0.00/0.15	YES
0.00/0.15	
0.00/0.15	DP problem for innermost termination.
0.00/0.15	P =
0.00/0.15	  eval#(x, y, z) -> eval#(x - 1, y - 1, z) [x > z && y > z]
0.00/0.15	R = 
0.00/0.15	  eval(x, y, z) -> eval(x - 1, y - 1, z) [x > z && y > z]
0.00/0.15	
0.00/0.15	We use the reverse value criterion with the projection function NU:
0.00/0.15	  NU[eval#(z1,z2,z3)] = z1 + -1 * z3
0.00/0.15	
0.00/0.15	This gives the following inequalities:
0.00/0.15	  x > z && y > z  ==>  x + -1 * z > x - 1 + -1 * z  with  x + -1 * z >= 0
0.00/0.15	
0.00/0.15	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.13	EOF