1.70/1.73	YES
1.70/1.73	
1.70/1.73	DP problem for innermost termination.
1.70/1.73	P =
1.70/1.73	  eval_2#(x, y) -> eval_1#(x - 1, y) [x > 0 && y >= 0 && y >= x]
1.70/1.73	  eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I0 > 0 && I1 >= 0 && I0 > I1]
1.70/1.73	  eval_1#(I2, I3) -> eval_2#(I2, 0) [I2 > 0]
1.70/1.73	R = 
1.70/1.73	  eval_2(x, y) -> eval_1(x - 1, y) [x > 0 && y >= 0 && y >= x]
1.70/1.73	  eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I0 > 0 && I1 >= 0 && I0 > I1]
1.70/1.73	  eval_1(I2, I3) -> eval_2(I2, 0) [I2 > 0]
1.70/1.73	
1.70/1.73	We use the reverse value criterion with the projection function NU:
1.70/1.73	  NU[eval_1#(z1,z2)] = z1
1.70/1.73	  NU[eval_2#(z1,z2)] = z1
1.70/1.73	
1.70/1.73	This gives the following inequalities:
1.70/1.73	  x > 0 && y >= 0 && y >= x  ==>  x > x - 1  with  x >= 0
1.70/1.73	  I0 > 0 && I1 >= 0 && I0 > I1  ==>  I0 >= I0
1.70/1.73	  I2 > 0  ==>  I2 >= I2
1.70/1.73	
1.70/1.73	We remove all the strictly oriented dependency pairs.
1.70/1.73	
1.70/1.73	DP problem for innermost termination.
1.70/1.73	P =
1.70/1.73	  eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I0 > 0 && I1 >= 0 && I0 > I1]
1.70/1.73	  eval_1#(I2, I3) -> eval_2#(I2, 0) [I2 > 0]
1.70/1.73	R = 
1.70/1.73	  eval_2(x, y) -> eval_1(x - 1, y) [x > 0 && y >= 0 && y >= x]
1.70/1.73	  eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I0 > 0 && I1 >= 0 && I0 > I1]
1.70/1.73	  eval_1(I2, I3) -> eval_2(I2, 0) [I2 > 0]
1.70/1.73	
1.70/1.73	The dependency graph for this problem is:
1.70/1.73	  1 -> 1
1.70/1.73	  2 -> 1
1.70/1.73	Where:
1.70/1.73	  1) eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I0 > 0 && I1 >= 0 && I0 > I1]
1.70/1.73	  2) eval_1#(I2, I3) -> eval_2#(I2, 0) [I2 > 0]
1.70/1.73	
1.70/1.73	We have the following SCCs.
1.70/1.73	  { 1 }
1.70/1.73	
1.70/1.73	DP problem for innermost termination.
1.70/1.73	P =
1.70/1.73	  eval_2#(I0, I1) -> eval_2#(I0, I1 + 1) [I0 > 0 && I1 >= 0 && I0 > I1]
1.70/1.73	R = 
1.70/1.73	  eval_2(x, y) -> eval_1(x - 1, y) [x > 0 && y >= 0 && y >= x]
1.70/1.73	  eval_2(I0, I1) -> eval_2(I0, I1 + 1) [I0 > 0 && I1 >= 0 && I0 > I1]
1.70/1.73	  eval_1(I2, I3) -> eval_2(I2, 0) [I2 > 0]
1.70/1.73	
1.70/1.73	We use the reverse value criterion with the projection function NU:
1.70/1.73	  NU[eval_2#(z1,z2)] = z1 + -1 * z2
1.70/1.73	
1.70/1.73	This gives the following inequalities:
1.70/1.73	  I0 > 0 && I1 >= 0 && I0 > I1  ==>  I0 + -1 * I1 > I0 + -1 * (I1 + 1)  with  I0 + -1 * I1 >= 0
1.70/1.73	
1.70/1.73	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.70/4.71	EOF