0.00/0.11	YES
0.00/0.11	
0.00/0.11	DP problem for innermost termination.
0.00/0.11	P =
0.00/0.11	  f#(true, x) -> f#(x > y && x >= 0, y) 
0.00/0.11	R = 
0.00/0.11	  f(true, x) -> f(x > y && x >= 0, y) 
0.00/0.11	
0.00/0.11	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.11	
0.00/0.11	DP problem for innermost termination.
0.00/0.11	P =
0.00/0.11	  f#(true, x) -> f_1#(x) 
0.00/0.11	  f_1#(x) -> f#(x > y && x >= 0, y) 
0.00/0.11	  f_1#(x) -> f_1#(y) [x > y && x >= 0]
0.00/0.11	R = 
0.00/0.11	  f(true, x) -> f(x > y && x >= 0, y) 
0.00/0.11	
0.00/0.11	The dependency graph for this problem is:
0.00/0.11	  0 -> 2, 1
0.00/0.11	  1 -> 
0.00/0.11	  2 -> 2, 1
0.00/0.11	Where:
0.00/0.11	  0) f#(true, x) -> f_1#(x) 
0.00/0.11	  1) f_1#(x) -> f#(x > y && x >= 0, y) 
0.00/0.11	  2) f_1#(x) -> f_1#(y) [x > y && x >= 0]
0.00/0.11	
0.00/0.11	We have the following SCCs.
0.00/0.11	  { 2 }
0.00/0.11	
0.00/0.11	DP problem for innermost termination.
0.00/0.11	P =
0.00/0.11	  f_1#(x) -> f_1#(y) [x > y && x >= 0]
0.00/0.11	R = 
0.00/0.11	  f(true, x) -> f(x > y && x >= 0, y) 
0.00/0.11	
0.00/0.11	We use the reverse value criterion with the projection function NU:
0.00/0.11	  NU[f_1#(z1)] = z1
0.00/0.11	
0.00/0.11	This gives the following inequalities:
0.00/0.11	  x > y && x >= 0  ==>  x > y  with  x >= 0
0.00/0.11	
0.00/0.11	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.09	EOF