0.00/0.14	YES
0.00/0.14	
0.00/0.14	DP problem for innermost termination.
0.00/0.14	P =
0.00/0.14	  sumto#(I0, I1) -> if#(sumto(I0 + 1, I1), I0, I1) [I1 >= I0]
0.00/0.14	  sumto#(I0, I1) -> sumto#(I0 + 1, I1) [I1 >= I0]
0.00/0.14	R = 
0.00/0.14	  if(wrap(z), x, y) -> wrap(x + z) 
0.00/0.14	  sumto(I0, I1) -> if(sumto(I0 + 1, I1), I0, I1) [I1 >= I0]
0.00/0.14	  sumto(I2, I3) -> wrap(0) [I2 > I3]
0.00/0.14	
0.00/0.14	The dependency graph for this problem is:
0.00/0.14	  0 -> 
0.00/0.14	  1 -> 0, 1
0.00/0.14	Where:
0.00/0.14	  0) sumto#(I0, I1) -> if#(sumto(I0 + 1, I1), I0, I1) [I1 >= I0]
0.00/0.14	  1) sumto#(I0, I1) -> sumto#(I0 + 1, I1) [I1 >= I0]
0.00/0.14	
0.00/0.14	We have the following SCCs.
0.00/0.14	  { 1 }
0.00/0.14	
0.00/0.14	DP problem for innermost termination.
0.00/0.14	P =
0.00/0.14	  sumto#(I0, I1) -> sumto#(I0 + 1, I1) [I1 >= I0]
0.00/0.14	R = 
0.00/0.14	  if(wrap(z), x, y) -> wrap(x + z) 
0.00/0.14	  sumto(I0, I1) -> if(sumto(I0 + 1, I1), I0, I1) [I1 >= I0]
0.00/0.14	  sumto(I2, I3) -> wrap(0) [I2 > I3]
0.00/0.14	
0.00/0.14	We use the reverse value criterion with the projection function NU:
0.00/0.14	  NU[sumto#(z1,z2)] = z2 + -1 * z1
0.00/0.14	
0.00/0.14	This gives the following inequalities:
0.00/0.14	  I1 >= I0  ==>  I1 + -1 * I0 > I1 + -1 * (I0 + 1)  with  I1 + -1 * I0 >= 0
0.00/0.14	
0.00/0.14	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.12	EOF