2.81/2.81	YES
2.81/2.81	
2.81/2.81	DP problem for innermost termination.
2.81/2.81	P =
2.81/2.81	  eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [x > z && z >= y]
2.81/2.81	  eval_2#(I0, I1, I2) -> eval_2#(I0, I1 - 1, I2) [I0 > I2 && I1 > I2]
2.81/2.81	  eval_1#(I3, I4, I5) -> eval_2#(I3, I4, I5) [I3 > I5]
2.81/2.81	R = 
2.81/2.81	  eval_2(x, y, z) -> eval_1(x - 1, y, z) [x > z && z >= y]
2.81/2.81	  eval_2(I0, I1, I2) -> eval_2(I0, I1 - 1, I2) [I0 > I2 && I1 > I2]
2.81/2.81	  eval_1(I3, I4, I5) -> eval_2(I3, I4, I5) [I3 > I5]
2.81/2.81	
2.81/2.81	We use the reverse value criterion with the projection function NU:
2.81/2.81	  NU[eval_1#(z1,z2,z3)] = z1 + -1 * z3
2.81/2.81	  NU[eval_2#(z1,z2,z3)] = z1 - 1 + -1 * z3
2.81/2.81	
2.81/2.81	This gives the following inequalities:
2.81/2.81	  x > z && z >= y  ==>  x - 1 + -1 * z >= x - 1 + -1 * z
2.81/2.81	  I0 > I2 && I1 > I2  ==>  I0 - 1 + -1 * I2 >= I0 - 1 + -1 * I2
2.81/2.81	  I3 > I5  ==>  I3 + -1 * I5 > I3 - 1 + -1 * I5  with  I3 + -1 * I5 >= 0
2.81/2.81	
2.81/2.81	We remove all the strictly oriented dependency pairs.
2.81/2.81	
2.81/2.81	DP problem for innermost termination.
2.81/2.81	P =
2.81/2.81	  eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [x > z && z >= y]
2.81/2.81	  eval_2#(I0, I1, I2) -> eval_2#(I0, I1 - 1, I2) [I0 > I2 && I1 > I2]
2.81/2.81	R = 
2.81/2.81	  eval_2(x, y, z) -> eval_1(x - 1, y, z) [x > z && z >= y]
2.81/2.81	  eval_2(I0, I1, I2) -> eval_2(I0, I1 - 1, I2) [I0 > I2 && I1 > I2]
2.81/2.81	  eval_1(I3, I4, I5) -> eval_2(I3, I4, I5) [I3 > I5]
2.81/2.81	
2.81/2.81	The dependency graph for this problem is:
2.81/2.81	  0 -> 
2.81/2.81	  1 -> 0, 1
2.81/2.81	Where:
2.81/2.81	  0) eval_2#(x, y, z) -> eval_1#(x - 1, y, z) [x > z && z >= y]
2.81/2.81	  1) eval_2#(I0, I1, I2) -> eval_2#(I0, I1 - 1, I2) [I0 > I2 && I1 > I2]
2.81/2.81	
2.81/2.81	We have the following SCCs.
2.81/2.81	  { 1 }
2.81/2.81	
2.81/2.81	DP problem for innermost termination.
2.81/2.81	P =
2.81/2.81	  eval_2#(I0, I1, I2) -> eval_2#(I0, I1 - 1, I2) [I0 > I2 && I1 > I2]
2.81/2.81	R = 
2.81/2.81	  eval_2(x, y, z) -> eval_1(x - 1, y, z) [x > z && z >= y]
2.81/2.81	  eval_2(I0, I1, I2) -> eval_2(I0, I1 - 1, I2) [I0 > I2 && I1 > I2]
2.81/2.81	  eval_1(I3, I4, I5) -> eval_2(I3, I4, I5) [I3 > I5]
2.81/2.81	
2.81/2.81	We use the reverse value criterion with the projection function NU:
2.81/2.81	  NU[eval_2#(z1,z2,z3)] = z2 + -1 * z3
2.81/2.81	
2.81/2.81	This gives the following inequalities:
2.81/2.81	  I0 > I2 && I1 > I2  ==>  I1 + -1 * I2 > I1 - 1 + -1 * I2  with  I1 + -1 * I2 >= 0
2.81/2.81	
2.81/2.81	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.81/5.79	EOF