0.61/0.66	YES
0.61/0.66	
0.61/0.66	DP problem for innermost termination.
0.61/0.66	P =
0.61/0.66	  init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
0.61/0.66	  f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1]
0.61/0.66	  f1#(I6, I7, I8, I9) -> f2#(I10, I11, I12, I13) [0 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I6 - 1 /\ -1 <= I13 - 1 /\ -1 <= I7 - 1 /\ -1 <= I12 - 1]
0.61/0.66	R = 
0.61/0.66	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
0.61/0.66	  f2(I0, I1, I2, I3) -> f2(I4, I5, I2, I3 + 1) [I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1]
0.61/0.66	  f1(I6, I7, I8, I9) -> f2(I10, I11, I12, I13) [0 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I6 - 1 /\ -1 <= I13 - 1 /\ -1 <= I7 - 1 /\ -1 <= I12 - 1]
0.61/0.66	
0.61/0.66	The dependency graph for this problem is:
0.61/0.66	  0 -> 2
0.61/0.66	  1 -> 1
0.61/0.66	  2 -> 1
0.61/0.66	Where:
0.61/0.66	  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
0.61/0.66	  1) f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1]
0.61/0.66	  2) f1#(I6, I7, I8, I9) -> f2#(I10, I11, I12, I13) [0 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I6 - 1 /\ -1 <= I13 - 1 /\ -1 <= I7 - 1 /\ -1 <= I12 - 1]
0.61/0.66	
0.61/0.66	We have the following SCCs.
0.61/0.66	  { 1 }
0.61/0.66	
0.61/0.66	DP problem for innermost termination.
0.61/0.66	P =
0.61/0.66	  f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1]
0.61/0.66	R = 
0.61/0.66	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
0.61/0.66	  f2(I0, I1, I2, I3) -> f2(I4, I5, I2, I3 + 1) [I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1]
0.61/0.66	  f1(I6, I7, I8, I9) -> f2(I10, I11, I12, I13) [0 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I6 - 1 /\ -1 <= I13 - 1 /\ -1 <= I7 - 1 /\ -1 <= I12 - 1]
0.61/0.66	
0.61/0.66	We use the reverse value criterion with the projection function NU:
0.61/0.66	  NU[f2#(z1,z2,z3,z4)] = z3 - 1 + -1 * z4
0.61/0.66	
0.61/0.66	This gives the following inequalities:
0.61/0.66	  I3 + 2 <= I1 /\ I2 + 2 <= I0 /\ 0 <= I5 - 1 /\ 0 <= I4 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I5 - 1 <= I1 /\ I4 <= I0 /\ I3 <= I2 - 1 /\ -1 <= I3 - 1  ==>  I2 - 1 + -1 * I3 > I2 - 1 + -1 * (I3 + 1)  with  I2 - 1 + -1 * I3 >= 0
0.61/0.66	
0.61/0.66	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.61/3.64	EOF