0.00/0.61	MAYBE
0.00/0.61	
0.00/0.61	DP problem for innermost termination.
0.00/0.61	P =
0.00/0.61	  max#(cons(I2, l)) -> if#(I2 > max(l), I2, max(l)) 
0.00/0.61	  max#(cons(I2, l)) -> max#(l) 
0.00/0.61	  max#(cons(I2, l)) -> max#(l) 
0.00/0.61	  member#(n, cons(m, B0)) -> member#(n, B0) 
0.00/0.61	  cond2#(false, I4, B1) -> st#(I4 + 1, B1) 
0.00/0.61	  cond1#(false, I6, B3) -> cond2#(I6 > max(B3), I6, B3) 
0.00/0.61	  cond1#(false, I6, B3) -> max#(B3) 
0.00/0.61	  cond1#(true, I7, B4) -> st#(I7 + 1, B4) 
0.00/0.61	  stNat#(true, I8, B5) -> cond1#(member(I8, B5), I8, B5) 
0.00/0.61	  stNat#(true, I8, B5) -> member#(I8, B5) 
0.00/0.61	  st#(I9, B6) -> stNat#(I9 >= 0, I9, B6) 
0.00/0.61	  sort#(B7) -> st#(0, B7) 
0.00/0.61	R = 
0.00/0.61	  if(false, u, v) -> v 
0.00/0.61	  if(true, I0, I1) -> I0 
0.00/0.61	  max(cons(I2, l)) -> if(I2 > max(l), I2, max(l)) 
0.00/0.61	  max(nil) -> 0 
0.00/0.61	  member(n, cons(m, B0)) -> n = m || member(n, B0) 
0.00/0.61	  member(I3, nil) -> false 
0.00/0.61	  cond2(false, I4, B1) -> st(I4 + 1, B1) 
0.00/0.61	  cond2(true, I5, B2) -> nil 
0.00/0.61	  cond1(false, I6, B3) -> cond2(I6 > max(B3), I6, B3) 
0.00/0.61	  cond1(true, I7, B4) -> cons(I7, st(I7 + 1, B4)) 
0.00/0.61	  stNat(true, I8, B5) -> cond1(member(I8, B5), I8, B5) 
0.00/0.61	  st(I9, B6) -> stNat(I9 >= 0, I9, B6) 
0.00/0.61	  sort(B7) -> st(0, B7) 
0.00/0.61	
0.00/0.61	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.61	
0.00/0.61	DP problem for innermost termination.
0.00/0.61	P =
0.00/0.61	  max#(cons(I2, l)) -> if#(I2 > max(l), I2, max(l)) 
0.00/0.61	  max#(cons(I2, l)) -> max#(l) 
0.00/0.61	  max#(cons(I2, l)) -> max#(l) 
0.00/0.61	  member#(n, cons(m, B0)) -> member#(n, B0) 
0.00/0.61	  cond2#(false, I4, B1) -> st#(I4 + 1, B1) 
0.00/0.61	  cond1#(false, I6, B3) -> cond2#(I6 > max(B3), I6, B3) 
0.00/0.61	  cond1#(false, I6, B3) -> max#(B3) 
0.00/0.61	  cond1#(true, I7, B4) -> st#(I7 + 1, B4) 
0.00/0.61	  stNat#(true, I8, B5) -> cond1#(member(I8, B5), I8, B5) 
0.00/0.61	  stNat#(true, I8, B5) -> member#(I8, B5) 
0.00/0.61	  st#(I9, B6) -> stNat#(I9 >= 0, I9, B6) 
0.00/0.61	  sort#(B7) -> st#(0, B7) 
0.00/0.61	  st#(I9, B6) -> cond1#(member(I9, B6), I9, B6) [I9 >= 0]
0.00/0.61	  st#(I9, B6) -> member#(I9, B6) [I9 >= 0]
0.00/0.61	R = 
0.00/0.61	  if(false, u, v) -> v 
0.00/0.61	  if(true, I0, I1) -> I0 
0.00/0.61	  max(cons(I2, l)) -> if(I2 > max(l), I2, max(l)) 
0.00/0.61	  max(nil) -> 0 
0.00/0.61	  member(n, cons(m, B0)) -> n = m || member(n, B0) 
0.00/0.61	  member(I3, nil) -> false 
0.00/0.61	  cond2(false, I4, B1) -> st(I4 + 1, B1) 
0.00/0.61	  cond2(true, I5, B2) -> nil 
0.00/0.61	  cond1(false, I6, B3) -> cond2(I6 > max(B3), I6, B3) 
0.00/0.61	  cond1(true, I7, B4) -> cons(I7, st(I7 + 1, B4)) 
0.00/0.61	  stNat(true, I8, B5) -> cond1(member(I8, B5), I8, B5) 
0.00/0.61	  st(I9, B6) -> stNat(I9 >= 0, I9, B6) 
0.00/0.61	  sort(B7) -> st(0, B7) 
0.00/0.61	
0.00/0.61	The dependency graph for this problem is:
0.00/0.61	  0 -> 
0.00/0.61	  1 -> 0, 1, 2
0.00/0.61	  2 -> 0, 1, 2
0.00/0.61	  3 -> 3
0.00/0.61	  4 -> 13, 12, 10
0.00/0.61	  5 -> 4
0.00/0.61	  6 -> 0, 1, 2
0.00/0.61	  7 -> 13, 12, 10
0.00/0.61	  8 -> 5, 6, 7
0.00/0.61	  9 -> 3
0.00/0.61	  10 -> 
0.00/0.61	  11 -> 13, 12, 10
0.00/0.61	  12 -> 7, 6, 5
0.00/0.61	  13 -> 3
0.00/0.61	Where:
0.00/0.61	  0) max#(cons(I2, l)) -> if#(I2 > max(l), I2, max(l)) 
0.00/0.61	  1) max#(cons(I2, l)) -> max#(l) 
0.00/0.61	  2) max#(cons(I2, l)) -> max#(l) 
0.00/0.61	  3) member#(n, cons(m, B0)) -> member#(n, B0) 
0.00/0.61	  4) cond2#(false, I4, B1) -> st#(I4 + 1, B1) 
0.00/0.61	  5) cond1#(false, I6, B3) -> cond2#(I6 > max(B3), I6, B3) 
0.00/0.61	  6) cond1#(false, I6, B3) -> max#(B3) 
0.00/0.61	  7) cond1#(true, I7, B4) -> st#(I7 + 1, B4) 
0.00/0.61	  8) stNat#(true, I8, B5) -> cond1#(member(I8, B5), I8, B5) 
0.00/0.61	  9) stNat#(true, I8, B5) -> member#(I8, B5) 
0.00/0.61	  10) st#(I9, B6) -> stNat#(I9 >= 0, I9, B6) 
0.00/0.61	  11) sort#(B7) -> st#(0, B7) 
0.00/0.61	  12) st#(I9, B6) -> cond1#(member(I9, B6), I9, B6) [I9 >= 0]
0.00/0.61	  13) st#(I9, B6) -> member#(I9, B6) [I9 >= 0]
0.00/0.61	
0.00/0.61	We have the following SCCs.
0.00/0.61	  { 4, 5, 7, 12 }
0.00/0.61	  { 1, 2 }
0.00/0.61	  { 3 }
0.00/0.61	
0.00/0.61	DP problem for innermost termination.
0.00/0.61	P =
0.00/0.61	  member#(n, cons(m, B0)) -> member#(n, B0) 
0.00/0.61	R = 
0.00/0.61	  if(false, u, v) -> v 
0.00/0.61	  if(true, I0, I1) -> I0 
0.00/0.61	  max(cons(I2, l)) -> if(I2 > max(l), I2, max(l)) 
0.00/0.61	  max(nil) -> 0 
0.00/0.61	  member(n, cons(m, B0)) -> n = m || member(n, B0) 
0.00/0.61	  member(I3, nil) -> false 
0.00/0.61	  cond2(false, I4, B1) -> st(I4 + 1, B1) 
0.00/0.61	  cond2(true, I5, B2) -> nil 
0.00/0.61	  cond1(false, I6, B3) -> cond2(I6 > max(B3), I6, B3) 
0.00/0.61	  cond1(true, I7, B4) -> cons(I7, st(I7 + 1, B4)) 
0.00/0.61	  stNat(true, I8, B5) -> cond1(member(I8, B5), I8, B5) 
0.00/0.61	  st(I9, B6) -> stNat(I9 >= 0, I9, B6) 
0.00/0.61	  sort(B7) -> st(0, B7) 
0.00/0.61	
0.00/0.61	We use the subterm criterion with the projection function NU:
0.00/0.61	  NU[member#(x0,x1)] = x1
0.00/0.61	
0.00/0.61	This gives the following inequalities:
0.00/0.61	  cons(m, B0) |> B0
0.00/0.61	
0.00/0.61	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.61	
0.00/0.61	DP problem for innermost termination.
0.00/0.61	P =
0.00/0.61	  max#(cons(I2, l)) -> max#(l) 
0.00/0.61	  max#(cons(I2, l)) -> max#(l) 
0.00/0.61	R = 
0.00/0.61	  if(false, u, v) -> v 
0.00/0.61	  if(true, I0, I1) -> I0 
0.00/0.61	  max(cons(I2, l)) -> if(I2 > max(l), I2, max(l)) 
0.00/0.61	  max(nil) -> 0 
0.00/0.61	  member(n, cons(m, B0)) -> n = m || member(n, B0) 
0.00/0.61	  member(I3, nil) -> false 
0.00/0.61	  cond2(false, I4, B1) -> st(I4 + 1, B1) 
0.00/0.61	  cond2(true, I5, B2) -> nil 
0.00/0.61	  cond1(false, I6, B3) -> cond2(I6 > max(B3), I6, B3) 
0.00/0.61	  cond1(true, I7, B4) -> cons(I7, st(I7 + 1, B4)) 
0.00/0.61	  stNat(true, I8, B5) -> cond1(member(I8, B5), I8, B5) 
0.00/0.61	  st(I9, B6) -> stNat(I9 >= 0, I9, B6) 
0.00/0.61	  sort(B7) -> st(0, B7) 
0.00/0.61	
0.00/0.61	We use the subterm criterion with the projection function NU:
0.00/0.61	  NU[max#(x0)] = x0
0.00/0.61	
0.00/0.61	This gives the following inequalities:
0.00/0.61	  cons(I2, l) |> l
0.00/0.61	  cons(I2, l) |> l
0.00/0.61	
0.00/0.61	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.61	
0.00/0.61	DP problem for innermost termination.
0.00/0.61	P =
0.00/0.61	  cond2#(false, I4, B1) -> st#(I4 + 1, B1) 
0.00/0.61	  cond1#(false, I6, B3) -> cond2#(I6 > max(B3), I6, B3) 
0.00/0.61	  cond1#(true, I7, B4) -> st#(I7 + 1, B4) 
0.00/0.61	  st#(I9, B6) -> cond1#(member(I9, B6), I9, B6) [I9 >= 0]
0.00/0.61	R = 
0.00/0.61	  if(false, u, v) -> v 
0.00/0.61	  if(true, I0, I1) -> I0 
0.00/0.61	  max(cons(I2, l)) -> if(I2 > max(l), I2, max(l)) 
0.00/0.61	  max(nil) -> 0 
0.00/0.61	  member(n, cons(m, B0)) -> n = m || member(n, B0) 
0.00/0.61	  member(I3, nil) -> false 
0.00/0.61	  cond2(false, I4, B1) -> st(I4 + 1, B1) 
0.00/0.61	  cond2(true, I5, B2) -> nil 
0.00/0.61	  cond1(false, I6, B3) -> cond2(I6 > max(B3), I6, B3) 
0.00/0.61	  cond1(true, I7, B4) -> cons(I7, st(I7 + 1, B4)) 
0.00/0.61	  stNat(true, I8, B5) -> cond1(member(I8, B5), I8, B5) 
0.00/0.61	  st(I9, B6) -> stNat(I9 >= 0, I9, B6) 
0.00/0.61	  sort(B7) -> st(0, B7) 
0.00/0.61	
0.00/3.59	EOF