Termination w.r.t. Q proof of TRS/secret06toList.trs

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 interpretation [21]:

POL(APPEND₂(x₁, x₂)) = 2·x₁
POL(cons₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(y) = 2

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.