0.00/0.19	YES
0.00/0.19	
0.00/0.19	DP problem for innermost termination.
0.00/0.19	P =
0.00/0.19	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.19	  f3#(I0, I1) -> f3#(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.19	  f2#(I3, I4) -> f3#(I5, I6) [0 <= I4 - 1 /\ -1 <= y1 - 1 /\ 0 <= I3 - 1 /\ y1 - 100 * y2 <= 99 /\ 0 <= y1 - 100 * y2 /\ y1 - 100 * y2 = I5]
0.00/0.19	  f1#(I7, I8) -> f2#(I7, I8) [0 <= I8 - 1 /\ -1 <= I9 - 1 /\ 0 <= I7 - 1]
0.00/0.19	R = 
0.00/0.19	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.19	  f3(I0, I1) -> f3(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.19	  f2(I3, I4) -> f3(I5, I6) [0 <= I4 - 1 /\ -1 <= y1 - 1 /\ 0 <= I3 - 1 /\ y1 - 100 * y2 <= 99 /\ 0 <= y1 - 100 * y2 /\ y1 - 100 * y2 = I5]
0.00/0.19	  f1(I7, I8) -> f2(I7, I8) [0 <= I8 - 1 /\ -1 <= I9 - 1 /\ 0 <= I7 - 1]
0.00/0.19	
0.00/0.19	The dependency graph for this problem is:
0.00/0.19	  0 -> 3
0.00/0.19	  1 -> 1
0.00/0.19	  2 -> 1
0.00/0.19	  3 -> 2
0.00/0.19	Where:
0.00/0.19	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.19	  1) f3#(I0, I1) -> f3#(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.19	  2) f2#(I3, I4) -> f3#(I5, I6) [0 <= I4 - 1 /\ -1 <= y1 - 1 /\ 0 <= I3 - 1 /\ y1 - 100 * y2 <= 99 /\ 0 <= y1 - 100 * y2 /\ y1 - 100 * y2 = I5]
0.00/0.19	  3) f1#(I7, I8) -> f2#(I7, I8) [0 <= I8 - 1 /\ -1 <= I9 - 1 /\ 0 <= I7 - 1]
0.00/0.19	
0.00/0.19	We have the following SCCs.
0.00/0.19	  { 1 }
0.00/0.19	
0.00/0.19	DP problem for innermost termination.
0.00/0.19	P =
0.00/0.19	  f3#(I0, I1) -> f3#(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.19	R = 
0.00/0.19	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.19	  f3(I0, I1) -> f3(I0 - 1, I2) [0 <= I0 - 1]
0.00/0.19	  f2(I3, I4) -> f3(I5, I6) [0 <= I4 - 1 /\ -1 <= y1 - 1 /\ 0 <= I3 - 1 /\ y1 - 100 * y2 <= 99 /\ 0 <= y1 - 100 * y2 /\ y1 - 100 * y2 = I5]
0.00/0.19	  f1(I7, I8) -> f2(I7, I8) [0 <= I8 - 1 /\ -1 <= I9 - 1 /\ 0 <= I7 - 1]
0.00/0.19	
0.00/0.19	We use the basic value criterion with the projection function NU:
0.00/0.19	  NU[f3#(z1,z2)] = z1
0.00/0.19	
0.00/0.19	This gives the following inequalities:
0.00/0.19	  0 <= I0 - 1  ==>  I0 >! I0 - 1
0.00/0.19	
0.00/0.19	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.18	EOF