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	  f#(true, x, y) -> f#(x > y, x + 1, y + 2) 
0.00/0.17	R = 
0.00/0.17	  f(true, x, y) -> f(x > y, x + 1, y + 2) 
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	  f#(true, x, y) -> f_1#(x, y) 
0.00/0.17	  f_1#(x, y) -> f#(x > y, x + 1, y + 2) 
0.00/0.17	  f_1#(x, y) -> f_1#(x + 1, y + 2) [x > y]
0.00/0.17	R = 
0.00/0.17	  f(true, x, y) -> f(x > y, x + 1, y + 2) 
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) f#(true, x, y) -> f_1#(x, y) 
0.00/0.17	  1) f_1#(x, y) -> f#(x > y, x + 1, y + 2) 
0.00/0.17	  2) f_1#(x, y) -> f_1#(x + 1, y + 2) [x > y]
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	  f_1#(x, y) -> f_1#(x + 1, y + 2) [x > y]
0.00/0.17	R = 
0.00/0.17	  f(true, x, y) -> f(x > y, x + 1, y + 2) 
0.00/0.17	
0.00/0.17	We use the reverse value criterion with the projection function NU:
0.00/0.17	  NU[f_1#(z1,z2)] = z1 + -1 * z2
0.00/0.17	
0.00/0.17	This gives the following inequalities:
0.00/0.17	  x > y  ==>  x + -1 * y > x + 1 + -1 * (y + 2)  with  x + -1 * y >= 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