0.00/0.36	YES
0.00/0.36	
0.00/0.36	DP problem for innermost termination.
0.00/0.36	P =
0.00/0.36	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.36	  f2#(I0, I1, I2) -> f2#(I3, I1 - 1, I2 + 1) [-1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1]
0.00/0.36	  f2#(I4, I5, I6) -> f2#(I4 - 1, -1, I6) [-1 = I5 /\ -1 <= I4 - 1]
0.00/0.36	  f1#(I7, I8, I9) -> f2#(I10, I11, 2) [0 <= I7 - 1 /\ -1 <= I11 - 1 /\ -1 <= I8 - 1 /\ -1 <= I10 - 1]
0.00/0.36	R = 
0.00/0.36	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.36	  f2(I0, I1, I2) -> f2(I3, I1 - 1, I2 + 1) [-1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1]
0.00/0.36	  f2(I4, I5, I6) -> f2(I4 - 1, -1, I6) [-1 = I5 /\ -1 <= I4 - 1]
0.00/0.36	  f1(I7, I8, I9) -> f2(I10, I11, 2) [0 <= I7 - 1 /\ -1 <= I11 - 1 /\ -1 <= I8 - 1 /\ -1 <= I10 - 1]
0.00/0.36	
0.00/0.36	The dependency graph for this problem is:
0.00/0.36	  0 -> 3
0.00/0.36	  1 -> 1, 2
0.00/0.36	  2 -> 2
0.00/0.36	  3 -> 1
0.00/0.36	Where:
0.00/0.36	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.00/0.36	  1) f2#(I0, I1, I2) -> f2#(I3, I1 - 1, I2 + 1) [-1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1]
0.00/0.36	  2) f2#(I4, I5, I6) -> f2#(I4 - 1, -1, I6) [-1 = I5 /\ -1 <= I4 - 1]
0.00/0.36	  3) f1#(I7, I8, I9) -> f2#(I10, I11, 2) [0 <= I7 - 1 /\ -1 <= I11 - 1 /\ -1 <= I8 - 1 /\ -1 <= I10 - 1]
0.00/0.36	
0.00/0.36	We have the following SCCs.
0.00/0.36	  { 1 }
0.00/0.36	  { 2 }
0.00/0.36	
0.00/0.36	DP problem for innermost termination.
0.00/0.36	P =
0.00/0.36	  f2#(I4, I5, I6) -> f2#(I4 - 1, -1, I6) [-1 = I5 /\ -1 <= I4 - 1]
0.00/0.36	R = 
0.00/0.36	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.36	  f2(I0, I1, I2) -> f2(I3, I1 - 1, I2 + 1) [-1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1]
0.00/0.36	  f2(I4, I5, I6) -> f2(I4 - 1, -1, I6) [-1 = I5 /\ -1 <= I4 - 1]
0.00/0.36	  f1(I7, I8, I9) -> f2(I10, I11, 2) [0 <= I7 - 1 /\ -1 <= I11 - 1 /\ -1 <= I8 - 1 /\ -1 <= I10 - 1]
0.00/0.36	
0.00/0.36	We use the basic value criterion with the projection function NU:
0.00/0.36	  NU[f2#(z1,z2,z3)] = z1
0.00/0.36	
0.00/0.36	This gives the following inequalities:
0.00/0.36	  -1 = I5 /\ -1 <= I4 - 1  ==>  I4 >! I4 - 1
0.00/0.36	
0.00/0.36	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.36	
0.00/0.36	DP problem for innermost termination.
0.00/0.36	P =
0.00/0.36	  f2#(I0, I1, I2) -> f2#(I3, I1 - 1, I2 + 1) [-1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1]
0.00/0.36	R = 
0.00/0.36	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.00/0.36	  f2(I0, I1, I2) -> f2(I3, I1 - 1, I2 + 1) [-1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1]
0.00/0.36	  f2(I4, I5, I6) -> f2(I4 - 1, -1, I6) [-1 = I5 /\ -1 <= I4 - 1]
0.00/0.36	  f1(I7, I8, I9) -> f2(I10, I11, 2) [0 <= I7 - 1 /\ -1 <= I11 - 1 /\ -1 <= I8 - 1 /\ -1 <= I10 - 1]
0.00/0.36	
0.00/0.36	We use the basic value criterion with the projection function NU:
0.00/0.36	  NU[f2#(z1,z2,z3)] = z2
0.00/0.36	
0.00/0.36	This gives the following inequalities:
0.00/0.36	  -1 <= I2 - 1 /\ -1 <= I1 - 1 /\ -1 <= I3 - 1  ==>  I1 >! I1 - 1
0.00/0.36	
0.00/0.36	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.34	EOF