0.00/0.45	YES
0.00/0.45	
0.00/0.45	DP problem for innermost termination.
0.00/0.45	P =
0.00/0.45	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.45	  f2#(I0, I1) -> f2#(I2, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.45	  f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.45	  f2#(I5, I6) -> f2#(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
0.00/0.45	  f1#(I7, I8) -> f2#(12, 10) 
0.00/0.45	R = 
0.00/0.45	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.45	  f2(I0, I1) -> f2(I2, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.45	  f2(I3, I4) -> f2(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.45	  f2(I5, I6) -> f2(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
0.00/0.45	  f1(I7, I8) -> f2(12, 10) 
0.00/0.45	
0.00/0.45	The dependency graph for this problem is:
0.00/0.45	  0 -> 4
0.00/0.45	  1 -> 1, 2, 3
0.00/0.45	  2 -> 1, 2, 3
0.00/0.45	  3 -> 1, 2
0.00/0.45	  4 -> 1, 2
0.00/0.45	Where:
0.00/0.45	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.00/0.45	  1) f2#(I0, I1) -> f2#(I2, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.45	  2) f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.45	  3) f2#(I5, I6) -> f2#(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
0.00/0.45	  4) f1#(I7, I8) -> f2#(12, 10) 
0.00/0.45	
0.00/0.45	We have the following SCCs.
0.00/0.45	  { 1, 2, 3 }
0.00/0.45	
0.00/0.45	DP problem for innermost termination.
0.00/0.45	P =
0.00/0.45	  f2#(I0, I1) -> f2#(I2, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.45	  f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.45	  f2#(I5, I6) -> f2#(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
0.00/0.45	R = 
0.00/0.45	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.45	  f2(I0, I1) -> f2(I2, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.45	  f2(I3, I4) -> f2(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.45	  f2(I5, I6) -> f2(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
0.00/0.45	  f1(I7, I8) -> f2(12, 10) 
0.00/0.45	
0.00/0.45	We use the basic value criterion with the projection function NU:
0.00/0.45	  NU[f2#(z1,z2)] = z2
0.00/0.45	
0.00/0.45	This gives the following inequalities:
0.00/0.45	  0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1  ==>  I1 >! I1 - 1
0.00/0.45	  0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1  ==>  I4 (>! \union =) I4
0.00/0.45	  I5 <= 0 /\ 0 <= I6 - 1  ==>  I6 >! I6 - 1
0.00/0.45	
0.00/0.45	We remove all the strictly oriented dependency pairs.
0.00/0.45	
0.00/0.45	DP problem for innermost termination.
0.00/0.45	P =
0.00/0.45	  f2#(I3, I4) -> f2#(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.45	R = 
0.00/0.45	  init(x1, x2) -> f1(rnd1, rnd2) 
0.00/0.45	  f2(I0, I1) -> f2(I2, I1 - 1) [0 <= I1 - 1 /\ I1 - 1 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.45	  f2(I3, I4) -> f2(I3 - 1, I4) [0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1]
0.00/0.45	  f2(I5, I6) -> f2(1, I6 - 1) [I5 <= 0 /\ 0 <= I6 - 1]
0.00/0.45	  f1(I7, I8) -> f2(12, 10) 
0.00/0.45	
0.00/0.45	We use the basic value criterion with the projection function NU:
0.00/0.45	  NU[f2#(z1,z2)] = z1
0.00/0.45	
0.00/0.45	This gives the following inequalities:
0.00/0.45	  0 <= I4 - 1 /\ I4 - 1 <= I4 - 1 /\ 0 <= I3 - 1  ==>  I3 >! I3 - 1
0.00/0.45	
0.00/0.45	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.44	EOF