0.00/0.19	YES
0.00/0.19	
0.00/0.19	DP problem for innermost termination.
0.00/0.19	P =
0.00/0.19	  eval_2#(x, y, z) -> eval_1#(x, y, z) [z >= y]
0.00/0.19	  eval_2#(I0, I1, I2) -> eval_2#(I0 - 1, I1 - 1, I2) [I1 > I2]
0.00/0.19	  eval_1#(I3, I4, I5) -> eval_2#(I3, I4, I5) [I3 = I4 && I3 > I5]
0.00/0.19	R = 
0.00/0.19	  eval_2(x, y, z) -> eval_1(x, y, z) [z >= y]
0.00/0.19	  eval_2(I0, I1, I2) -> eval_2(I0 - 1, I1 - 1, I2) [I1 > I2]
0.00/0.19	  eval_1(I3, I4, I5) -> eval_2(I3, I4, I5) [I3 = I4 && I3 > I5]
0.00/0.19	
0.00/0.19	The dependency graph for this problem is:
0.00/0.19	  0 -> 
0.00/0.19	  1 -> 0, 1
0.00/0.19	  2 -> 1
0.00/0.19	Where:
0.00/0.19	  0) eval_2#(x, y, z) -> eval_1#(x, y, z) [z >= y]
0.00/0.19	  1) eval_2#(I0, I1, I2) -> eval_2#(I0 - 1, I1 - 1, I2) [I1 > I2]
0.00/0.19	  2) eval_1#(I3, I4, I5) -> eval_2#(I3, I4, I5) [I3 = I4 && I3 > I5]
0.00/0.19	
0.00/0.19	We have the following SCCs.
0.00/0.19	  { 1 }
0.00/0.19	
0.00/0.19	DP problem for innermost termination.
0.00/0.19	P =
0.00/0.19	  eval_2#(I0, I1, I2) -> eval_2#(I0 - 1, I1 - 1, I2) [I1 > I2]
0.00/0.19	R = 
0.00/0.19	  eval_2(x, y, z) -> eval_1(x, y, z) [z >= y]
0.00/0.19	  eval_2(I0, I1, I2) -> eval_2(I0 - 1, I1 - 1, I2) [I1 > I2]
0.00/0.19	  eval_1(I3, I4, I5) -> eval_2(I3, I4, I5) [I3 = I4 && I3 > I5]
0.00/0.19	
0.00/0.19	We use the reverse value criterion with the projection function NU:
0.00/0.19	  NU[eval_2#(z1,z2,z3)] = z2 + -1 * z3
0.00/0.19	
0.00/0.19	This gives the following inequalities:
0.00/0.19	  I1 > I2  ==>  I1 + -1 * I2 > I1 - 1 + -1 * I2  with  I1 + -1 * I2 >= 0
0.00/0.19	
0.00/0.19	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.17	EOF