Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

isEmpty1(empty) -> true
isEmpty1(node3(l, x, r)) -> false
left1(empty) -> empty
left1(node3(l, x, r)) -> l
right1(empty) -> empty
right1(node3(l, x, r)) -> r
elem1(node3(l, x, r)) -> x
append2(nil, x) -> cons2(x, nil)
append2(cons2(y, ys), x) -> cons2(y, append2(ys, x))
listify2(n, xs) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node3(left1(left1(n)), elem1(left1(n)), node3(right1(left1(n)), elem1(n), right1(n))), xs, append2(xs, n))
if6(true, b, n, m, xs, ys) -> xs
if6(false, false, n, m, xs, ys) -> listify2(m, xs)
if6(false, true, n, m, xs, ys) -> listify2(n, ys)
toList1(n) -> listify2(n, nil)

Q is empty.


QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

isEmpty1(empty) -> true
isEmpty1(node3(l, x, r)) -> false
left1(empty) -> empty
left1(node3(l, x, r)) -> l
right1(empty) -> empty
right1(node3(l, x, r)) -> r
elem1(node3(l, x, r)) -> x
append2(nil, x) -> cons2(x, nil)
append2(cons2(y, ys), x) -> cons2(y, append2(ys, x))
listify2(n, xs) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node3(left1(left1(n)), elem1(left1(n)), node3(right1(left1(n)), elem1(n), right1(n))), xs, append2(xs, n))
if6(true, b, n, m, xs, ys) -> xs
if6(false, false, n, m, xs, ys) -> listify2(m, xs)
if6(false, true, n, m, xs, ys) -> listify2(n, ys)
toList1(n) -> listify2(n, nil)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

LISTIFY2(n, xs) -> IF6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node3(left1(left1(n)), elem1(left1(n)), node3(right1(left1(n)), elem1(n), right1(n))), xs, append2(xs, n))
LISTIFY2(n, xs) -> ELEM1(n)
IF6(false, false, n, m, xs, ys) -> LISTIFY2(m, xs)
TOLIST1(n) -> LISTIFY2(n, nil)
LISTIFY2(n, xs) -> RIGHT1(n)
LISTIFY2(n, xs) -> LEFT1(left1(n))
IF6(false, true, n, m, xs, ys) -> LISTIFY2(n, ys)
APPEND2(cons2(y, ys), x) -> APPEND2(ys, x)
LISTIFY2(n, xs) -> LEFT1(n)
LISTIFY2(n, xs) -> RIGHT1(left1(n))
LISTIFY2(n, xs) -> ISEMPTY1(n)
LISTIFY2(n, xs) -> ISEMPTY1(left1(n))
LISTIFY2(n, xs) -> APPEND2(xs, n)
LISTIFY2(n, xs) -> ELEM1(left1(n))

The TRS R consists of the following rules:

isEmpty1(empty) -> true
isEmpty1(node3(l, x, r)) -> false
left1(empty) -> empty
left1(node3(l, x, r)) -> l
right1(empty) -> empty
right1(node3(l, x, r)) -> r
elem1(node3(l, x, r)) -> x
append2(nil, x) -> cons2(x, nil)
append2(cons2(y, ys), x) -> cons2(y, append2(ys, x))
listify2(n, xs) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node3(left1(left1(n)), elem1(left1(n)), node3(right1(left1(n)), elem1(n), right1(n))), xs, append2(xs, n))
if6(true, b, n, m, xs, ys) -> xs
if6(false, false, n, m, xs, ys) -> listify2(m, xs)
if6(false, true, n, m, xs, ys) -> listify2(n, ys)
toList1(n) -> listify2(n, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

LISTIFY2(n, xs) -> IF6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node3(left1(left1(n)), elem1(left1(n)), node3(right1(left1(n)), elem1(n), right1(n))), xs, append2(xs, n))
LISTIFY2(n, xs) -> ELEM1(n)
IF6(false, false, n, m, xs, ys) -> LISTIFY2(m, xs)
TOLIST1(n) -> LISTIFY2(n, nil)
LISTIFY2(n, xs) -> RIGHT1(n)
LISTIFY2(n, xs) -> LEFT1(left1(n))
IF6(false, true, n, m, xs, ys) -> LISTIFY2(n, ys)
APPEND2(cons2(y, ys), x) -> APPEND2(ys, x)
LISTIFY2(n, xs) -> LEFT1(n)
LISTIFY2(n, xs) -> RIGHT1(left1(n))
LISTIFY2(n, xs) -> ISEMPTY1(n)
LISTIFY2(n, xs) -> ISEMPTY1(left1(n))
LISTIFY2(n, xs) -> APPEND2(xs, n)
LISTIFY2(n, xs) -> ELEM1(left1(n))

The TRS R consists of the following rules:

isEmpty1(empty) -> true
isEmpty1(node3(l, x, r)) -> false
left1(empty) -> empty
left1(node3(l, x, r)) -> l
right1(empty) -> empty
right1(node3(l, x, r)) -> r
elem1(node3(l, x, r)) -> x
append2(nil, x) -> cons2(x, nil)
append2(cons2(y, ys), x) -> cons2(y, append2(ys, x))
listify2(n, xs) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node3(left1(left1(n)), elem1(left1(n)), node3(right1(left1(n)), elem1(n), right1(n))), xs, append2(xs, n))
if6(true, b, n, m, xs, ys) -> xs
if6(false, false, n, m, xs, ys) -> listify2(m, xs)
if6(false, true, n, m, xs, ys) -> listify2(n, ys)
toList1(n) -> listify2(n, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 10 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APPEND2(cons2(y, ys), x) -> APPEND2(ys, x)

The TRS R consists of the following rules:

isEmpty1(empty) -> true
isEmpty1(node3(l, x, r)) -> false
left1(empty) -> empty
left1(node3(l, x, r)) -> l
right1(empty) -> empty
right1(node3(l, x, r)) -> r
elem1(node3(l, x, r)) -> x
append2(nil, x) -> cons2(x, nil)
append2(cons2(y, ys), x) -> cons2(y, append2(ys, x))
listify2(n, xs) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node3(left1(left1(n)), elem1(left1(n)), node3(right1(left1(n)), elem1(n), right1(n))), xs, append2(xs, n))
if6(true, b, n, m, xs, ys) -> xs
if6(false, false, n, m, xs, ys) -> listify2(m, xs)
if6(false, true, n, m, xs, ys) -> listify2(n, ys)
toList1(n) -> listify2(n, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


APPEND2(cons2(y, ys), x) -> APPEND2(ys, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( cons2(x1, x2) ) = x1 + x2 + 1


POL( APPEND2(x1, x2) ) = max{0, x1 + x2 - 1}


POL( y ) = 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

isEmpty1(empty) -> true
isEmpty1(node3(l, x, r)) -> false
left1(empty) -> empty
left1(node3(l, x, r)) -> l
right1(empty) -> empty
right1(node3(l, x, r)) -> r
elem1(node3(l, x, r)) -> x
append2(nil, x) -> cons2(x, nil)
append2(cons2(y, ys), x) -> cons2(y, append2(ys, x))
listify2(n, xs) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node3(left1(left1(n)), elem1(left1(n)), node3(right1(left1(n)), elem1(n), right1(n))), xs, append2(xs, n))
if6(true, b, n, m, xs, ys) -> xs
if6(false, false, n, m, xs, ys) -> listify2(m, xs)
if6(false, true, n, m, xs, ys) -> listify2(n, ys)
toList1(n) -> listify2(n, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP

Q DP problem:
The TRS P consists of the following rules:

LISTIFY2(n, xs) -> IF6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node3(left1(left1(n)), elem1(left1(n)), node3(right1(left1(n)), elem1(n), right1(n))), xs, append2(xs, n))
IF6(false, true, n, m, xs, ys) -> LISTIFY2(n, ys)
IF6(false, false, n, m, xs, ys) -> LISTIFY2(m, xs)

The TRS R consists of the following rules:

isEmpty1(empty) -> true
isEmpty1(node3(l, x, r)) -> false
left1(empty) -> empty
left1(node3(l, x, r)) -> l
right1(empty) -> empty
right1(node3(l, x, r)) -> r
elem1(node3(l, x, r)) -> x
append2(nil, x) -> cons2(x, nil)
append2(cons2(y, ys), x) -> cons2(y, append2(ys, x))
listify2(n, xs) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node3(left1(left1(n)), elem1(left1(n)), node3(right1(left1(n)), elem1(n), right1(n))), xs, append2(xs, n))
if6(true, b, n, m, xs, ys) -> xs
if6(false, false, n, m, xs, ys) -> listify2(m, xs)
if6(false, true, n, m, xs, ys) -> listify2(n, ys)
toList1(n) -> listify2(n, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.