0.00/0.22	YES
0.00/0.22	
0.00/0.22	DP problem for innermost termination.
0.00/0.22	P =
0.00/0.22	  log#(x, y) -> log#((x - y) / y, y) [x >= 2 && y >= 2]
0.00/0.22	R = 
0.00/0.22	  log(x, y) -> 1 + log((x - y) / y, y) [x >= 2 && y >= 2]
0.00/0.22	  log(1, I0) -> 0 [I0 >= 2]
0.00/0.22	
0.00/0.22	We use the reverse value criterion with the projection function NU:
0.00/0.22	  NU[log#(z1,z2)] = z1
0.00/0.22	
0.00/0.22	This gives the following inequalities:
0.00/0.22	  x >= 2 && y >= 2  ==>  x > (x - y) / y  with  x >= 0
0.00/0.22	
0.00/0.22	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.20	EOF