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	  sif#(true, I0, I1) -> sum#(I0, I1 + 1) 
0.00/0.17	  sum#(I2, I3) -> sif#(I2 >= I3, I2, I3) 
0.00/0.17	R = 
0.00/0.17	  sif(false, x, y) -> 0 
0.00/0.17	  sif(true, I0, I1) -> I1 + sum(I0, I1 + 1) 
0.00/0.17	  sum(I2, I3) -> sif(I2 >= I3, I2, I3) 
0.00/0.17	
0.00/0.17	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.17	
0.00/0.17	DP problem for innermost termination.
0.00/0.17	P =
0.00/0.17	  sif#(true, I0, I1) -> sum#(I0, I1 + 1) 
0.00/0.17	  sum#(I2, I3) -> sif#(I2 >= I3, I2, I3) 
0.00/0.17	  sum#(I2, I3) -> sum#(I2, I3 + 1) [I2 >= I3]
0.00/0.17	R = 
0.00/0.17	  sif(false, x, y) -> 0 
0.00/0.17	  sif(true, I0, I1) -> I1 + sum(I0, I1 + 1) 
0.00/0.17	  sum(I2, I3) -> sif(I2 >= I3, I2, I3) 
0.00/0.17	
0.00/0.17	The dependency graph for this problem is:
0.00/0.17	  0 -> 2, 1
0.00/0.17	  1 -> 
0.00/0.17	  2 -> 2, 1
0.00/0.17	Where:
0.00/0.17	  0) sif#(true, I0, I1) -> sum#(I0, I1 + 1) 
0.00/0.17	  1) sum#(I2, I3) -> sif#(I2 >= I3, I2, I3) 
0.00/0.17	  2) sum#(I2, I3) -> sum#(I2, I3 + 1) [I2 >= I3]
0.00/0.17	
0.00/0.17	We have the following SCCs.
0.00/0.17	  { 2 }
0.00/0.17	
0.00/0.17	DP problem for innermost termination.
0.00/0.17	P =
0.00/0.17	  sum#(I2, I3) -> sum#(I2, I3 + 1) [I2 >= I3]
0.00/0.17	R = 
0.00/0.17	  sif(false, x, y) -> 0 
0.00/0.17	  sif(true, I0, I1) -> I1 + sum(I0, I1 + 1) 
0.00/0.17	  sum(I2, I3) -> sif(I2 >= I3, I2, I3) 
0.00/0.17	
0.00/0.17	We use the reverse value criterion with the projection function NU:
0.00/0.17	  NU[sum#(z1,z2)] = z1 + -1 * z2
0.00/0.17	
0.00/0.17	This gives the following inequalities:
0.00/0.17	  I2 >= I3  ==>  I2 + -1 * I3 > I2 + -1 * (I3 + 1)  with  I2 + -1 * I3 >= 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