0.00/0.51	YES
0.00/0.51	
0.00/0.51	DP problem for innermost termination.
0.00/0.51	P =
0.00/0.51	  eval#(x, y, z) -> eval#(x, y, z + 1) [y > x && x >= z]
0.00/0.51	  eval#(I0, I1, I2) -> eval#(I0 + 1, I1, I2) [I1 > I0 && I2 > I0]
0.00/0.51	R = 
0.00/0.51	  eval(x, y, z) -> eval(x, y, z + 1) [y > x && x >= z]
0.00/0.51	  eval(I0, I1, I2) -> eval(I0 + 1, I1, I2) [I1 > I0 && I2 > I0]
0.00/0.51	
0.00/0.51	We use the reverse value criterion with the projection function NU:
0.00/0.51	  NU[eval#(z1,z2,z3)] = z2 + -1 * z1
0.00/0.51	
0.00/0.51	This gives the following inequalities:
0.00/0.51	  y > x && x >= z  ==>  y + -1 * x >= y + -1 * x
0.00/0.51	  I1 > I0 && I2 > I0  ==>  I1 + -1 * I0 > I1 + -1 * (I0 + 1)  with  I1 + -1 * I0 >= 0
0.00/0.51	
0.00/0.51	We remove all the strictly oriented dependency pairs.
0.00/0.51	
0.00/0.51	DP problem for innermost termination.
0.00/0.51	P =
0.00/0.51	  eval#(x, y, z) -> eval#(x, y, z + 1) [y > x && x >= z]
0.00/0.51	R = 
0.00/0.51	  eval(x, y, z) -> eval(x, y, z + 1) [y > x && x >= z]
0.00/0.51	  eval(I0, I1, I2) -> eval(I0 + 1, I1, I2) [I1 > I0 && I2 > I0]
0.00/0.51	
0.00/0.51	We use the reverse value criterion with the projection function NU:
0.00/0.51	  NU[eval#(z1,z2,z3)] = z1 + -1 * z3
0.00/0.51	
0.00/0.51	This gives the following inequalities:
0.00/0.51	  y > x && x >= z  ==>  x + -1 * z > x + -1 * (z + 1)  with  x + -1 * z >= 0
0.00/0.51	
0.00/0.51	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.49	EOF