0.00/0.28	YES
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  eval_3#(x, y) -> eval_1#(x, y) [0 >= y]
0.00/0.28	  eval_3#(I0, I1) -> eval_3#(I0, I1 - 1) [I1 > 0]
0.00/0.28	  eval_2#(I2, I3) -> eval_1#(I2, I3) [0 >= I2]
0.00/0.28	  eval_2#(I4, I5) -> eval_2#(I4 - 1, I5) [I4 > 0]
0.00/0.28	  eval_1#(I6, I7) -> eval_3#(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6]
0.00/0.28	  eval_1#(I8, I9) -> eval_2#(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9]
0.00/0.28	R = 
0.00/0.28	  eval_3(x, y) -> eval_1(x, y) [0 >= y]
0.00/0.28	  eval_3(I0, I1) -> eval_3(I0, I1 - 1) [I1 > 0]
0.00/0.28	  eval_2(I2, I3) -> eval_1(I2, I3) [0 >= I2]
0.00/0.28	  eval_2(I4, I5) -> eval_2(I4 - 1, I5) [I4 > 0]
0.00/0.28	  eval_1(I6, I7) -> eval_3(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6]
0.00/0.28	  eval_1(I8, I9) -> eval_2(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9]
0.00/0.28	
0.00/0.28	The dependency graph for this problem is:
0.00/0.28	  0 -> 
0.00/0.28	  1 -> 0, 1
0.00/0.28	  2 -> 
0.00/0.28	  3 -> 2, 3
0.00/0.28	  4 -> 1
0.00/0.28	  5 -> 3
0.00/0.28	Where:
0.00/0.28	  0) eval_3#(x, y) -> eval_1#(x, y) [0 >= y]
0.00/0.28	  1) eval_3#(I0, I1) -> eval_3#(I0, I1 - 1) [I1 > 0]
0.00/0.28	  2) eval_2#(I2, I3) -> eval_1#(I2, I3) [0 >= I2]
0.00/0.28	  3) eval_2#(I4, I5) -> eval_2#(I4 - 1, I5) [I4 > 0]
0.00/0.28	  4) eval_1#(I6, I7) -> eval_3#(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6]
0.00/0.28	  5) eval_1#(I8, I9) -> eval_2#(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9]
0.00/0.28	
0.00/0.28	We have the following SCCs.
0.00/0.28	  { 1 }
0.00/0.28	  { 3 }
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  eval_2#(I4, I5) -> eval_2#(I4 - 1, I5) [I4 > 0]
0.00/0.28	R = 
0.00/0.28	  eval_3(x, y) -> eval_1(x, y) [0 >= y]
0.00/0.28	  eval_3(I0, I1) -> eval_3(I0, I1 - 1) [I1 > 0]
0.00/0.28	  eval_2(I2, I3) -> eval_1(I2, I3) [0 >= I2]
0.00/0.28	  eval_2(I4, I5) -> eval_2(I4 - 1, I5) [I4 > 0]
0.00/0.28	  eval_1(I6, I7) -> eval_3(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6]
0.00/0.28	  eval_1(I8, I9) -> eval_2(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9]
0.00/0.28	
0.00/0.28	We use the reverse value criterion with the projection function NU:
0.00/0.28	  NU[eval_2#(z1,z2)] = z1
0.00/0.28	
0.00/0.28	This gives the following inequalities:
0.00/0.28	  I4 > 0  ==>  I4 > I4 - 1  with  I4 >= 0
0.00/0.28	
0.00/0.28	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  eval_3#(I0, I1) -> eval_3#(I0, I1 - 1) [I1 > 0]
0.00/0.28	R = 
0.00/0.28	  eval_3(x, y) -> eval_1(x, y) [0 >= y]
0.00/0.28	  eval_3(I0, I1) -> eval_3(I0, I1 - 1) [I1 > 0]
0.00/0.28	  eval_2(I2, I3) -> eval_1(I2, I3) [0 >= I2]
0.00/0.28	  eval_2(I4, I5) -> eval_2(I4 - 1, I5) [I4 > 0]
0.00/0.28	  eval_1(I6, I7) -> eval_3(I6, I7) [I6 > 0 && I7 > 0 && I7 >= I6]
0.00/0.28	  eval_1(I8, I9) -> eval_2(I8, I9) [I8 > 0 && I9 > 0 && I8 > I9]
0.00/0.28	
0.00/0.28	We use the reverse value criterion with the projection function NU:
0.00/0.28	  NU[eval_3#(z1,z2)] = z2 + -1 * 0
0.00/0.28	
0.00/0.28	This gives the following inequalities:
0.00/0.28	  I1 > 0  ==>  I1 + -1 * 0 > I1 - 1 + -1 * 0  with  I1 + -1 * 0 >= 0
0.00/0.28	
0.00/0.28	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.26	EOF