1.89/2.25	YES
1.89/2.25	
1.89/2.25	DP problem for innermost termination.
1.89/2.25	P =
1.89/2.25	  init#(x1, x2) -> f1#(rnd1, rnd2) 
1.89/2.25	  f3#(I0, I1) -> f2#(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1]
1.89/2.25	  f2#(I2, I3) -> f3#(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1]
1.89/2.25	  f3#(I5, I6) -> f2#(I5, I6 + 1) [I6 + 1 - 2 * I7 = 0 /\ I6 <= I5 - 1 /\ -1 <= I6 - 1 /\ I6 + 1 - 2 * I7 <= 1 /\ 0 <= I6 + 1 - 2 * I7]
1.89/2.25	  f2#(I8, I9) -> f3#(I8, I9) [I9 + 1 - 2 * I10 = 0 /\ -1 <= I9 - 1 /\ I9 <= I8 - 1]
1.89/2.25	  f1#(I11, I12) -> f2#(I13, I14) [0 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1]
1.89/2.25	R = 
1.89/2.25	  init(x1, x2) -> f1(rnd1, rnd2) 
1.89/2.25	  f3(I0, I1) -> f2(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1]
1.89/2.25	  f2(I2, I3) -> f3(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1]
1.89/2.25	  f3(I5, I6) -> f2(I5, I6 + 1) [I6 + 1 - 2 * I7 = 0 /\ I6 <= I5 - 1 /\ -1 <= I6 - 1 /\ I6 + 1 - 2 * I7 <= 1 /\ 0 <= I6 + 1 - 2 * I7]
1.89/2.25	  f2(I8, I9) -> f3(I8, I9) [I9 + 1 - 2 * I10 = 0 /\ -1 <= I9 - 1 /\ I9 <= I8 - 1]
1.89/2.25	  f1(I11, I12) -> f2(I13, I14) [0 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1]
1.89/2.25	
1.89/2.25	The dependency graph for this problem is:
1.89/2.25	  0 -> 5
1.89/2.25	  1 -> 2
1.89/2.25	  2 -> 1
1.89/2.25	  3 -> 2
1.89/2.25	  4 -> 3
1.89/2.25	  5 -> 2, 4
1.89/2.25	Where:
1.89/2.25	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
1.89/2.25	  1) f3#(I0, I1) -> f2#(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1]
1.89/2.25	  2) f2#(I2, I3) -> f3#(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1]
1.89/2.25	  3) f3#(I5, I6) -> f2#(I5, I6 + 1) [I6 + 1 - 2 * I7 = 0 /\ I6 <= I5 - 1 /\ -1 <= I6 - 1 /\ I6 + 1 - 2 * I7 <= 1 /\ 0 <= I6 + 1 - 2 * I7]
1.89/2.25	  4) f2#(I8, I9) -> f3#(I8, I9) [I9 + 1 - 2 * I10 = 0 /\ -1 <= I9 - 1 /\ I9 <= I8 - 1]
1.89/2.25	  5) f1#(I11, I12) -> f2#(I13, I14) [0 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1]
1.89/2.25	
1.89/2.25	We have the following SCCs.
1.89/2.25	  { 1, 2 }
1.89/2.25	
1.89/2.25	DP problem for innermost termination.
1.89/2.25	P =
1.89/2.25	  f3#(I0, I1) -> f2#(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1]
1.89/2.25	  f2#(I2, I3) -> f3#(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1]
1.89/2.25	R = 
1.89/2.25	  init(x1, x2) -> f1(rnd1, rnd2) 
1.89/2.25	  f3(I0, I1) -> f2(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1]
1.89/2.25	  f2(I2, I3) -> f3(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1]
1.89/2.25	  f3(I5, I6) -> f2(I5, I6 + 1) [I6 + 1 - 2 * I7 = 0 /\ I6 <= I5 - 1 /\ -1 <= I6 - 1 /\ I6 + 1 - 2 * I7 <= 1 /\ 0 <= I6 + 1 - 2 * I7]
1.89/2.25	  f2(I8, I9) -> f3(I8, I9) [I9 + 1 - 2 * I10 = 0 /\ -1 <= I9 - 1 /\ I9 <= I8 - 1]
1.89/2.25	  f1(I11, I12) -> f2(I13, I14) [0 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1]
1.89/2.25	
1.89/2.25	We use the reverse value criterion with the projection function NU:
1.89/2.25	  NU[f2#(z1,z2)] = z1 - 1 + -1 * z2
1.89/2.25	  NU[f3#(z1,z2)] = z1 - 1 + -1 * (z2 + 2)
1.89/2.25	
1.89/2.25	This gives the following inequalities:
1.89/2.25	  -1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1  ==>  I0 - 1 + -1 * (I1 + 2) >= I0 - 1 + -1 * (I1 + 2)
1.89/2.25	  -1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1  ==>  I2 - 1 + -1 * I3 > I2 - 1 + -1 * (I3 + 2)  with  I2 - 1 + -1 * I3 >= 0
1.89/2.25	
1.89/2.25	We remove all the strictly oriented dependency pairs.
1.89/2.25	
1.89/2.25	DP problem for innermost termination.
1.89/2.25	P =
1.89/2.25	  f3#(I0, I1) -> f2#(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1]
1.89/2.25	R = 
1.89/2.25	  init(x1, x2) -> f1(rnd1, rnd2) 
1.89/2.25	  f3(I0, I1) -> f2(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1]
1.89/2.25	  f2(I2, I3) -> f3(I2, I3) [-1 <= I3 - 1 /\ I3 + 1 - 2 * I4 = 1 /\ I3 <= I2 - 1]
1.89/2.25	  f3(I5, I6) -> f2(I5, I6 + 1) [I6 + 1 - 2 * I7 = 0 /\ I6 <= I5 - 1 /\ -1 <= I6 - 1 /\ I6 + 1 - 2 * I7 <= 1 /\ 0 <= I6 + 1 - 2 * I7]
1.89/2.25	  f2(I8, I9) -> f3(I8, I9) [I9 + 1 - 2 * I10 = 0 /\ -1 <= I9 - 1 /\ I9 <= I8 - 1]
1.89/2.25	  f1(I11, I12) -> f2(I13, I14) [0 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I12 - 1 /\ -1 <= I14 - 1]
1.89/2.25	
1.89/2.25	The dependency graph for this problem is:
1.89/2.25	  1 -> 
1.89/2.25	Where:
1.89/2.25	  1) f3#(I0, I1) -> f2#(I0, I1 + 2) [-1 <= I1 - 1 /\ I1 <= I0 - 1 /\ I1 + 1 - 2 * y1 = 1 /\ I1 + 1 - 2 * y1 <= 1 /\ 0 <= I1 + 1 - 2 * y1]
1.89/2.25	
1.89/2.25	We have the following SCCs.
1.89/2.25	  
1.89/5.23	EOF