0.00/0.27	YES
0.00/0.27	
0.00/0.27	DP problem for innermost termination.
0.00/0.27	P =
0.00/0.27	  eval#(x) -> eval#(x / 2) [x > 0 && not(x = 0) && x % 2 = 0]
0.00/0.27	  eval#(I0) -> eval#(I0 - 1) [I0 > 0 && not(I0 = 0) && I0 % 2 > 0]
0.00/0.27	R = 
0.00/0.27	  eval(x) -> eval(x / 2) [x > 0 && not(x = 0) && x % 2 = 0]
0.00/0.27	  eval(I0) -> eval(I0 - 1) [I0 > 0 && not(I0 = 0) && I0 % 2 > 0]
0.00/0.27	
0.00/0.27	We use the reverse value criterion with the projection function NU:
0.00/0.27	  NU[eval#(z1)] = z1
0.00/0.27	
0.00/0.27	This gives the following inequalities:
0.00/0.27	  x > 0 && not(x = 0) && x % 2 = 0  ==>  x > x / 2  with  x >= 0
0.00/0.27	  I0 > 0 && not(I0 = 0) && I0 % 2 > 0  ==>  I0 > I0 - 1  with  I0 >= 0
0.00/0.27	
0.00/0.27	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.25	EOF