0.64/0.70	YES
0.64/0.70	
0.64/0.70	DP problem for innermost termination.
0.64/0.70	P =
0.64/0.70	  init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.64/0.70	  f2#(I0, I1, I2, I3, I4) -> f2#(I5, I6, I2 + 1, I3 + 1, I4) [I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1]
0.64/0.70	  f1#(I7, I8, I9, I10, I11) -> f2#(I12, I13, 0, I14, I15) [0 <= I13 - 1 /\ 0 <= I12 - 1 /\ 0 <= I7 - 1 /\ I13 <= I7 /\ -1 <= I14 - 1 /\ 1 <= I8 - 1 /\ -1 <= I15 - 1]
0.64/0.70	R = 
0.64/0.70	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.64/0.70	  f2(I0, I1, I2, I3, I4) -> f2(I5, I6, I2 + 1, I3 + 1, I4) [I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1]
0.64/0.70	  f1(I7, I8, I9, I10, I11) -> f2(I12, I13, 0, I14, I15) [0 <= I13 - 1 /\ 0 <= I12 - 1 /\ 0 <= I7 - 1 /\ I13 <= I7 /\ -1 <= I14 - 1 /\ 1 <= I8 - 1 /\ -1 <= I15 - 1]
0.64/0.70	
0.64/0.70	The dependency graph for this problem is:
0.64/0.70	  0 -> 2
0.64/0.70	  1 -> 1
0.64/0.70	  2 -> 1
0.64/0.70	Where:
0.64/0.70	  0) init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.64/0.70	  1) f2#(I0, I1, I2, I3, I4) -> f2#(I5, I6, I2 + 1, I3 + 1, I4) [I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1]
0.64/0.70	  2) f1#(I7, I8, I9, I10, I11) -> f2#(I12, I13, 0, I14, I15) [0 <= I13 - 1 /\ 0 <= I12 - 1 /\ 0 <= I7 - 1 /\ I13 <= I7 /\ -1 <= I14 - 1 /\ 1 <= I8 - 1 /\ -1 <= I15 - 1]
0.64/0.70	
0.64/0.70	We have the following SCCs.
0.64/0.70	  { 1 }
0.64/0.70	
0.64/0.70	DP problem for innermost termination.
0.64/0.70	P =
0.64/0.70	  f2#(I0, I1, I2, I3, I4) -> f2#(I5, I6, I2 + 1, I3 + 1, I4) [I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1]
0.64/0.70	R = 
0.64/0.70	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
0.64/0.70	  f2(I0, I1, I2, I3, I4) -> f2(I5, I6, I2 + 1, I3 + 1, I4) [I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1]
0.64/0.70	  f1(I7, I8, I9, I10, I11) -> f2(I12, I13, 0, I14, I15) [0 <= I13 - 1 /\ 0 <= I12 - 1 /\ 0 <= I7 - 1 /\ I13 <= I7 /\ -1 <= I14 - 1 /\ 1 <= I8 - 1 /\ -1 <= I15 - 1]
0.64/0.70	
0.64/0.70	We use the reverse value criterion with the projection function NU:
0.64/0.70	  NU[f2#(z1,z2,z3,z4,z5)] = z5 - 1 + -1 * z3
0.64/0.70	
0.64/0.70	This gives the following inequalities:
0.64/0.70	  I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1  ==>  I4 - 1 + -1 * I2 > I4 - 1 + -1 * (I2 + 1)  with  I4 - 1 + -1 * I2 >= 0
0.64/0.70	
0.64/0.70	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.64/3.68	EOF