0.66/0.91	YES
0.66/0.91	
0.66/0.91	DP problem for innermost termination.
0.66/0.91	P =
0.66/0.91	  init#(x1, x2) -> f1#(rnd1, rnd2) 
0.66/0.91	  f2#(I0, I1) -> f2#(I2, I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0]
0.66/0.91	  f2#(I4, I5) -> f2#(I6, I7) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4]
0.66/0.91	  f2#(I8, I9) -> f2#(I10, I11) [-1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8]
0.66/0.91	  f1#(I12, I13) -> f2#(I14, I15) [4 <= I14 - 1 /\ 0 <= I15 - 1]
0.66/0.91	R = 
0.66/0.91	  init(x1, x2) -> f1(rnd1, rnd2) 
0.66/0.91	  f2(I0, I1) -> f2(I2, I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0]
0.66/0.91	  f2(I4, I5) -> f2(I6, I7) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4]
0.66/0.91	  f2(I8, I9) -> f2(I10, I11) [-1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8]
0.66/0.91	  f1(I12, I13) -> f2(I14, I15) [4 <= I14 - 1 /\ 0 <= I15 - 1]
0.66/0.91	
0.66/0.91	The dependency graph for this problem is:
0.66/0.91	  0 -> 4
0.66/0.91	  1 -> 1, 2, 3
0.66/0.91	  2 -> 1, 2, 3
0.66/0.91	  3 -> 1, 2, 3
0.66/0.91	  4 -> 1, 2, 3
0.66/0.91	Where:
0.66/0.91	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
0.66/0.91	  1) f2#(I0, I1) -> f2#(I2, I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0]
0.66/0.91	  2) f2#(I4, I5) -> f2#(I6, I7) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4]
0.66/0.91	  3) f2#(I8, I9) -> f2#(I10, I11) [-1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8]
0.66/0.91	  4) f1#(I12, I13) -> f2#(I14, I15) [4 <= I14 - 1 /\ 0 <= I15 - 1]
0.66/0.91	
0.66/0.91	We have the following SCCs.
0.66/0.91	  { 1, 2, 3 }
0.66/0.91	
0.66/0.91	DP problem for innermost termination.
0.66/0.91	P =
0.66/0.91	  f2#(I0, I1) -> f2#(I2, I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0]
0.66/0.91	  f2#(I4, I5) -> f2#(I6, I7) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4]
0.66/0.91	  f2#(I8, I9) -> f2#(I10, I11) [-1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8]
0.66/0.91	R = 
0.66/0.91	  init(x1, x2) -> f1(rnd1, rnd2) 
0.66/0.91	  f2(I0, I1) -> f2(I2, I3) [-1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0]
0.66/0.91	  f2(I4, I5) -> f2(I6, I7) [-1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4]
0.66/0.91	  f2(I8, I9) -> f2(I10, I11) [-1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8]
0.66/0.91	  f1(I12, I13) -> f2(I14, I15) [4 <= I14 - 1 /\ 0 <= I15 - 1]
0.66/0.91	
0.66/0.91	We use the basic value criterion with the projection function NU:
0.66/0.91	  NU[f2#(z1,z2)] = z1
0.66/0.91	
0.66/0.91	This gives the following inequalities:
0.66/0.91	  -1 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I1 - 1 /\ 2 <= I0 - 1 /\ I3 + 3 <= I0 /\ I2 + 2 <= I0  ==>  I0 >! I2
0.66/0.91	  -1 <= I7 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 2 <= I4 - 1 /\ I7 + 3 <= I4 /\ I6 + 2 <= I4  ==>  I4 >! I6
0.66/0.91	  -1 <= I11 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ 2 <= I8 - 1 /\ I11 + 3 <= I8 /\ I10 + 2 <= I8  ==>  I8 >! I10
0.66/0.91	
0.66/0.91	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.66/3.66	EOF