0.00/0.23	YES
0.00/0.23	
0.00/0.23	DP problem for innermost termination.
0.00/0.23	P =
0.00/0.23	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.23	  f2#(I0, I1, I2) -> f2#(I0 + 1, I1 - I2, I2) [-1 <= I0 - 1 /\ I2 <= I1 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1]
0.00/0.23	  f1#(I3, I4, I5) -> f2#(0, I6, I7) [0 <= I3 - 1 /\ -1 <= I6 - 1 /\ -1 <= I4 - 1 /\ -1 <= I7 - 1]
0.00/0.23	R = 
0.00/0.23	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.23	  f2(I0, I1, I2) -> f2(I0 + 1, I1 - I2, I2) [-1 <= I0 - 1 /\ I2 <= I1 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1]
0.00/0.23	  f1(I3, I4, I5) -> f2(0, I6, I7) [0 <= I3 - 1 /\ -1 <= I6 - 1 /\ -1 <= I4 - 1 /\ -1 <= I7 - 1]
0.00/0.23	
0.00/0.23	The dependency graph for this problem is:
0.00/0.23	  0 -> 2
0.00/0.23	  1 -> 1
0.00/0.23	  2 -> 1
0.00/0.23	Where:
0.00/0.23	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.23	  1) f2#(I0, I1, I2) -> f2#(I0 + 1, I1 - I2, I2) [-1 <= I0 - 1 /\ I2 <= I1 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1]
0.00/0.23	  2) f1#(I3, I4, I5) -> f2#(0, I6, I7) [0 <= I3 - 1 /\ -1 <= I6 - 1 /\ -1 <= I4 - 1 /\ -1 <= I7 - 1]
0.00/0.23	
0.00/0.23	We have the following SCCs.
0.00/0.23	  { 1 }
0.00/0.23	
0.00/0.23	DP problem for innermost termination.
0.00/0.23	P =
0.00/0.23	  f2#(I0, I1, I2) -> f2#(I0 + 1, I1 - I2, I2) [-1 <= I0 - 1 /\ I2 <= I1 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1]
0.00/0.23	R = 
0.00/0.23	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.23	  f2(I0, I1, I2) -> f2(I0 + 1, I1 - I2, I2) [-1 <= I0 - 1 /\ I2 <= I1 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1]
0.00/0.23	  f1(I3, I4, I5) -> f2(0, I6, I7) [0 <= I3 - 1 /\ -1 <= I6 - 1 /\ -1 <= I4 - 1 /\ -1 <= I7 - 1]
0.00/0.23	
0.00/0.23	We use the basic value criterion with the projection function NU:
0.00/0.23	  NU[f2#(z1,z2,z3)] = z2
0.00/0.23	
0.00/0.23	This gives the following inequalities:
0.00/0.23	  -1 <= I0 - 1 /\ I2 <= I1 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1  ==>  I1 >! I1 - I2
0.00/0.23	
0.00/0.23	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.21	EOF