0.00/0.06	YES
0.00/0.06	
0.00/0.06	DP problem for innermost termination.
0.00/0.06	P =
0.00/0.06	  eval#(x, y) -> eval#(x - 1, y - 1) [x > 0 && y > 0]
0.00/0.06	R = 
0.00/0.06	  eval(x, y) -> eval(x - 1, y - 1) [x > 0 && y > 0]
0.00/0.06	
0.00/0.06	We use the reverse value criterion with the projection function NU:
0.00/0.06	  NU[eval#(z1,z2)] = z1
0.00/0.06	
0.00/0.06	This gives the following inequalities:
0.00/0.06	  x > 0 && y > 0  ==>  x > x - 1  with  x >= 0
0.00/0.06	
0.00/0.06	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.04	EOF