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) -> f1#(rnd1, rnd2) 
0.00/0.23	  f2#(I0, I1) -> f2#(I0 - 1, 0) [0 = I1 /\ 0 <= I0 - 1]
0.00/0.23	  f2#(I2, I3) -> f2#(I2, I3 - 1) [0 <= I3 - 1]
0.00/0.23	  f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ -1 <= I7 - 1 /\ -1 <= I5 - 1 /\ -1 <= I6 - 1]
0.00/0.23	R = 
0.00/0.23	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.23	  f2(I0, I1) -> f2(I0 - 1, 0) [0 = I1 /\ 0 <= I0 - 1]
0.00/0.23	  f2(I2, I3) -> f2(I2, I3 - 1) [0 <= I3 - 1]
0.00/0.23	  f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I7 - 1 /\ -1 <= I5 - 1 /\ -1 <= I6 - 1]
0.00/0.23	
0.00/0.23	The dependency graph for this problem is:
0.00/0.23	  0 -> 3
0.00/0.23	  1 -> 1
0.00/0.23	  2 -> 1, 2
0.00/0.23	  3 -> 1, 2
0.00/0.23	Where:
0.00/0.23	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.23	  1) f2#(I0, I1) -> f2#(I0 - 1, 0) [0 = I1 /\ 0 <= I0 - 1]
0.00/0.23	  2) f2#(I2, I3) -> f2#(I2, I3 - 1) [0 <= I3 - 1]
0.00/0.23	  3) f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ -1 <= I7 - 1 /\ -1 <= I5 - 1 /\ -1 <= I6 - 1]
0.00/0.23	
0.00/0.23	We have the following SCCs.
0.00/0.23	  { 2 }
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) -> f2#(I0 - 1, 0) [0 = I1 /\ 0 <= I0 - 1]
0.00/0.23	R = 
0.00/0.23	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.23	  f2(I0, I1) -> f2(I0 - 1, 0) [0 = I1 /\ 0 <= I0 - 1]
0.00/0.23	  f2(I2, I3) -> f2(I2, I3 - 1) [0 <= I3 - 1]
0.00/0.23	  f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I7 - 1 /\ -1 <= I5 - 1 /\ -1 <= I6 - 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)] = z1
0.00/0.23	
0.00/0.23	This gives the following inequalities:
0.00/0.23	  0 = I1 /\ 0 <= I0 - 1  ==>  I0 >! I0 - 1
0.00/0.23	
0.00/0.23	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.23	
0.00/0.23	DP problem for innermost termination.
0.00/0.23	P =
0.00/0.23	  f2#(I2, I3) -> f2#(I2, I3 - 1) [0 <= I3 - 1]
0.00/0.23	R = 
0.00/0.23	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.23	  f2(I0, I1) -> f2(I0 - 1, 0) [0 = I1 /\ 0 <= I0 - 1]
0.00/0.23	  f2(I2, I3) -> f2(I2, I3 - 1) [0 <= I3 - 1]
0.00/0.23	  f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I7 - 1 /\ -1 <= I5 - 1 /\ -1 <= I6 - 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)] = z2
0.00/0.23	
0.00/0.23	This gives the following inequalities:
0.00/0.23	  0 <= I3 - 1  ==>  I3 >! I3 - 1
0.00/0.23	
0.00/0.23	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.22	EOF