0.00/0.22	YES
0.00/0.22	
0.00/0.22	DP problem for innermost termination.
0.00/0.22	P =
0.00/0.22	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.22	  f3#(I0, I1, I2) -> f3#(I0 - 1, I1 - 1, I1 - 1) [I1 = I2 /\ 0 <= I1 - 1]
0.00/0.22	  f3#(I3, I4, I5) -> f2#(I3, 0, I6) [0 = I5 /\ 0 = I4]
0.00/0.22	  f2#(I7, I8, I9) -> f3#(I7, I7, I7) [I7 = I8 /\ 0 <= I7 - 1]
0.00/0.22	  f1#(I10, I11, I12) -> f2#(I13, I14, I15) [0 <= I10 - 1 /\ -1 <= I13 - 1 /\ -1 <= I11 - 1 /\ -1 <= I14 - 1]
0.00/0.22	R = 
0.00/0.22	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.22	  f3(I0, I1, I2) -> f3(I0 - 1, I1 - 1, I1 - 1) [I1 = I2 /\ 0 <= I1 - 1]
0.00/0.22	  f3(I3, I4, I5) -> f2(I3, 0, I6) [0 = I5 /\ 0 = I4]
0.00/0.22	  f2(I7, I8, I9) -> f3(I7, I7, I7) [I7 = I8 /\ 0 <= I7 - 1]
0.00/0.22	  f1(I10, I11, I12) -> f2(I13, I14, I15) [0 <= I10 - 1 /\ -1 <= I13 - 1 /\ -1 <= I11 - 1 /\ -1 <= I14 - 1]
0.00/0.22	
0.00/0.22	The dependency graph for this problem is:
0.00/0.22	  0 -> 4
0.00/0.22	  1 -> 1, 2
0.00/0.22	  2 -> 
0.00/0.22	  3 -> 1
0.00/0.22	  4 -> 3
0.00/0.22	Where:
0.00/0.22	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.22	  1) f3#(I0, I1, I2) -> f3#(I0 - 1, I1 - 1, I1 - 1) [I1 = I2 /\ 0 <= I1 - 1]
0.00/0.22	  2) f3#(I3, I4, I5) -> f2#(I3, 0, I6) [0 = I5 /\ 0 = I4]
0.00/0.22	  3) f2#(I7, I8, I9) -> f3#(I7, I7, I7) [I7 = I8 /\ 0 <= I7 - 1]
0.00/0.22	  4) f1#(I10, I11, I12) -> f2#(I13, I14, I15) [0 <= I10 - 1 /\ -1 <= I13 - 1 /\ -1 <= I11 - 1 /\ -1 <= I14 - 1]
0.00/0.22	
0.00/0.22	We have the following SCCs.
0.00/0.22	  { 1 }
0.00/0.22	
0.00/0.22	DP problem for innermost termination.
0.00/0.22	P =
0.00/0.22	  f3#(I0, I1, I2) -> f3#(I0 - 1, I1 - 1, I1 - 1) [I1 = I2 /\ 0 <= I1 - 1]
0.00/0.22	R = 
0.00/0.22	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.22	  f3(I0, I1, I2) -> f3(I0 - 1, I1 - 1, I1 - 1) [I1 = I2 /\ 0 <= I1 - 1]
0.00/0.22	  f3(I3, I4, I5) -> f2(I3, 0, I6) [0 = I5 /\ 0 = I4]
0.00/0.22	  f2(I7, I8, I9) -> f3(I7, I7, I7) [I7 = I8 /\ 0 <= I7 - 1]
0.00/0.22	  f1(I10, I11, I12) -> f2(I13, I14, I15) [0 <= I10 - 1 /\ -1 <= I13 - 1 /\ -1 <= I11 - 1 /\ -1 <= I14 - 1]
0.00/0.22	
0.00/0.22	We use the basic value criterion with the projection function NU:
0.00/0.22	  NU[f3#(z1,z2,z3)] = z3
0.00/0.22	
0.00/0.22	This gives the following inequalities:
0.00/0.22	  I1 = I2 /\ 0 <= I1 - 1  ==>  I2 >! I1 - 1
0.00/0.22	
0.00/0.22	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.20	EOF