0.00/0.54	YES
0.00/0.54	
0.00/0.54	DP problem for innermost termination.
0.00/0.54	P =
0.00/0.54	  eval#(x, y, z) -> eval#(x, y, z) [y > z && z >= x && z >= y]
0.00/0.54	  eval#(I0, I1, I2) -> eval#(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1]
0.00/0.54	  eval#(I3, I4, I5) -> eval#(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
0.00/0.54	  eval#(I6, I7, I8) -> eval#(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8]
0.00/0.54	  eval#(I9, I10, I11) -> eval#(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
0.00/0.54	  eval#(I12, I13, I14) -> eval#(I12 - 1, I13, I14) [I12 > I14]
0.00/0.54	R = 
0.00/0.54	  eval(x, y, z) -> eval(x, y, z) [y > z && z >= x && z >= y]
0.00/0.54	  eval(I0, I1, I2) -> eval(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1]
0.00/0.54	  eval(I3, I4, I5) -> eval(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
0.00/0.54	  eval(I6, I7, I8) -> eval(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8]
0.00/0.54	  eval(I9, I10, I11) -> eval(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
0.00/0.54	  eval(I12, I13, I14) -> eval(I12 - 1, I13, I14) [I12 > I14]
0.00/0.54	
0.00/0.54	The dependency graph for this problem is:
0.00/0.54	  0 -> 
0.00/0.54	  1 -> 
0.00/0.54	  2 -> 2
0.00/0.54	  3 -> 
0.00/0.54	  4 -> 2, 4, 5
0.00/0.54	  5 -> 2, 4, 5
0.00/0.54	Where:
0.00/0.54	  0) eval#(x, y, z) -> eval#(x, y, z) [y > z && z >= x && z >= y]
0.00/0.54	  1) eval#(I0, I1, I2) -> eval#(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1]
0.00/0.54	  2) eval#(I3, I4, I5) -> eval#(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
0.00/0.54	  3) eval#(I6, I7, I8) -> eval#(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8]
0.00/0.54	  4) eval#(I9, I10, I11) -> eval#(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
0.00/0.54	  5) eval#(I12, I13, I14) -> eval#(I12 - 1, I13, I14) [I12 > I14]
0.00/0.54	
0.00/0.54	We have the following SCCs.
0.00/0.54	  { 4, 5 }
0.00/0.54	  { 2 }
0.00/0.54	
0.00/0.54	DP problem for innermost termination.
0.00/0.54	P =
0.00/0.54	  eval#(I3, I4, I5) -> eval#(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
0.00/0.54	R = 
0.00/0.54	  eval(x, y, z) -> eval(x, y, z) [y > z && z >= x && z >= y]
0.00/0.54	  eval(I0, I1, I2) -> eval(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1]
0.00/0.54	  eval(I3, I4, I5) -> eval(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
0.00/0.54	  eval(I6, I7, I8) -> eval(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8]
0.00/0.54	  eval(I9, I10, I11) -> eval(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
0.00/0.54	  eval(I12, I13, I14) -> eval(I12 - 1, I13, I14) [I12 > I14]
0.00/0.54	
0.00/0.54	We use the reverse value criterion with the projection function NU:
0.00/0.54	  NU[eval#(z1,z2,z3)] = z2 + -1 * z3
0.00/0.54	
0.00/0.54	This gives the following inequalities:
0.00/0.54	  I4 > I5 && I5 >= I3  ==>  I4 + -1 * I5 > I4 - 1 + -1 * I5  with  I4 + -1 * I5 >= 0
0.00/0.54	
0.00/0.54	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.54	
0.00/0.54	DP problem for innermost termination.
0.00/0.54	P =
0.00/0.54	  eval#(I9, I10, I11) -> eval#(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
0.00/0.54	  eval#(I12, I13, I14) -> eval#(I12 - 1, I13, I14) [I12 > I14]
0.00/0.54	R = 
0.00/0.54	  eval(x, y, z) -> eval(x, y, z) [y > z && z >= x && z >= y]
0.00/0.54	  eval(I0, I1, I2) -> eval(I0, I1, I2) [I0 > I2 && I2 >= I0 && I2 >= I1]
0.00/0.54	  eval(I3, I4, I5) -> eval(I3, I4 - 1, I5) [I4 > I5 && I5 >= I3]
0.00/0.54	  eval(I6, I7, I8) -> eval(I6, I7 - 1, I8) [I6 > I8 && I8 >= I6 && I7 > I8]
0.00/0.54	  eval(I9, I10, I11) -> eval(I9 - 1, I10, I11) [I10 > I11 && I9 > I11]
0.00/0.54	  eval(I12, I13, I14) -> eval(I12 - 1, I13, I14) [I12 > I14]
0.00/0.54	
0.00/0.54	We use the reverse value criterion with the projection function NU:
0.00/0.54	  NU[eval#(z1,z2,z3)] = z1 + -1 * z3
0.00/0.54	
0.00/0.54	This gives the following inequalities:
0.00/0.54	  I10 > I11 && I9 > I11  ==>  I9 + -1 * I11 > I9 - 1 + -1 * I11  with  I9 + -1 * I11 >= 0
0.00/0.54	  I12 > I14  ==>  I12 + -1 * I14 > I12 - 1 + -1 * I14  with  I12 + -1 * I14 >= 0
0.00/0.54	
0.00/0.54	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.52	EOF