0.00/0.31	YES
0.00/0.31	
0.00/0.31	DP problem for innermost termination.
0.00/0.31	P =
0.00/0.31	  eval#(x, y) -> eval#(x, y - 1) [y >= 0]
0.00/0.31	  eval#(I0, I1) -> eval#(I0 - 1, z) [I0 >= 0]
0.00/0.31	R = 
0.00/0.31	  eval(x, y) -> eval(x, y - 1) [y >= 0]
0.00/0.31	  eval(I0, I1) -> eval(I0 - 1, z) [I0 >= 0]
0.00/0.31	
0.00/0.31	We use the reverse value criterion with the projection function NU:
0.00/0.31	  NU[eval#(z1,z2)] = z1
0.00/0.31	
0.00/0.31	This gives the following inequalities:
0.00/0.31	  y >= 0  ==>  x >= x
0.00/0.31	  I0 >= 0  ==>  I0 > I0 - 1  with  I0 >= 0
0.00/0.31	
0.00/0.31	We remove all the strictly oriented dependency pairs.
0.00/0.31	
0.00/0.31	DP problem for innermost termination.
0.00/0.31	P =
0.00/0.31	  eval#(x, y) -> eval#(x, y - 1) [y >= 0]
0.00/0.31	R = 
0.00/0.31	  eval(x, y) -> eval(x, y - 1) [y >= 0]
0.00/0.31	  eval(I0, I1) -> eval(I0 - 1, z) [I0 >= 0]
0.00/0.31	
0.00/0.31	We use the reverse value criterion with the projection function NU:
0.00/0.31	  NU[eval#(z1,z2)] = z2 + -1 * 0
0.00/0.31	
0.00/0.31	This gives the following inequalities:
0.00/0.31	  y >= 0  ==>  y + -1 * 0 > y - 1 + -1 * 0  with  y + -1 * 0 >= 0
0.00/0.31	
0.00/0.31	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.29	EOF