2.43/2.60	YES
2.43/2.60	
2.43/2.60	DP problem for innermost termination.
2.43/2.60	P =
2.43/2.60	  eval_2#(x, y) -> eval_1#(x - 2, y) [x >= 0 && y > 0 && y > x]
2.43/2.60	  eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I0 >= 0 && I1 > 0 && I0 >= I1]
2.43/2.60	  eval_1#(I2, I3) -> eval_2#(I2 + 1, 1) [I2 >= 0]
2.43/2.60	R = 
2.43/2.60	  eval_2(x, y) -> eval_1(x - 2, y) [x >= 0 && y > 0 && y > x]
2.43/2.60	  eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I0 >= 0 && I1 > 0 && I0 >= I1]
2.43/2.60	  eval_1(I2, I3) -> eval_2(I2 + 1, 1) [I2 >= 0]
2.43/2.60	
2.43/2.60	We use the reverse value criterion with the projection function NU:
2.43/2.60	  NU[eval_1#(z1,z2)] = z1 + 1 + -1 * 0
2.43/2.60	  NU[eval_2#(z1,z2)] = z1
2.43/2.60	
2.43/2.60	This gives the following inequalities:
2.43/2.60	  x >= 0 && y > 0 && y > x  ==>  x > x - 2 + 1 + -1 * 0  with  x >= 0
2.43/2.60	  I0 >= 0 && I1 > 0 && I0 >= I1  ==>  I0 >= I0
2.43/2.60	  I2 >= 0  ==>  I2 + 1 + -1 * 0 >= I2 + 1
2.43/2.60	
2.43/2.60	We remove all the strictly oriented dependency pairs.
2.43/2.60	
2.43/2.60	DP problem for innermost termination.
2.43/2.60	P =
2.43/2.60	  eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I0 >= 0 && I1 > 0 && I0 >= I1]
2.43/2.60	  eval_1#(I2, I3) -> eval_2#(I2 + 1, 1) [I2 >= 0]
2.43/2.60	R = 
2.43/2.60	  eval_2(x, y) -> eval_1(x - 2, y) [x >= 0 && y > 0 && y > x]
2.43/2.60	  eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I0 >= 0 && I1 > 0 && I0 >= I1]
2.43/2.60	  eval_1(I2, I3) -> eval_2(I2 + 1, 1) [I2 >= 0]
2.43/2.60	
2.43/2.60	The dependency graph for this problem is:
2.43/2.60	  1 -> 1
2.43/2.60	  2 -> 1
2.43/2.60	Where:
2.43/2.60	  1) eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I0 >= 0 && I1 > 0 && I0 >= I1]
2.43/2.60	  2) eval_1#(I2, I3) -> eval_2#(I2 + 1, 1) [I2 >= 0]
2.43/2.60	
2.43/2.60	We have the following SCCs.
2.43/2.60	  { 1 }
2.43/2.60	
2.43/2.60	DP problem for innermost termination.
2.43/2.60	P =
2.43/2.60	  eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I0 >= 0 && I1 > 0 && I0 >= I1]
2.43/2.60	R = 
2.43/2.60	  eval_2(x, y) -> eval_1(x - 2, y) [x >= 0 && y > 0 && y > x]
2.43/2.60	  eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I0 >= 0 && I1 > 0 && I0 >= I1]
2.43/2.60	  eval_1(I2, I3) -> eval_2(I2 + 1, 1) [I2 >= 0]
2.43/2.60	
2.43/2.60	We use the reverse value criterion with the projection function NU:
2.43/2.60	  NU[eval_2#(z1,z2)] = z1 + -1 * z2
2.43/2.60	
2.43/2.60	This gives the following inequalities:
2.43/2.60	  I0 >= 0 && I1 > 0 && I0 >= I1  ==>  I0 + -1 * I1 > I0 + -1 * (I1 + 1)  with  I0 + -1 * I1 >= 0
2.43/2.60	
2.43/2.60	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.43/5.58	EOF