0.00/0.10	YES
0.00/0.10	
0.00/0.10	DP problem for innermost termination.
0.00/0.10	P =
0.00/0.10	  eval#(x, y) -> eval#(x - 1, y) [x > y]
0.00/0.10	R = 
0.00/0.10	  eval(x, y) -> eval(x - 1, y) [x > y]
0.00/0.10	
0.00/0.10	We use the reverse value criterion with the projection function NU:
0.00/0.10	  NU[eval#(z1,z2)] = z1 + -1 * z2
0.00/0.10	
0.00/0.10	This gives the following inequalities:
0.00/0.10	  x > y  ==>  x + -1 * y > x - 1 + -1 * y  with  x + -1 * y >= 0
0.00/0.10	
0.00/0.10	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.09	EOF