0.00/0.48	YES
0.00/0.48	
0.00/0.48	DP problem for innermost termination.
0.00/0.48	P =
0.00/0.48	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.48	  f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1]
0.00/0.48	  f2#(I2, I3) -> f2#(I2 + 1, I3) [I2 <= I3 - 1]
0.00/0.48	  f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1]
0.00/0.48	R = 
0.00/0.48	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.48	  f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1]
0.00/0.48	  f2(I2, I3) -> f2(I2 + 1, I3) [I2 <= I3 - 1]
0.00/0.48	  f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1]
0.00/0.48	
0.00/0.48	The dependency graph for this problem is:
0.00/0.48	  0 -> 3
0.00/0.48	  1 -> 1
0.00/0.48	  2 -> 2
0.00/0.48	  3 -> 1, 2
0.00/0.48	Where:
0.00/0.48	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.48	  1) f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1]
0.00/0.48	  2) f2#(I2, I3) -> f2#(I2 + 1, I3) [I2 <= I3 - 1]
0.00/0.48	  3) f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1]
0.00/0.48	
0.00/0.48	We have the following SCCs.
0.00/0.48	  { 2 }
0.00/0.48	  { 1 }
0.00/0.48	
0.00/0.48	DP problem for innermost termination.
0.00/0.48	P =
0.00/0.48	  f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1]
0.00/0.48	R = 
0.00/0.48	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.48	  f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1]
0.00/0.48	  f2(I2, I3) -> f2(I2 + 1, I3) [I2 <= I3 - 1]
0.00/0.48	  f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1]
0.00/0.48	
0.00/0.48	We use the reverse value criterion with the projection function NU:
0.00/0.48	  NU[f2#(z1,z2)] = z1 - 1 + -1 * z2
0.00/0.48	
0.00/0.48	This gives the following inequalities:
0.00/0.48	  I1 <= I0 - 1  ==>  I0 - 1 + -1 * I1 > I0 - 1 + -1 * (I1 + 1)  with  I0 - 1 + -1 * I1 >= 0
0.00/0.48	
0.00/0.48	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.48	
0.00/0.48	DP problem for innermost termination.
0.00/0.48	P =
0.00/0.48	  f2#(I2, I3) -> f2#(I2 + 1, I3) [I2 <= I3 - 1]
0.00/0.48	R = 
0.00/0.48	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.48	  f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1]
0.00/0.48	  f2(I2, I3) -> f2(I2 + 1, I3) [I2 <= I3 - 1]
0.00/0.48	  f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1]
0.00/0.48	
0.00/0.48	We use the reverse value criterion with the projection function NU:
0.00/0.48	  NU[f2#(z1,z2)] = z2 - 1 + -1 * z1
0.00/0.48	
0.00/0.48	This gives the following inequalities:
0.00/0.48	  I2 <= I3 - 1  ==>  I3 - 1 + -1 * I2 > I3 - 1 + -1 * (I2 + 1)  with  I3 - 1 + -1 * I2 >= 0
0.00/0.48	
0.00/0.48	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.46	EOF