0.00/0.11	YES
0.00/0.11	
0.00/0.11	DP problem for innermost termination.
0.00/0.11	P =
0.00/0.11	  f#(x, y) -> f#(x, y + 1) [x > y]
0.00/0.11	R = 
0.00/0.11	  f(x, y) -> f(x, y + 1) [x > y]
0.00/0.11	
0.00/0.11	We use the reverse value criterion with the projection function NU:
0.00/0.11	  NU[f#(z1,z2)] = z1 + -1 * z2
0.00/0.11	
0.00/0.11	This gives the following inequalities:
0.00/0.11	  x > y  ==>  x + -1 * y > x + -1 * (y + 1)  with  x + -1 * y >= 0
0.00/0.11	
0.00/0.11	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.09	EOF