0.00/0.12	YES
0.00/0.12	
0.00/0.12	DP problem for innermost termination.
0.00/0.12	P =
0.00/0.12	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.12	  f2#(I0, I1) -> f2#(I0 - 1, I1 - 1) [I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1]
0.00/0.12	  f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1]
0.00/0.12	R = 
0.00/0.12	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.12	  f2(I0, I1) -> f2(I0 - 1, I1 - 1) [I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1]
0.00/0.12	  f1(I2, I3) -> f2(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1]
0.00/0.12	
0.00/0.12	The dependency graph for this problem is:
0.00/0.12	  0 -> 2
0.00/0.12	  1 -> 1
0.00/0.12	  2 -> 1
0.00/0.12	Where:
0.00/0.12	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.12	  1) f2#(I0, I1) -> f2#(I0 - 1, I1 - 1) [I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1]
0.00/0.12	  2) f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1]
0.00/0.12	
0.00/0.12	We have the following SCCs.
0.00/0.12	  { 1 }
0.00/0.12	
0.00/0.12	DP problem for innermost termination.
0.00/0.12	P =
0.00/0.12	  f2#(I0, I1) -> f2#(I0 - 1, I1 - 1) [I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1]
0.00/0.12	R = 
0.00/0.12	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.12	  f2(I0, I1) -> f2(I0 - 1, I1 - 1) [I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1]
0.00/0.12	  f1(I2, I3) -> f2(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1]
0.00/0.12	
0.00/0.12	We use the basic value criterion with the projection function NU:
0.00/0.12	  NU[f2#(z1,z2)] = z2
0.00/0.12	
0.00/0.12	This gives the following inequalities:
0.00/0.12	  I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1  ==>  I1 >! I1 - 1
0.00/0.12	
0.00/0.12	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.10	EOF