0.00/0.35	YES
0.00/0.35	
0.00/0.35	DP problem for innermost termination.
0.00/0.35	P =
0.00/0.35	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.35	  f2#(I0, I1) -> f2#(I0, I1 * I1) [I1 <= I0 - 1 /\ 1 <= I1 - 1]
0.00/0.35	  f1#(I2, I3) -> f2#(I4, 2) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ -1 <= I4 - 1]
0.00/0.35	R = 
0.00/0.35	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.35	  f2(I0, I1) -> f2(I0, I1 * I1) [I1 <= I0 - 1 /\ 1 <= I1 - 1]
0.00/0.35	  f1(I2, I3) -> f2(I4, 2) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ -1 <= I4 - 1]
0.00/0.35	
0.00/0.35	The dependency graph for this problem is:
0.00/0.35	  0 -> 2
0.00/0.35	  1 -> 1
0.00/0.35	  2 -> 1
0.00/0.35	Where:
0.00/0.35	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.35	  1) f2#(I0, I1) -> f2#(I0, I1 * I1) [I1 <= I0 - 1 /\ 1 <= I1 - 1]
0.00/0.35	  2) f1#(I2, I3) -> f2#(I4, 2) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ -1 <= I4 - 1]
0.00/0.35	
0.00/0.35	We have the following SCCs.
0.00/0.35	  { 1 }
0.00/0.35	
0.00/0.35	DP problem for innermost termination.
0.00/0.35	P =
0.00/0.35	  f2#(I0, I1) -> f2#(I0, I1 * I1) [I1 <= I0 - 1 /\ 1 <= I1 - 1]
0.00/0.35	R = 
0.00/0.35	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.35	  f2(I0, I1) -> f2(I0, I1 * I1) [I1 <= I0 - 1 /\ 1 <= I1 - 1]
0.00/0.35	  f1(I2, I3) -> f2(I4, 2) [0 <= I2 - 1 /\ -1 <= I3 - 1 /\ -1 <= I4 - 1]
0.00/0.35	
0.00/0.35	We use the reverse value criterion with the projection function NU:
0.00/0.35	  NU[f2#(z1,z2)] = z1 - 1 + -1 * z2
0.00/0.35	
0.00/0.35	This gives the following inequalities:
0.00/0.35	  I1 <= I0 - 1 /\ 1 <= I1 - 1  ==>  I0 - 1 + -1 * I1 > I0 - 1 + -1 * (I1 * I1)  with  I0 - 1 + -1 * I1 >= 0
0.00/0.35	
0.00/0.35	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.34	EOF