0.00/0.16	YES
0.00/0.16	
0.00/0.16	DP problem for innermost termination.
0.00/0.16	P =
0.00/0.16	  eval#(x, y) -> eval#(x - 1, y) [x > 0 && not(x = 0) && x <= y]
0.00/0.16	  eval#(I0, I1) -> eval#(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1]
0.00/0.16	R = 
0.00/0.16	  eval(x, y) -> eval(x - 1, y) [x > 0 && not(x = 0) && x <= y]
0.00/0.16	  eval(I0, I1) -> eval(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1]
0.00/0.16	
0.00/0.16	The dependency graph for this problem is:
0.00/0.16	  0 -> 0
0.00/0.16	  1 -> 0
0.00/0.16	Where:
0.00/0.16	  0) eval#(x, y) -> eval#(x - 1, y) [x > 0 && not(x = 0) && x <= y]
0.00/0.16	  1) eval#(I0, I1) -> eval#(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1]
0.00/0.16	
0.00/0.16	We have the following SCCs.
0.00/0.16	  { 0 }
0.00/0.16	
0.00/0.16	DP problem for innermost termination.
0.00/0.16	P =
0.00/0.16	  eval#(x, y) -> eval#(x - 1, y) [x > 0 && not(x = 0) && x <= y]
0.00/0.16	R = 
0.00/0.16	  eval(x, y) -> eval(x - 1, y) [x > 0 && not(x = 0) && x <= y]
0.00/0.16	  eval(I0, I1) -> eval(I1, I1) [I0 > 0 && not(I0 = 0) && I0 > I1]
0.00/0.16	
0.00/0.16	We use the reverse value criterion with the projection function NU:
0.00/0.16	  NU[eval#(z1,z2)] = z1
0.00/0.16	
0.00/0.16	This gives the following inequalities:
0.00/0.16	  x > 0 && not(x = 0) && x <= y  ==>  x > x - 1  with  x >= 0
0.00/0.16	
0.00/0.16	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.14	EOF