0.00/0.12	YES
0.00/0.12	
0.00/0.12	DP problem for innermost termination.
0.00/0.12	P =
0.00/0.12	  eval#(x, y) -> eval#(x, y + 1) [x >= y + 1]
0.00/0.12	R = 
0.00/0.12	  eval(x, y) -> eval(x, y + 1) [x >= y + 1]
0.00/0.12	
0.00/0.12	We use the reverse value criterion with the projection function NU:
0.00/0.12	  NU[eval#(z1,z2)] = z1 + -1 * (z2 + 1)
0.00/0.12	
0.00/0.12	This gives the following inequalities:
0.00/0.12	  x >= y + 1  ==>  x + -1 * (y + 1) > x + -1 * (y + 1 + 1)  with  x + -1 * (y + 1) >= 0
0.00/0.12	
0.00/0.12	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.10	EOF