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#(i, j) -> eval#(i - nat, j + pos) [i - j >= 1 && nat >= 0 && pos > 0]
0.00/0.17	R = 
0.00/0.17	  eval(i, j) -> eval(i - nat, j + pos) [i - j >= 1 && nat >= 0 && pos > 0]
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 - z2 + -1 * 1
0.00/0.17	
0.00/0.17	This gives the following inequalities:
0.00/0.17	  i - j >= 1 && nat >= 0 && pos > 0  ==>  i - j + -1 * 1 > i - nat - (j + pos) + -1 * 1  with  i - j + -1 * 1 >= 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