0.00/0.13	YES
0.00/0.13	
0.00/0.13	DP problem for innermost termination.
0.00/0.13	P =
0.00/0.13	  eval_2#(x, y) -> eval_1#(x, y) [0 >= y]
0.00/0.13	  eval_2#(I0, I1) -> eval_2#(I0 - 1, I1 - 1) [I1 > 0]
0.00/0.13	  eval_1#(I2, I3) -> eval_2#(I2, I3) [I2 = I3 && I2 > 0]
0.00/0.13	R = 
0.00/0.13	  eval_2(x, y) -> eval_1(x, y) [0 >= y]
0.00/0.13	  eval_2(I0, I1) -> eval_2(I0 - 1, I1 - 1) [I1 > 0]
0.00/0.13	  eval_1(I2, I3) -> eval_2(I2, I3) [I2 = I3 && I2 > 0]
0.00/0.13	
0.00/0.13	The dependency graph for this problem is:
0.00/0.13	  0 -> 
0.00/0.13	  1 -> 0, 1
0.00/0.13	  2 -> 1
0.00/0.13	Where:
0.00/0.13	  0) eval_2#(x, y) -> eval_1#(x, y) [0 >= y]
0.00/0.13	  1) eval_2#(I0, I1) -> eval_2#(I0 - 1, I1 - 1) [I1 > 0]
0.00/0.13	  2) eval_1#(I2, I3) -> eval_2#(I2, I3) [I2 = I3 && I2 > 0]
0.00/0.13	
0.00/0.13	We have the following SCCs.
0.00/0.13	  { 1 }
0.00/0.13	
0.00/0.13	DP problem for innermost termination.
0.00/0.13	P =
0.00/0.13	  eval_2#(I0, I1) -> eval_2#(I0 - 1, I1 - 1) [I1 > 0]
0.00/0.13	R = 
0.00/0.13	  eval_2(x, y) -> eval_1(x, y) [0 >= y]
0.00/0.13	  eval_2(I0, I1) -> eval_2(I0 - 1, I1 - 1) [I1 > 0]
0.00/0.13	  eval_1(I2, I3) -> eval_2(I2, I3) [I2 = I3 && I2 > 0]
0.00/0.13	
0.00/0.13	We use the reverse value criterion with the projection function NU:
0.00/0.13	  NU[eval_2#(z1,z2)] = z2 + -1 * 0
0.00/0.13	
0.00/0.13	This gives the following inequalities:
0.00/0.13	  I1 > 0  ==>  I1 + -1 * 0 > I1 - 1 + -1 * 0  with  I1 + -1 * 0 >= 0
0.00/0.13	
0.00/0.13	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.11	EOF