0.00/0.24	YES
0.00/0.24	
0.00/0.24	DP problem for innermost termination.
0.00/0.24	P =
0.00/0.24	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.24	  f2#(I0, I1, I2) -> f2#(I3, I1 + 10, I2 - 1) [I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1]
0.00/0.24	  f1#(I4, I5, I6) -> f2#(I7, 0, 10) [13 <= I7 - 1]
0.00/0.24	R = 
0.00/0.24	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.24	  f2(I0, I1, I2) -> f2(I3, I1 + 10, I2 - 1) [I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1]
0.00/0.24	  f1(I4, I5, I6) -> f2(I7, 0, 10) [13 <= I7 - 1]
0.00/0.24	
0.00/0.24	The dependency graph for this problem is:
0.00/0.24	  0 -> 2
0.00/0.24	  1 -> 1
0.00/0.24	  2 -> 1
0.00/0.24	Where:
0.00/0.24	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.24	  1) f2#(I0, I1, I2) -> f2#(I3, I1 + 10, I2 - 1) [I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1]
0.00/0.24	  2) f1#(I4, I5, I6) -> f2#(I7, 0, 10) [13 <= I7 - 1]
0.00/0.24	
0.00/0.24	We have the following SCCs.
0.00/0.24	  { 1 }
0.00/0.24	
0.00/0.24	DP problem for innermost termination.
0.00/0.24	P =
0.00/0.24	  f2#(I0, I1, I2) -> f2#(I3, I1 + 10, I2 - 1) [I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1]
0.00/0.24	R = 
0.00/0.24	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.24	  f2(I0, I1, I2) -> f2(I3, I1 + 10, I2 - 1) [I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1]
0.00/0.24	  f1(I4, I5, I6) -> f2(I7, 0, 10) [13 <= I7 - 1]
0.00/0.24	
0.00/0.24	We use the basic value criterion with the projection function NU:
0.00/0.24	  NU[f2#(z1,z2,z3)] = z3
0.00/0.24	
0.00/0.24	This gives the following inequalities:
0.00/0.24	  I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1  ==>  I2 >! I2 - 1
0.00/0.24	
0.00/0.24	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.22	EOF