0.00/0.13	YES
0.00/0.13	
0.00/0.13	DP problem for innermost termination.
0.00/0.13	P =
0.00/0.13	  f#(true, x) -> f#(1000 >= x, x + 1) 
0.00/0.13	R = 
0.00/0.13	  f(true, x) -> f(1000 >= x, x + 1) 
0.00/0.13	
0.00/0.13	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.13	
0.00/0.13	DP problem for innermost termination.
0.00/0.13	P =
0.00/0.13	  f#(true, x) -> f_1#(x) 
0.00/0.13	  f_1#(x) -> f#(1000 >= x, x + 1) 
0.00/0.13	  f_1#(x) -> f_1#(x + 1) [1000 >= x]
0.00/0.13	R = 
0.00/0.13	  f(true, x) -> f(1000 >= x, x + 1) 
0.00/0.13	
0.00/0.13	The dependency graph for this problem is:
0.00/0.13	  0 -> 2, 1
0.00/0.13	  1 -> 
0.00/0.13	  2 -> 2, 1
0.00/0.13	Where:
0.00/0.13	  0) f#(true, x) -> f_1#(x) 
0.00/0.13	  1) f_1#(x) -> f#(1000 >= x, x + 1) 
0.00/0.13	  2) f_1#(x) -> f_1#(x + 1) [1000 >= x]
0.00/0.13	
0.00/0.13	We have the following SCCs.
0.00/0.13	  { 2 }
0.00/0.13	
0.00/0.13	DP problem for innermost termination.
0.00/0.13	P =
0.00/0.13	  f_1#(x) -> f_1#(x + 1) [1000 >= x]
0.00/0.13	R = 
0.00/0.13	  f(true, x) -> f(1000 >= x, x + 1) 
0.00/0.13	
0.00/0.13	We use the reverse value criterion with the projection function NU:
0.00/0.13	  NU[f_1#(z1)] = 1000 + -1 * z1
0.00/0.13	
0.00/0.13	This gives the following inequalities:
0.00/0.13	  1000 >= x  ==>  1000 + -1 * x > 1000 + -1 * (x + 1)  with  1000 + -1 * x >= 0
0.00/0.13	
0.00/0.13	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.11	EOF