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	  ft#(true, x, y) -> ft#(y >= x, x + 1, y) 
0.00/0.17	  f#(true, I0) -> ft#(true, I0, I1) 
0.00/0.17	R = 
0.00/0.17	  ft(true, x, y) -> ft(y >= x, x + 1, y) 
0.00/0.17	  f(true, I0) -> ft(true, I0, I1) 
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	  ft#(true, x, y) -> ft_1#(x, y) 
0.00/0.17	  f#(true, I0) -> ft#(true, I0, I1) 
0.00/0.17	  ft_1#(x, y) -> ft#(y >= x, x + 1, y) 
0.00/0.17	  ft_1#(x, y) -> ft_1#(x + 1, y) [y >= x]
0.00/0.17	  f#(true, I0) -> ft_1#(I0, I1) [true]
0.00/0.17	R = 
0.00/0.17	  ft(true, x, y) -> ft(y >= x, x + 1, y) 
0.00/0.17	  f(true, I0) -> ft(true, I0, I1) 
0.00/0.17	
0.00/0.17	The dependency graph for this problem is:
0.00/0.17	  0 -> 3, 2
0.00/0.17	  1 -> 
0.00/0.17	  2 -> 
0.00/0.17	  3 -> 3, 2
0.00/0.17	  4 -> 2, 3
0.00/0.17	Where:
0.00/0.17	  0) ft#(true, x, y) -> ft_1#(x, y) 
0.00/0.17	  1) f#(true, I0) -> ft#(true, I0, I1) 
0.00/0.17	  2) ft_1#(x, y) -> ft#(y >= x, x + 1, y) 
0.00/0.17	  3) ft_1#(x, y) -> ft_1#(x + 1, y) [y >= x]
0.00/0.17	  4) f#(true, I0) -> ft_1#(I0, I1) [true]
0.00/0.17	
0.00/0.17	We have the following SCCs.
0.00/0.17	  { 3 }
0.00/0.17	
0.00/0.17	DP problem for innermost termination.
0.00/0.17	P =
0.00/0.17	  ft_1#(x, y) -> ft_1#(x + 1, y) [y >= x]
0.00/0.17	R = 
0.00/0.17	  ft(true, x, y) -> ft(y >= x, x + 1, y) 
0.00/0.17	  f(true, I0) -> ft(true, I0, I1) 
0.00/0.17	
0.00/0.17	We use the reverse value criterion with the projection function NU:
0.00/0.17	  NU[ft_1#(z1,z2)] = z2 + -1 * z1
0.00/0.17	
0.00/0.17	This gives the following inequalities:
0.00/0.17	  y >= x  ==>  y + -1 * x > y + -1 * (x + 1)  with  y + -1 * x >= 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