0.00/0.40	YES
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  eval#(x, y, z) -> eval#(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y]
0.00/0.40	  eval#(I0, I1, I2) -> eval#(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0]
0.00/0.40	  eval#(I3, I4, I5) -> eval#(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0]
0.00/0.40	R = 
0.00/0.40	  eval(x, y, z) -> eval(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y]
0.00/0.40	  eval(I0, I1, I2) -> eval(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0]
0.00/0.40	  eval(I3, I4, I5) -> eval(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0]
0.00/0.40	
0.00/0.40	The dependency graph for this problem is:
0.00/0.40	  0 -> 
0.00/0.40	  1 -> 1
0.00/0.40	  2 -> 1, 2
0.00/0.40	Where:
0.00/0.40	  0) eval#(x, y, z) -> eval#(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y]
0.00/0.40	  1) eval#(I0, I1, I2) -> eval#(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0]
0.00/0.40	  2) eval#(I3, I4, I5) -> eval#(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0]
0.00/0.40	
0.00/0.40	We have the following SCCs.
0.00/0.40	  { 2 }
0.00/0.40	  { 1 }
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  eval#(I0, I1, I2) -> eval#(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0]
0.00/0.40	R = 
0.00/0.40	  eval(x, y, z) -> eval(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y]
0.00/0.40	  eval(I0, I1, I2) -> eval(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0]
0.00/0.40	  eval(I3, I4, I5) -> eval(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0]
0.00/0.40	
0.00/0.40	We use the reverse value criterion with the projection function NU:
0.00/0.40	  NU[eval#(z1,z2,z3)] = z2
0.00/0.40	
0.00/0.40	This gives the following inequalities:
0.00/0.40	  I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0  ==>  I1 > I1 - 1  with  I1 >= 0
0.00/0.40	
0.00/0.40	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.40	
0.00/0.40	DP problem for innermost termination.
0.00/0.40	P =
0.00/0.40	  eval#(I3, I4, I5) -> eval#(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0]
0.00/0.40	R = 
0.00/0.40	  eval(x, y, z) -> eval(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y]
0.00/0.40	  eval(I0, I1, I2) -> eval(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0]
0.00/0.40	  eval(I3, I4, I5) -> eval(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0]
0.00/0.40	
0.00/0.40	We use the reverse value criterion with the projection function NU:
0.00/0.40	  NU[eval#(z1,z2,z3)] = z1 + z2 + -1 * z3
0.00/0.40	
0.00/0.40	This gives the following inequalities:
0.00/0.40	  I3 + I4 > I5 && I5 >= 0 && I3 > 0  ==>  I3 + I4 + -1 * I5 > I3 - 1 + I4 + -1 * I5  with  I3 + I4 + -1 * I5 >= 0
0.00/0.40	
0.00/0.40	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.38	EOF