0.00/0.17	YES
0.00/0.17	
0.00/0.17	DP problem for innermost termination.
0.00/0.17	P =
0.00/0.17	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.17	  f2#(I0, I1) -> f2#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1]
0.00/0.17	  f1#(I3, I4) -> f2#(I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.17	R = 
0.00/0.17	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.17	  f2(I0, I1) -> f2(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1]
0.00/0.17	  f1(I3, I4) -> f2(I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.17	
0.00/0.17	The dependency graph for this problem is:
0.00/0.17	  0 -> 2
0.00/0.17	  1 -> 1
0.00/0.17	  2 -> 1
0.00/0.17	Where:
0.00/0.17	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.17	  1) f2#(I0, I1) -> f2#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1]
0.00/0.17	  2) f1#(I3, I4) -> f2#(I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.17	
0.00/0.17	We have the following SCCs.
0.00/0.17	  { 1 }
0.00/0.17	
0.00/0.17	DP problem for innermost termination.
0.00/0.17	P =
0.00/0.17	  f2#(I0, I1) -> f2#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1]
0.00/0.17	R = 
0.00/0.17	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.17	  f2(I0, I1) -> f2(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1]
0.00/0.17	  f1(I3, I4) -> f2(I4, I5) [-1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.17	
0.00/0.17	We use the basic value criterion with the projection function NU:
0.00/0.17	  NU[f2#(z1,z2)] = z1
0.00/0.17	
0.00/0.17	This gives the following inequalities:
0.00/0.17	  I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1  ==>  I0 >! I0 - 1
0.00/0.17	
0.00/0.17	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.15	EOF