0.00/0.40	YES
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.40	  f2#(I0, I1, I2) -> f2#(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1]
0.00/0.40	  f2#(I3, I4, I5) -> f2#(0, 0, 0) [0 = I4 /\ 0 = I3 /\ 0 <= I5 - 1]
0.00/0.40	  f2#(I6, I7, I8) -> f2#(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1]
0.00/0.40	  f1#(I9, I10, I11) -> f2#(I12, I13, I14) [I12 + I13 = I14 /\ 0 <= I9 - 1 /\ -1 <= I13 - 1 /\ -1 <= I10 - 1 /\ -1 <= I12 - 1]
0.00/0.40	R = 
0.00/0.40	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.40	  f2(I0, I1, I2) -> f2(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1]
0.00/0.40	  f2(I3, I4, I5) -> f2(0, 0, 0) [0 = I4 /\ 0 = I3 /\ 0 <= I5 - 1]
0.00/0.40	  f2(I6, I7, I8) -> f2(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1]
0.00/0.40	  f1(I9, I10, I11) -> f2(I12, I13, I14) [I12 + I13 = I14 /\ 0 <= I9 - 1 /\ -1 <= I13 - 1 /\ -1 <= I10 - 1 /\ -1 <= I12 - 1]
0.00/0.40	
0.00/0.40	The dependency graph for this problem is:
0.00/0.40	  0 -> 4
0.00/0.40	  1 -> 1
0.00/0.40	  2 -> 
0.00/0.40	  3 -> 1, 3
0.00/0.40	  4 -> 1, 3
0.00/0.40	Where:
0.00/0.40	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.40	  1) f2#(I0, I1, I2) -> f2#(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1]
0.00/0.40	  2) f2#(I3, I4, I5) -> f2#(0, 0, 0) [0 = I4 /\ 0 = I3 /\ 0 <= I5 - 1]
0.00/0.40	  3) f2#(I6, I7, I8) -> f2#(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1]
0.00/0.40	  4) f1#(I9, I10, I11) -> f2#(I12, I13, I14) [I12 + I13 = I14 /\ 0 <= I9 - 1 /\ -1 <= I13 - 1 /\ -1 <= I10 - 1 /\ -1 <= I12 - 1]
0.00/0.40	
0.00/0.40	We have the following SCCs.
0.00/0.40	  { 3 }
0.00/0.40	  { 1 }
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  f2#(I0, I1, I2) -> f2#(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1]
0.00/0.40	R = 
0.00/0.40	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.40	  f2(I0, I1, I2) -> f2(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1]
0.00/0.40	  f2(I3, I4, I5) -> f2(0, 0, 0) [0 = I4 /\ 0 = I3 /\ 0 <= I5 - 1]
0.00/0.40	  f2(I6, I7, I8) -> f2(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1]
0.00/0.40	  f1(I9, I10, I11) -> f2(I12, I13, I14) [I12 + I13 = I14 /\ 0 <= I9 - 1 /\ -1 <= I13 - 1 /\ -1 <= I10 - 1 /\ -1 <= I12 - 1]
0.00/0.40	
0.00/0.40	We use the basic value criterion with the projection function NU:
0.00/0.40	  NU[f2#(z1,z2,z3)] = z2
0.00/0.40	
0.00/0.40	This gives the following inequalities:
0.00/0.40	  0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1  ==>  I1 >! I1 - 1
0.00/0.40	
0.00/0.40	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  f2#(I6, I7, I8) -> f2#(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1]
0.00/0.40	R = 
0.00/0.40	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.40	  f2(I0, I1, I2) -> f2(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1]
0.00/0.40	  f2(I3, I4, I5) -> f2(0, 0, 0) [0 = I4 /\ 0 = I3 /\ 0 <= I5 - 1]
0.00/0.40	  f2(I6, I7, I8) -> f2(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1]
0.00/0.40	  f1(I9, I10, I11) -> f2(I12, I13, I14) [I12 + I13 = I14 /\ 0 <= I9 - 1 /\ -1 <= I13 - 1 /\ -1 <= I10 - 1 /\ -1 <= I12 - 1]
0.00/0.40	
0.00/0.40	We use the basic value criterion with the projection function NU:
0.00/0.40	  NU[f2#(z1,z2,z3)] = z1
0.00/0.40	
0.00/0.40	This gives the following inequalities:
0.00/0.40	  -1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1  ==>  I6 >! I6 - 1
0.00/0.40	
0.00/0.40	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.39	EOF