0.00/0.54	MAYBE
0.00/0.54	
0.00/0.54	DP problem for innermost termination.
0.00/0.54	P =
0.00/0.54	  sort#(cons(x, xs)) -> max#(cons(x, xs)) 
0.00/0.54	  sort#(cons(x, xs)) -> sort#(del(max(cons(x, xs)), cons(x, xs))) 
0.00/0.54	  sort#(cons(x, xs)) -> del#(max(cons(x, xs)), cons(x, xs)) 
0.00/0.54	  sort#(cons(x, xs)) -> max#(cons(x, xs)) 
0.00/0.54	  if2#(false, I0, y, B0) -> del#(I0, B0) 
0.00/0.54	  del#(I3, cons(I4, B2)) -> if2#(I3 = I4, I3, I4, B2) 
0.00/0.54	  if1#(false, I6, I7, B3) -> max#(cons(I7, B3)) 
0.00/0.54	  if1#(true, I8, I9, B4) -> max#(cons(I8, B4)) 
0.00/0.54	  max#(cons(I10, cons(I11, B5))) -> if1#(I10 >= I11, I10, I11, B5) 
0.00/0.54	R = 
0.00/0.54	  sort(cons(x, xs)) -> cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs)))) 
0.00/0.54	  sort(nil) -> nil 
0.00/0.54	  if2(false, I0, y, B0) -> cons(y, del(I0, B0)) 
0.00/0.54	  if2(true, I1, I2, B1) -> B1 
0.00/0.54	  del(I3, cons(I4, B2)) -> if2(I3 = I4, I3, I4, B2) 
0.00/0.54	  del(I5, nil) -> nil 
0.00/0.54	  if1(false, I6, I7, B3) -> max(cons(I7, B3)) 
0.00/0.54	  if1(true, I8, I9, B4) -> max(cons(I8, B4)) 
0.00/0.54	  max(cons(I10, cons(I11, B5))) -> if1(I10 >= I11, I10, I11, B5) 
0.00/0.54	  max(cons(I12, nil)) -> I12 
0.00/0.54	  max(nil) -> 0 
0.00/0.54	
0.00/0.54	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.54	
0.00/0.54	DP problem for innermost termination.
0.00/0.54	P =
0.00/0.54	  sort#(cons(x, xs)) -> max#(cons(x, xs)) 
0.00/0.54	  sort#(cons(x, xs)) -> sort#(del(max(cons(x, xs)), cons(x, xs))) 
0.00/0.54	  sort#(cons(x, xs)) -> del#(max(cons(x, xs)), cons(x, xs)) 
0.00/0.54	  sort#(cons(x, xs)) -> max#(cons(x, xs)) 
0.00/0.54	  if2#(false, I0, y, B0) -> del#(I0, B0) 
0.00/0.54	  del#(I3, cons(I4, B2)) -> if2#(I3 = I4, I3, I4, B2) 
0.00/0.54	  if1#(false, I6, I7, B3) -> max#(cons(I7, B3)) 
0.00/0.54	  if1#(true, I8, I9, B4) -> max#(cons(I8, B4)) 
0.00/0.54	  max#(cons(I10, cons(I11, B5))) -> if1#(I10 >= I11, I10, I11, B5) 
0.00/0.54	  max#(cons(I10, cons(I11, B5))) -> max#(cons(I11, B5)) [not(I10 >= I11)]
0.00/0.54	  max#(cons(I10, cons(I11, B5))) -> max#(cons(I10, B5)) [I10 >= I11]
0.00/0.54	  del#(I3, cons(I4, B2)) -> del#(I3, B2) [not(I3 = I4)]
0.00/0.54	R = 
0.00/0.54	  sort(cons(x, xs)) -> cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs)))) 
0.00/0.54	  sort(nil) -> nil 
0.00/0.54	  if2(false, I0, y, B0) -> cons(y, del(I0, B0)) 
0.00/0.54	  if2(true, I1, I2, B1) -> B1 
0.00/0.54	  del(I3, cons(I4, B2)) -> if2(I3 = I4, I3, I4, B2) 
0.00/0.54	  del(I5, nil) -> nil 
0.00/0.54	  if1(false, I6, I7, B3) -> max(cons(I7, B3)) 
0.00/0.54	  if1(true, I8, I9, B4) -> max(cons(I8, B4)) 
0.00/0.54	  max(cons(I10, cons(I11, B5))) -> if1(I10 >= I11, I10, I11, B5) 
0.00/0.54	  max(cons(I12, nil)) -> I12 
0.00/0.54	  max(nil) -> 0 
0.00/0.54	
0.00/0.54	The dependency graph for this problem is:
0.00/0.54	  0 -> 10, 9, 8
0.00/0.54	  1 -> 0, 1, 2, 3
0.00/0.54	  2 -> 11, 5
0.00/0.54	  3 -> 10, 9, 8
0.00/0.54	  4 -> 11, 5
0.00/0.54	  5 -> 
0.00/0.54	  6 -> 10, 9, 8
0.00/0.54	  7 -> 10, 9, 8
0.00/0.54	  8 -> 
0.00/0.54	  9 -> 10, 9, 8
0.00/0.54	  10 -> 10, 8, 9
0.00/0.54	  11 -> 11, 5
0.00/0.54	Where:
0.00/0.54	  0) sort#(cons(x, xs)) -> max#(cons(x, xs)) 
0.00/0.54	  1) sort#(cons(x, xs)) -> sort#(del(max(cons(x, xs)), cons(x, xs))) 
0.00/0.54	  2) sort#(cons(x, xs)) -> del#(max(cons(x, xs)), cons(x, xs)) 
0.00/0.54	  3) sort#(cons(x, xs)) -> max#(cons(x, xs)) 
0.00/0.54	  4) if2#(false, I0, y, B0) -> del#(I0, B0) 
0.00/0.54	  5) del#(I3, cons(I4, B2)) -> if2#(I3 = I4, I3, I4, B2) 
0.00/0.54	  6) if1#(false, I6, I7, B3) -> max#(cons(I7, B3)) 
0.00/0.54	  7) if1#(true, I8, I9, B4) -> max#(cons(I8, B4)) 
0.00/0.54	  8) max#(cons(I10, cons(I11, B5))) -> if1#(I10 >= I11, I10, I11, B5) 
0.00/0.54	  9) max#(cons(I10, cons(I11, B5))) -> max#(cons(I11, B5)) [not(I10 >= I11)]
0.00/0.54	  10) max#(cons(I10, cons(I11, B5))) -> max#(cons(I10, B5)) [I10 >= I11]
0.00/0.54	  11) del#(I3, cons(I4, B2)) -> del#(I3, B2) [not(I3 = I4)]
0.00/0.54	
0.00/0.54	We have the following SCCs.
0.00/0.54	  { 1 }
0.00/0.54	  { 9, 10 }
0.00/0.54	  { 11 }
0.00/0.54	
0.00/0.54	DP problem for innermost termination.
0.00/0.54	P =
0.00/0.54	  del#(I3, cons(I4, B2)) -> del#(I3, B2) [not(I3 = I4)]
0.00/0.54	R = 
0.00/0.54	  sort(cons(x, xs)) -> cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs)))) 
0.00/0.54	  sort(nil) -> nil 
0.00/0.54	  if2(false, I0, y, B0) -> cons(y, del(I0, B0)) 
0.00/0.54	  if2(true, I1, I2, B1) -> B1 
0.00/0.54	  del(I3, cons(I4, B2)) -> if2(I3 = I4, I3, I4, B2) 
0.00/0.54	  del(I5, nil) -> nil 
0.00/0.54	  if1(false, I6, I7, B3) -> max(cons(I7, B3)) 
0.00/0.54	  if1(true, I8, I9, B4) -> max(cons(I8, B4)) 
0.00/0.54	  max(cons(I10, cons(I11, B5))) -> if1(I10 >= I11, I10, I11, B5) 
0.00/0.54	  max(cons(I12, nil)) -> I12 
0.00/0.54	  max(nil) -> 0 
0.00/0.54	
0.00/0.54	We use the subterm criterion with the projection function NU:
0.00/0.54	  NU[del#(x0,x1)] = x1
0.00/0.54	
0.00/0.54	This gives the following inequalities:
0.00/0.54	  cons(I4, B2) |> B2
0.00/0.54	
0.00/0.54	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.54	
0.00/0.54	DP problem for innermost termination.
0.00/0.54	P =
0.00/0.54	  max#(cons(I10, cons(I11, B5))) -> max#(cons(I11, B5)) [not(I10 >= I11)]
0.00/0.54	  max#(cons(I10, cons(I11, B5))) -> max#(cons(I10, B5)) [I10 >= I11]
0.00/0.54	R = 
0.00/0.54	  sort(cons(x, xs)) -> cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs)))) 
0.00/0.54	  sort(nil) -> nil 
0.00/0.54	  if2(false, I0, y, B0) -> cons(y, del(I0, B0)) 
0.00/0.54	  if2(true, I1, I2, B1) -> B1 
0.00/0.54	  del(I3, cons(I4, B2)) -> if2(I3 = I4, I3, I4, B2) 
0.00/0.54	  del(I5, nil) -> nil 
0.00/0.54	  if1(false, I6, I7, B3) -> max(cons(I7, B3)) 
0.00/0.54	  if1(true, I8, I9, B4) -> max(cons(I8, B4)) 
0.00/0.54	  max(cons(I10, cons(I11, B5))) -> if1(I10 >= I11, I10, I11, B5) 
0.00/0.54	  max(cons(I12, nil)) -> I12 
0.00/0.54	  max(nil) -> 0 
0.00/0.54	
0.00/3.52	EOF