0.00/0.22	YES
0.00/0.22	
0.00/0.22	DP problem for innermost termination.
0.00/0.22	P =
0.00/0.22	  eval#(x, y) -> eval#(x + 1, y) [y > x && y >= x]
0.00/0.22	  eval#(I0, I1) -> eval#(I0 + 1, I1) [I0 > I1 && I1 >= I0]
0.00/0.22	  eval#(I2, I3) -> eval#(I2, I3 + 1) [I3 > I2 && I2 > I3]
0.00/0.22	  eval#(I4, I5) -> eval#(I4, I5 + 1) [I4 > I5]
0.00/0.22	R = 
0.00/0.22	  eval(x, y) -> eval(x + 1, y) [y > x && y >= x]
0.00/0.22	  eval(I0, I1) -> eval(I0 + 1, I1) [I0 > I1 && I1 >= I0]
0.00/0.22	  eval(I2, I3) -> eval(I2, I3 + 1) [I3 > I2 && I2 > I3]
0.00/0.22	  eval(I4, I5) -> eval(I4, I5 + 1) [I4 > I5]
0.00/0.22	
0.00/0.22	The dependency graph for this problem is:
0.00/0.22	  0 -> 0
0.00/0.22	  1 -> 
0.00/0.22	  2 -> 
0.00/0.22	  3 -> 3
0.00/0.22	Where:
0.00/0.22	  0) eval#(x, y) -> eval#(x + 1, y) [y > x && y >= x]
0.00/0.22	  1) eval#(I0, I1) -> eval#(I0 + 1, I1) [I0 > I1 && I1 >= I0]
0.00/0.22	  2) eval#(I2, I3) -> eval#(I2, I3 + 1) [I3 > I2 && I2 > I3]
0.00/0.22	  3) eval#(I4, I5) -> eval#(I4, I5 + 1) [I4 > I5]
0.00/0.22	
0.00/0.22	We have the following SCCs.
0.00/0.22	  { 0 }
0.00/0.22	  { 3 }
0.00/0.22	
0.00/0.22	DP problem for innermost termination.
0.00/0.22	P =
0.00/0.22	  eval#(I4, I5) -> eval#(I4, I5 + 1) [I4 > I5]
0.00/0.22	R = 
0.00/0.22	  eval(x, y) -> eval(x + 1, y) [y > x && y >= x]
0.00/0.22	  eval(I0, I1) -> eval(I0 + 1, I1) [I0 > I1 && I1 >= I0]
0.00/0.22	  eval(I2, I3) -> eval(I2, I3 + 1) [I3 > I2 && I2 > I3]
0.00/0.22	  eval(I4, I5) -> eval(I4, I5 + 1) [I4 > I5]
0.00/0.22	
0.00/0.22	We use the reverse value criterion with the projection function NU:
0.00/0.22	  NU[eval#(z1,z2)] = z1 + -1 * z2
0.00/0.22	
0.00/0.22	This gives the following inequalities:
0.00/0.22	  I4 > I5  ==>  I4 + -1 * I5 > I4 + -1 * (I5 + 1)  with  I4 + -1 * I5 >= 0
0.00/0.22	
0.00/0.22	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.22	
0.00/0.22	DP problem for innermost termination.
0.00/0.22	P =
0.00/0.22	  eval#(x, y) -> eval#(x + 1, y) [y > x && y >= x]
0.00/0.22	R = 
0.00/0.22	  eval(x, y) -> eval(x + 1, y) [y > x && y >= x]
0.00/0.22	  eval(I0, I1) -> eval(I0 + 1, I1) [I0 > I1 && I1 >= I0]
0.00/0.22	  eval(I2, I3) -> eval(I2, I3 + 1) [I3 > I2 && I2 > I3]
0.00/0.22	  eval(I4, I5) -> eval(I4, I5 + 1) [I4 > I5]
0.00/0.22	
0.00/0.22	We use the reverse value criterion with the projection function NU:
0.00/0.22	  NU[eval#(z1,z2)] = z2 + -1 * z1
0.00/0.22	
0.00/0.22	This gives the following inequalities:
0.00/0.22	  y > x && y >= x  ==>  y + -1 * x > y + -1 * (x + 1)  with  y + -1 * x >= 0
0.00/0.22	
0.00/0.22	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.20	EOF