0.00/0.17	YES
0.00/0.17	
0.00/0.17	DP problem for innermost termination.
0.00/0.17	P =
0.00/0.17	  eval#(x, y) -> eval#(y, y) [x > 0 && x > y]
0.00/0.17	  eval#(I0, I1) -> eval#(I0 - 1, I1) [I0 > 0 && I1 >= I0]
0.00/0.17	R = 
0.00/0.17	  eval(x, y) -> eval(y, y) [x > 0 && x > y]
0.00/0.17	  eval(I0, I1) -> eval(I0 - 1, I1) [I0 > 0 && I1 >= I0]
0.00/0.17	
0.00/0.17	The dependency graph for this problem is:
0.00/0.17	  0 -> 1
0.00/0.17	  1 -> 1
0.00/0.17	Where:
0.00/0.17	  0) eval#(x, y) -> eval#(y, y) [x > 0 && x > y]
0.00/0.17	  1) eval#(I0, I1) -> eval#(I0 - 1, I1) [I0 > 0 && I1 >= I0]
0.00/0.17	
0.00/0.17	We have the following SCCs.
0.00/0.17	  { 1 }
0.00/0.17	
0.00/0.17	DP problem for innermost termination.
0.00/0.17	P =
0.00/0.17	  eval#(I0, I1) -> eval#(I0 - 1, I1) [I0 > 0 && I1 >= I0]
0.00/0.17	R = 
0.00/0.17	  eval(x, y) -> eval(y, y) [x > 0 && x > y]
0.00/0.17	  eval(I0, I1) -> eval(I0 - 1, I1) [I0 > 0 && I1 >= I0]
0.00/0.17	
0.00/0.17	We use the reverse value criterion with the projection function NU:
0.00/0.17	  NU[eval#(z1,z2)] = z1
0.00/0.17	
0.00/0.17	This gives the following inequalities:
0.00/0.17	  I0 > 0 && I1 >= I0  ==>  I0 > I0 - 1  with  I0 >= 0
0.00/0.17	
0.00/0.17	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.15	EOF