2.80/2.78	YES
2.80/2.78	
2.80/2.78	DP problem for innermost termination.
2.80/2.78	P =
2.80/2.78	  init#(x1, x2) -> f1#(rnd1, rnd2) 
2.80/2.78	  f3#(I0, I1) -> f3#(I0, 2 * I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1]
2.80/2.78	  f3#(I2, I3) -> f2#(I2 - 1, I4) [I2 <= I3]
2.80/2.78	  f2#(I5, I6) -> f3#(I5, 1) [-1 <= I5 - 1]
2.80/2.78	  f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
2.80/2.78	R = 
2.80/2.78	  init(x1, x2) -> f1(rnd1, rnd2) 
2.80/2.78	  f3(I0, I1) -> f3(I0, 2 * I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1]
2.80/2.78	  f3(I2, I3) -> f2(I2 - 1, I4) [I2 <= I3]
2.80/2.78	  f2(I5, I6) -> f3(I5, 1) [-1 <= I5 - 1]
2.80/2.78	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
2.80/2.78	
2.80/2.78	The dependency graph for this problem is:
2.80/2.78	  0 -> 4
2.80/2.78	  1 -> 1, 2
2.80/2.78	  2 -> 3
2.80/2.78	  3 -> 1, 2
2.80/2.78	  4 -> 3
2.80/2.78	Where:
2.80/2.78	  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
2.80/2.78	  1) f3#(I0, I1) -> f3#(I0, 2 * I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1]
2.80/2.78	  2) f3#(I2, I3) -> f2#(I2 - 1, I4) [I2 <= I3]
2.80/2.78	  3) f2#(I5, I6) -> f3#(I5, 1) [-1 <= I5 - 1]
2.80/2.78	  4) f1#(I7, I8) -> f2#(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
2.80/2.78	
2.80/2.78	We have the following SCCs.
2.80/2.78	  { 1, 2, 3 }
2.80/2.78	
2.80/2.78	DP problem for innermost termination.
2.80/2.79	P =
2.80/2.79	  f3#(I0, I1) -> f3#(I0, 2 * I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1]
2.80/2.79	  f3#(I2, I3) -> f2#(I2 - 1, I4) [I2 <= I3]
2.80/2.79	  f2#(I5, I6) -> f3#(I5, 1) [-1 <= I5 - 1]
2.80/2.79	R = 
2.80/2.79	  init(x1, x2) -> f1(rnd1, rnd2) 
2.80/2.79	  f3(I0, I1) -> f3(I0, 2 * I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1]
2.80/2.79	  f3(I2, I3) -> f2(I2 - 1, I4) [I2 <= I3]
2.80/2.79	  f2(I5, I6) -> f3(I5, 1) [-1 <= I5 - 1]
2.80/2.79	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
2.80/2.79	
2.80/2.79	We use the reverse value criterion with the projection function NU:
2.80/2.79	  NU[f2#(z1,z2)] = z1
2.80/2.79	  NU[f3#(z1,z2)] = z1 - 1 - 1 + -1 * -1
2.80/2.79	
2.80/2.79	This gives the following inequalities:
2.80/2.79	  I1 <= I0 - 1 /\ 0 <= I1 - 1  ==>  I0 - 1 - 1 + -1 * -1 >= I0 - 1 - 1 + -1 * -1
2.80/2.79	  I2 <= I3  ==>  I2 - 1 - 1 + -1 * -1 >= I2 - 1
2.80/2.79	  -1 <= I5 - 1  ==>  I5 > I5 - 1 - 1 + -1 * -1  with  I5 >= 0
2.80/2.79	
2.80/2.79	We remove all the strictly oriented dependency pairs.
2.80/2.79	
2.80/2.79	DP problem for innermost termination.
2.80/2.79	P =
2.80/2.79	  f3#(I0, I1) -> f3#(I0, 2 * I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1]
2.80/2.79	  f3#(I2, I3) -> f2#(I2 - 1, I4) [I2 <= I3]
2.80/2.79	R = 
2.80/2.79	  init(x1, x2) -> f1(rnd1, rnd2) 
2.80/2.79	  f3(I0, I1) -> f3(I0, 2 * I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1]
2.80/2.79	  f3(I2, I3) -> f2(I2 - 1, I4) [I2 <= I3]
2.80/2.79	  f2(I5, I6) -> f3(I5, 1) [-1 <= I5 - 1]
2.80/2.79	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
2.80/2.79	
2.80/2.79	The dependency graph for this problem is:
2.80/2.79	  1 -> 1, 2
2.80/2.79	  2 -> 
2.80/2.79	Where:
2.80/2.79	  1) f3#(I0, I1) -> f3#(I0, 2 * I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1]
2.80/2.79	  2) f3#(I2, I3) -> f2#(I2 - 1, I4) [I2 <= I3]
2.80/2.79	
2.80/2.79	We have the following SCCs.
2.80/2.79	  { 1 }
2.80/2.79	
2.80/2.79	DP problem for innermost termination.
2.80/2.79	P =
2.80/2.79	  f3#(I0, I1) -> f3#(I0, 2 * I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1]
2.80/2.79	R = 
2.80/2.79	  init(x1, x2) -> f1(rnd1, rnd2) 
2.80/2.79	  f3(I0, I1) -> f3(I0, 2 * I1) [I1 <= I0 - 1 /\ 0 <= I1 - 1]
2.80/2.79	  f3(I2, I3) -> f2(I2 - 1, I4) [I2 <= I3]
2.80/2.79	  f2(I5, I6) -> f3(I5, 1) [-1 <= I5 - 1]
2.80/2.79	  f1(I7, I8) -> f2(I9, I10) [0 <= I7 - 1 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1]
2.80/2.79	
2.80/2.79	We use the reverse value criterion with the projection function NU:
2.80/2.79	  NU[f3#(z1,z2)] = z1 - 1 + -1 * z2
2.80/2.79	
2.80/2.79	This gives the following inequalities:
2.80/2.79	  I1 <= I0 - 1 /\ 0 <= I1 - 1  ==>  I0 - 1 + -1 * I1 > I0 - 1 + -1 * (2 * I1)  with  I0 - 1 + -1 * I1 >= 0
2.80/2.79	
2.80/2.79	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.80/5.77	EOF