0.00/0.44	YES
0.00/0.44	
0.00/0.44	DP problem for innermost termination.
0.00/0.44	P =
0.00/0.44	  eval#(x, y) -> eval#(x, y) [y > 0 && 0 >= x && 0 >= y]
0.00/0.44	  eval#(I0, I1) -> eval#(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1]
0.00/0.44	  eval#(I2, I3) -> eval#(I2, I3 - 1) [I3 > 0 && 0 >= I2]
0.00/0.44	  eval#(I4, I5) -> eval#(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0]
0.00/0.44	  eval#(I6, I7) -> eval#(I6 - 1, I7) [I7 > 0 && I6 > 0]
0.00/0.44	  eval#(I8, I9) -> eval#(I8 - 1, I9) [I8 > 0]
0.00/0.44	R = 
0.00/0.44	  eval(x, y) -> eval(x, y) [y > 0 && 0 >= x && 0 >= y]
0.00/0.44	  eval(I0, I1) -> eval(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1]
0.00/0.44	  eval(I2, I3) -> eval(I2, I3 - 1) [I3 > 0 && 0 >= I2]
0.00/0.44	  eval(I4, I5) -> eval(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0]
0.00/0.44	  eval(I6, I7) -> eval(I6 - 1, I7) [I7 > 0 && I6 > 0]
0.00/0.44	  eval(I8, I9) -> eval(I8 - 1, I9) [I8 > 0]
0.00/0.44	
0.00/0.44	The dependency graph for this problem is:
0.00/0.44	  0 -> 
0.00/0.44	  1 -> 
0.00/0.44	  2 -> 2
0.00/0.44	  3 -> 
0.00/0.44	  4 -> 2, 4, 5
0.00/0.44	  5 -> 2, 4, 5
0.00/0.44	Where:
0.00/0.44	  0) eval#(x, y) -> eval#(x, y) [y > 0 && 0 >= x && 0 >= y]
0.00/0.44	  1) eval#(I0, I1) -> eval#(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1]
0.00/0.44	  2) eval#(I2, I3) -> eval#(I2, I3 - 1) [I3 > 0 && 0 >= I2]
0.00/0.44	  3) eval#(I4, I5) -> eval#(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0]
0.00/0.44	  4) eval#(I6, I7) -> eval#(I6 - 1, I7) [I7 > 0 && I6 > 0]
0.00/0.44	  5) eval#(I8, I9) -> eval#(I8 - 1, I9) [I8 > 0]
0.00/0.44	
0.00/0.44	We have the following SCCs.
0.00/0.44	  { 4, 5 }
0.00/0.44	  { 2 }
0.00/0.44	
0.00/0.44	DP problem for innermost termination.
0.00/0.44	P =
0.00/0.44	  eval#(I2, I3) -> eval#(I2, I3 - 1) [I3 > 0 && 0 >= I2]
0.00/0.44	R = 
0.00/0.44	  eval(x, y) -> eval(x, y) [y > 0 && 0 >= x && 0 >= y]
0.00/0.44	  eval(I0, I1) -> eval(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1]
0.00/0.44	  eval(I2, I3) -> eval(I2, I3 - 1) [I3 > 0 && 0 >= I2]
0.00/0.44	  eval(I4, I5) -> eval(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0]
0.00/0.44	  eval(I6, I7) -> eval(I6 - 1, I7) [I7 > 0 && I6 > 0]
0.00/0.44	  eval(I8, I9) -> eval(I8 - 1, I9) [I8 > 0]
0.00/0.44	
0.00/0.44	We use the reverse value criterion with the projection function NU:
0.00/0.44	  NU[eval#(z1,z2)] = z2 + -1 * 0
0.00/0.44	
0.00/0.44	This gives the following inequalities:
0.00/0.44	  I3 > 0 && 0 >= I2  ==>  I3 + -1 * 0 > I3 - 1 + -1 * 0  with  I3 + -1 * 0 >= 0
0.00/0.44	
0.00/0.44	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.44	
0.00/0.44	DP problem for innermost termination.
0.00/0.44	P =
0.00/0.44	  eval#(I6, I7) -> eval#(I6 - 1, I7) [I7 > 0 && I6 > 0]
0.00/0.44	  eval#(I8, I9) -> eval#(I8 - 1, I9) [I8 > 0]
0.00/0.44	R = 
0.00/0.44	  eval(x, y) -> eval(x, y) [y > 0 && 0 >= x && 0 >= y]
0.00/0.44	  eval(I0, I1) -> eval(I0, I1) [I0 > 0 && 0 >= I0 && 0 >= I1]
0.00/0.44	  eval(I2, I3) -> eval(I2, I3 - 1) [I3 > 0 && 0 >= I2]
0.00/0.44	  eval(I4, I5) -> eval(I4, I5 - 1) [I4 > 0 && 0 >= I4 && I5 > 0]
0.00/0.44	  eval(I6, I7) -> eval(I6 - 1, I7) [I7 > 0 && I6 > 0]
0.00/0.44	  eval(I8, I9) -> eval(I8 - 1, I9) [I8 > 0]
0.00/0.44	
0.00/0.44	We use the reverse value criterion with the projection function NU:
0.00/0.44	  NU[eval#(z1,z2)] = z1
0.00/0.44	
0.00/0.44	This gives the following inequalities:
0.00/0.44	  I7 > 0 && I6 > 0  ==>  I6 > I6 - 1  with  I6 >= 0
0.00/0.44	  I8 > 0  ==>  I8 > I8 - 1  with  I8 >= 0
0.00/0.44	
0.00/0.44	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.42	EOF