	  f1#(I5, I6) -> f2#(I5, I6) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1]
0.00/0.18	R = 
0.00/0.18	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.18	  f3(I0, I1) -> f3(I0 - 1, I1) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I1 - 1]
0.00/0.18	  f2(I2, I3) -> f3(I4, I3) [-1 <= y1 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ y1 - 100 * y2 <= 99 /\ 0 <= y1 - 100 * y2 /\ y1 - 100 * y2 = I4]
0.00/0.18	  f1(I5, I6) -> f2(I5, I6) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1]
0.00/0.18	
0.00/0.18	The dependency graph for this problem is:
0.00/0.18	  0 -> 3
0.00/0.18	  1 -> 1
0.00/0.18	  2 -> 1
0.00/0.18	  3 -> 2
0.00/0.18	Where:
0.00/0.18	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.18	  1) f3#(I0, I1) -> f3#(I0 - 1, I1) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I1 - 1]
0.00/0.18	  2) f2#(I2, I3) -> f3#(I4, I3) [-1 <= y1 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ y1 - 100 * y2 <= 99 /\ 0 <= y1 - 100 * y2 /\ y1 - 100 * y2 = I4]
0.00/0.18	  3) f1#(I5, I6) -> f2#(I5, I6) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1]
0.00/0.18	
0.00/0.18	We have the following SCCs.
0.00/0.18	  { 1 }
0.00/0.18	
0.00/0.18	DP problem for innermost termination.
0.00/0.18	P =
0.00/0.18	  f3#(I0, I1) -> f3#(I0 - 1, I1) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I1 - 1]
0.00/0.18	R = 
0.00/0.18	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.18	  f3(I0, I1) -> f3(I0 - 1, I1) [0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I1 - 1]
0.00/0.18	  f2(I2, I3) -> f3(I4, I3) [-1 <= y1 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ y1 - 100 * y2 <= 99 /\ 0 <= y1 - 100 * y2 /\ y1 - 100 * y2 = I4]
0.00/0.18	  f1(I5, I6) -> f2(I5, I6) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1]
0.00/0.18	
0.00/0.18	We use the basic value criterion with the projection function NU:
0.00/0.18	  NU[f3#(z1,z2)] = z1
0.00/0.18	
0.00/0.18	This gives the following inequalities:
0.00/0.18	  0 <= I0 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I1 - 1  ==>  I0 >! I0 - 1
0.00/0.18	
0.00/0.18	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.16	EOF