0.00/0.33	YES
0.00/0.33	
0.00/0.33	DP problem for innermost termination.
0.00/0.33	P =
0.00/0.33	  eval#(x, y) -> eval#(x, y - 1) [x > 0 && y > 0]
0.00/0.33	  eval#(I0, I1) -> eval#(I0 - 1, z) [I0 > 0 && I1 > 0]
0.00/0.33	R = 
0.00/0.33	  eval(x, y) -> eval(x, y - 1) [x > 0 && y > 0]
0.00/0.33	  eval(I0, I1) -> eval(I0 - 1, z) [I0 > 0 && I1 > 0]
0.00/0.33	
0.00/0.33	We use the reverse value criterion with the projection function NU:
0.00/0.33	  NU[eval#(z1,z2)] = z1
0.00/0.33	
0.00/0.33	This gives the following inequalities:
0.00/0.33	  x > 0 && y > 0  ==>  x >= x
0.00/0.33	  I0 > 0 && I1 > 0  ==>  I0 > I0 - 1  with  I0 >= 0
0.00/0.33	
0.00/0.33	We remove all the strictly oriented dependency pairs.
0.00/0.33	
0.00/0.33	DP problem for innermost termination.
0.00/0.33	P =
0.00/0.33	  eval#(x, y) -> eval#(x, y - 1) [x > 0 && y > 0]
0.00/0.33	R = 
0.00/0.33	  eval(x, y) -> eval(x, y - 1) [x > 0 && y > 0]
0.00/0.33	  eval(I0, I1) -> eval(I0 - 1, z) [I0 > 0 && I1 > 0]
0.00/0.33	
0.00/0.33	We use the reverse value criterion with the projection function NU:
0.00/0.33	  NU[eval#(z1,z2)] = z2
0.00/0.33	
0.00/0.33	This gives the following inequalities:
0.00/0.33	  x > 0 && y > 0  ==>  y > y - 1  with  y >= 0
0.00/0.33	
0.00/0.33	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.31	EOF