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	  conv#(x) -> conv#(x / 2) [x > 0]
0.00/0.17	  conv#(x) -> lastbit#(x) [x > 0]
0.00/0.17	  lastbit#(I0) -> lastbit#(I0 - 2) [I0 > 1]
0.00/0.17	R = 
0.00/0.17	  conv(x) -> cons(conv(x / 2), lastbit(x)) [x > 0]
0.00/0.17	  conv(0) -> cons(nil, 0) 
0.00/0.17	  lastbit(I0) -> lastbit(I0 - 2) [I0 > 1]
0.00/0.17	  lastbit(1) -> 1 
0.00/0.17	  lastbit(0) -> 0 
0.00/0.17	
0.00/0.17	The dependency graph for this problem is:
0.00/0.17	  0 -> 0, 1
0.00/0.17	  1 -> 2
0.00/0.17	  2 -> 2
0.00/0.17	Where:
0.00/0.17	  0) conv#(x) -> conv#(x / 2) [x > 0]
0.00/0.17	  1) conv#(x) -> lastbit#(x) [x > 0]
0.00/0.17	  2) lastbit#(I0) -> lastbit#(I0 - 2) [I0 > 1]
0.00/0.17	
0.00/0.17	We have the following SCCs.
0.00/0.17	  { 0 }
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	  lastbit#(I0) -> lastbit#(I0 - 2) [I0 > 1]
0.00/0.17	R = 
0.00/0.17	  conv(x) -> cons(conv(x / 2), lastbit(x)) [x > 0]
0.00/0.17	  conv(0) -> cons(nil, 0) 
0.00/0.17	  lastbit(I0) -> lastbit(I0 - 2) [I0 > 1]
0.00/0.17	  lastbit(1) -> 1 
0.00/0.17	  lastbit(0) -> 0 
0.00/0.17	
0.00/0.17	We use the reverse value criterion with the projection function NU:
0.00/0.17	  NU[lastbit#(z1)] = z1
0.00/0.17	
0.00/0.17	This gives the following inequalities:
0.00/0.17	  I0 > 1  ==>  I0 > I0 - 2  with  I0 >= 0
0.00/0.17	
0.00/0.17	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.17	
0.00/0.17	DP problem for innermost termination.
0.00/0.17	P =
0.00/0.17	  conv#(x) -> conv#(x / 2) [x > 0]
0.00/0.17	R = 
0.00/0.17	  conv(x) -> cons(conv(x / 2), lastbit(x)) [x > 0]
0.00/0.17	  conv(0) -> cons(nil, 0) 
0.00/0.17	  lastbit(I0) -> lastbit(I0 - 2) [I0 > 1]
0.00/0.17	  lastbit(1) -> 1 
0.00/0.17	  lastbit(0) -> 0 
0.00/0.17	
0.00/0.17	We use the reverse value criterion with the projection function NU:
0.00/0.17	  NU[conv#(z1)] = z1
0.00/0.17	
0.00/0.17	This gives the following inequalities:
0.00/0.17	  x > 0  ==>  x > x / 2  with  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