0.00/0.39	YES
0.00/0.39	
0.00/0.39	DP problem for innermost termination.
0.00/0.39	P =
0.00/0.39	  eval#(x, y) -> eval#(x, y - 1) [x + y > 0 && y >= x && x > y]
0.00/0.39	  eval#(I0, I1) -> eval#(I0, I1 - 1) [I0 + I1 > 0 && I1 >= I0 && I1 > I0]
0.00/0.39	  eval#(I2, I3) -> eval#(I2 - 1, I3) [I2 + I3 > 0 && I3 >= I2 && I2 = I3]
0.00/0.39	  eval#(I4, I5) -> eval#(I4 - 1, I5) [I4 + I5 > 0 && I4 > I5]
0.00/0.39	R = 
0.00/0.39	  eval(x, y) -> eval(x, y - 1) [x + y > 0 && y >= x && x > y]
0.00/0.39	  eval(I0, I1) -> eval(I0, I1 - 1) [I0 + I1 > 0 && I1 >= I0 && I1 > I0]
0.00/0.39	  eval(I2, I3) -> eval(I2 - 1, I3) [I2 + I3 > 0 && I3 >= I2 && I2 = I3]
0.00/0.39	  eval(I4, I5) -> eval(I4 - 1, I5) [I4 + I5 > 0 && I4 > I5]
0.00/0.39	
0.00/0.39	The dependency graph for this problem is:
0.00/0.39	  0 -> 
0.00/0.39	  1 -> 1, 2
0.00/0.39	  2 -> 1
0.00/0.39	  3 -> 2, 3
0.00/0.39	Where:
0.00/0.39	  0) eval#(x, y) -> eval#(x, y - 1) [x + y > 0 && y >= x && x > y]
0.00/0.39	  1) eval#(I0, I1) -> eval#(I0, I1 - 1) [I0 + I1 > 0 && I1 >= I0 && I1 > I0]
0.00/0.39	  2) eval#(I2, I3) -> eval#(I2 - 1, I3) [I2 + I3 > 0 && I3 >= I2 && I2 = I3]
0.00/0.39	  3) eval#(I4, I5) -> eval#(I4 - 1, I5) [I4 + I5 > 0 && I4 > I5]
0.00/0.39	
0.00/0.39	We have the following SCCs.
0.00/0.39	  { 3 }
0.00/0.39	  { 1, 2 }
0.00/0.39	
0.00/0.39	DP problem for innermost termination.
0.00/0.39	P =
0.00/0.39	  eval#(I0, I1) -> eval#(I0, I1 - 1) [I0 + I1 > 0 && I1 >= I0 && I1 > I0]
0.00/0.39	  eval#(I2, I3) -> eval#(I2 - 1, I3) [I2 + I3 > 0 && I3 >= I2 && I2 = I3]
0.00/0.39	R = 
0.00/0.39	  eval(x, y) -> eval(x, y - 1) [x + y > 0 && y >= x && x > y]
0.00/0.39	  eval(I0, I1) -> eval(I0, I1 - 1) [I0 + I1 > 0 && I1 >= I0 && I1 > I0]
0.00/0.39	  eval(I2, I3) -> eval(I2 - 1, I3) [I2 + I3 > 0 && I3 >= I2 && I2 = I3]
0.00/0.39	  eval(I4, I5) -> eval(I4 - 1, I5) [I4 + I5 > 0 && I4 > I5]
0.00/0.39	
0.00/0.39	We use the reverse value criterion with the projection function NU:
0.00/0.39	  NU[eval#(z1,z2)] = z1 + z2 + -1 * 0
0.00/0.39	
0.00/0.39	This gives the following inequalities:
0.00/0.39	  I0 + I1 > 0 && I1 >= I0 && I1 > I0  ==>  I0 + I1 + -1 * 0 > I0 + (I1 - 1) + -1 * 0  with  I0 + I1 + -1 * 0 >= 0
0.00/0.39	  I2 + I3 > 0 && I3 >= I2 && I2 = I3  ==>  I2 + I3 + -1 * 0 > I2 - 1 + I3 + -1 * 0  with  I2 + I3 + -1 * 0 >= 0
0.00/0.39	
0.00/0.39	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.39	
0.00/0.39	DP problem for innermost termination.
0.00/0.39	P =
0.00/0.39	  eval#(I4, I5) -> eval#(I4 - 1, I5) [I4 + I5 > 0 && I4 > I5]
0.00/0.39	R = 
0.00/0.39	  eval(x, y) -> eval(x, y - 1) [x + y > 0 && y >= x && x > y]
0.00/0.39	  eval(I0, I1) -> eval(I0, I1 - 1) [I0 + I1 > 0 && I1 >= I0 && I1 > I0]
0.00/0.39	  eval(I2, I3) -> eval(I2 - 1, I3) [I2 + I3 > 0 && I3 >= I2 && I2 = I3]
0.00/0.39	  eval(I4, I5) -> eval(I4 - 1, I5) [I4 + I5 > 0 && I4 > I5]
0.00/0.39	
0.00/0.39	We use the reverse value criterion with the projection function NU:
0.00/0.39	  NU[eval#(z1,z2)] = z1
0.00/0.39	
0.00/0.39	This gives the following inequalities:
0.00/0.39	  I4 + I5 > 0 && I4 > I5  ==>  I4 > I4 - 1  with  I4 >= 0
0.00/0.39	
0.00/0.39	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.37	EOF