0.00/0.32	YES
0.00/0.32	
0.00/0.32	DP problem for innermost termination.
0.00/0.32	P =
0.00/0.32	  eval#(l, u) -> eval#(l, (l + u) / 2) [l >= 0 && u >= 0 && l < u]
0.00/0.32	  eval#(I0, I1) -> eval#((I0 + I1) / 2 + 1, I1) [I0 >= 0 && I1 >= 0 && I0 < I1]
0.00/0.32	R = 
0.00/0.32	  eval(l, u) -> eval(l, (l + u) / 2) [l >= 0 && u >= 0 && l < u]
0.00/0.32	  eval(I0, I1) -> eval((I0 + I1) / 2 + 1, I1) [I0 >= 0 && I1 >= 0 && I0 < I1]
0.00/0.32	
0.00/0.32	We use the reverse value criterion with the projection function NU:
0.00/0.32	  NU[eval#(z1,z2)] = z2 + -1 * z1
0.00/0.32	
0.00/0.32	This gives the following inequalities:
0.00/0.32	  l >= 0 && u >= 0 && l < u  ==>  u + -1 * l > (l + u) / 2 + -1 * l  with  u + -1 * l >= 0
0.00/0.32	  I0 >= 0 && I1 >= 0 && I0 < I1  ==>  I1 + -1 * I0 > I1 + -1 * ((I0 + I1) / 2 + 1)  with  I1 + -1 * I0 >= 0
0.00/0.32	
0.00/0.32	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.30	EOF