0.00/0.16	YES
0.00/0.16	
0.00/0.16	DP problem for innermost termination.
0.00/0.16	P =
0.00/0.16	  eval#(x, y, z) -> eval#(x, y, z + y) [x >= z && y > 0]
0.00/0.16	R = 
0.00/0.16	  eval(x, y, z) -> eval(x, y, z + y) [x >= z && y > 0]
0.00/0.16	
0.00/0.16	We use the reverse value criterion with the projection function NU:
0.00/0.16	  NU[eval#(z1,z2,z3)] = z1 + -1 * z3
0.00/0.16	
0.00/0.16	This gives the following inequalities:
0.00/0.16	  x >= z && y > 0  ==>  x + -1 * z > x + -1 * (z + y)  with  x + -1 * z >= 0
0.00/0.16	
0.00/0.16	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.14	EOF