0.00/0.24	YES
0.00/0.24	
0.00/0.24	DP problem for innermost termination.
0.00/0.24	P =
0.00/0.24	  if#(true, x, y) -> div#(x - y, y) 
0.00/0.24	  div#(I0, I1) -> if#(I0 >= I1 && I1 > 0, I0, I1) 
0.00/0.24	R = 
0.00/0.24	  if(true, x, y) -> div(x - y, y) + 1 
0.00/0.24	  div(I0, I1) -> if(I0 >= I1 && I1 > 0, I0, I1) 
0.00/0.24	
0.00/0.24	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.24	
0.00/0.24	DP problem for innermost termination.
0.00/0.24	P =
0.00/0.24	  if#(true, x, y) -> div#(x - y, y) 
0.00/0.24	  div#(I0, I1) -> if#(I0 >= I1 && I1 > 0, I0, I1) 
0.00/0.24	  div#(I0, I1) -> div#(I0 - I1, I1) [I0 >= I1 && I1 > 0]
0.00/0.24	R = 
0.00/0.24	  if(true, x, y) -> div(x - y, y) + 1 
0.00/0.24	  div(I0, I1) -> if(I0 >= I1 && I1 > 0, I0, I1) 
0.00/0.24	
0.00/0.24	The dependency graph for this problem is:
0.00/0.24	  0 -> 2, 1
0.00/0.24	  1 -> 
0.00/0.24	  2 -> 2, 1
0.00/0.24	Where:
0.00/0.24	  0) if#(true, x, y) -> div#(x - y, y) 
0.00/0.24	  1) div#(I0, I1) -> if#(I0 >= I1 && I1 > 0, I0, I1) 
0.00/0.24	  2) div#(I0, I1) -> div#(I0 - I1, I1) [I0 >= I1 && I1 > 0]
0.00/0.24	
0.00/0.24	We have the following SCCs.
0.00/0.24	  { 2 }
0.00/0.24	
0.00/0.24	DP problem for innermost termination.
0.00/0.24	P =
0.00/0.24	  div#(I0, I1) -> div#(I0 - I1, I1) [I0 >= I1 && I1 > 0]
0.00/0.24	R = 
0.00/0.24	  if(true, x, y) -> div(x - y, y) + 1 
0.00/0.24	  div(I0, I1) -> if(I0 >= I1 && I1 > 0, I0, I1) 
0.00/0.24	
0.00/0.24	We use the reverse value criterion with the projection function NU:
0.00/0.24	  NU[div#(z1,z2)] = z1
0.00/0.24	
0.00/0.24	This gives the following inequalities:
0.00/0.24	  I0 >= I1 && I1 > 0  ==>  I0 > I0 - I1  with  I0 >= 0
0.00/0.24	
0.00/0.24	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.23	EOF