0.94/1.31	YES
0.94/1.31	
0.94/1.31	DP problem for innermost termination.
0.94/1.31	P =
0.94/1.31	  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.94/1.31	  f2#(I0, I1, I2) -> f2#(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1]
0.94/1.31	  f2#(I3, I4, I5) -> f2#(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1]
0.94/1.31	  f1#(I6, I7, I8) -> f2#(I9, I10, I11) [0 <= I6 - 1 /\ -1 <= I11 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I10 - 1]
0.94/1.31	R = 
0.94/1.31	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.94/1.31	  f2(I0, I1, I2) -> f2(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1]
0.94/1.31	  f2(I3, I4, I5) -> f2(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1]
0.94/1.31	  f1(I6, I7, I8) -> f2(I9, I10, I11) [0 <= I6 - 1 /\ -1 <= I11 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I10 - 1]
0.94/1.31	
0.94/1.31	The dependency graph for this problem is:
0.94/1.31	  0 -> 3
0.94/1.31	  1 -> 1, 2
0.94/1.31	  2 -> 1, 2
0.94/1.31	  3 -> 1, 2
0.94/1.31	Where:
0.94/1.31	  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
0.94/1.31	  1) f2#(I0, I1, I2) -> f2#(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1]
0.94/1.31	  2) f2#(I3, I4, I5) -> f2#(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1]
0.94/1.31	  3) f1#(I6, I7, I8) -> f2#(I9, I10, I11) [0 <= I6 - 1 /\ -1 <= I11 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I10 - 1]
0.94/1.31	
0.94/1.31	We have the following SCCs.
0.94/1.31	  { 1, 2 }
0.94/1.31	
0.94/1.31	DP problem for innermost termination.
0.94/1.31	P =
0.94/1.31	  f2#(I0, I1, I2) -> f2#(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1]
0.94/1.31	  f2#(I3, I4, I5) -> f2#(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1]
0.94/1.31	R = 
0.94/1.31	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.94/1.31	  f2(I0, I1, I2) -> f2(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1]
0.94/1.31	  f2(I3, I4, I5) -> f2(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1]
0.94/1.31	  f1(I6, I7, I8) -> f2(I9, I10, I11) [0 <= I6 - 1 /\ -1 <= I11 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I10 - 1]
0.94/1.31	
0.94/1.31	We use the reverse value criterion with the projection function NU:
0.94/1.31	  NU[f2#(z1,z2,z3)] = z3 - 1 + -1 * z2
0.94/1.31	
0.94/1.31	This gives the following inequalities:
0.94/1.31	  I1 <= I0 - 1 /\ I1 <= I2 - 1  ==>  I2 - 1 + -1 * I1 > I2 - 1 + -1 * (I1 + 1)  with  I2 - 1 + -1 * I1 >= 0
0.94/1.31	  I3 <= I4 /\ I4 <= I5 - 1  ==>  I5 - 1 + -1 * I4 >= I5 - 1 + -1 * I4
0.94/1.31	
0.94/1.31	We remove all the strictly oriented dependency pairs.
0.94/1.31	
0.94/1.31	DP problem for innermost termination.
0.94/1.31	P =
0.94/1.31	  f2#(I3, I4, I5) -> f2#(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1]
0.94/1.31	R = 
0.94/1.31	  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
0.94/1.31	  f2(I0, I1, I2) -> f2(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1]
0.94/1.31	  f2(I3, I4, I5) -> f2(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1]
0.94/1.31	  f1(I6, I7, I8) -> f2(I9, I10, I11) [0 <= I6 - 1 /\ -1 <= I11 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I10 - 1]
0.94/1.31	
0.94/1.31	We use the reverse value criterion with the projection function NU:
0.94/1.31	  NU[f2#(z1,z2,z3)] = z2 + -1 * z1
0.94/1.31	
0.94/1.31	This gives the following inequalities:
0.94/1.31	  I3 <= I4 /\ I4 <= I5 - 1  ==>  I4 + -1 * I3 > I4 + -1 * (I3 + 1)  with  I4 + -1 * I3 >= 0
0.94/1.31	
0.94/1.31	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.94/4.29	EOF