Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/AG01SLASHHASH3.17a.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:

app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
sum₁(plus₂(cons₂(0, x), cons₂(y, l))) -> pred₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
pred₁(cons₂(s₁(x), nil)) -> cons₂(x, nil)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
sum₁(plus₂(cons₂(0, x), cons₂(y, l))) -> pred₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
pred₁(cons₂(s₁(x), nil)) -> cons₂(x, 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:

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> APP₂(l, sum₁(cons₂(x, cons₂(y, k))))
SUM₁(cons₂(x, cons₂(y, l))) -> PLUS₂(x, y)
SUM₁(plus₂(cons₂(0, x), cons₂(y, l))) -> SUM₁(cons₂(s₁(x), cons₂(y, l)))
SUM₁(cons₂(x, cons₂(y, l))) -> SUM₁(cons₂(plus₂(x, y), l))
SUM₁(plus₂(cons₂(0, x), cons₂(y, l))) -> PRED₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> SUM₁(cons₂(x, cons₂(y, k)))
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> SUM₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
APP₂(cons₂(x, l), k) -> APP₂(l, k)

The TRS R consists of the following rules:

app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
sum₁(plus₂(cons₂(0, x), cons₂(y, l))) -> pred₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
pred₁(cons₂(s₁(x), nil)) -> cons₂(x, 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:

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> APP₂(l, sum₁(cons₂(x, cons₂(y, k))))
SUM₁(cons₂(x, cons₂(y, l))) -> PLUS₂(x, y)
SUM₁(plus₂(cons₂(0, x), cons₂(y, l))) -> SUM₁(cons₂(s₁(x), cons₂(y, l)))
SUM₁(cons₂(x, cons₂(y, l))) -> SUM₁(cons₂(plus₂(x, y), l))
SUM₁(plus₂(cons₂(0, x), cons₂(y, l))) -> PRED₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> SUM₁(cons₂(x, cons₂(y, k)))
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> SUM₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
APP₂(cons₂(x, l), k) -> APP₂(l, k)

The TRS R consists of the following rules:

app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
sum₁(plus₂(cons₂(0, x), cons₂(y, l))) -> pred₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
pred₁(cons₂(s₁(x), nil)) -> cons₂(x, nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The TRS R consists of the following rules:

app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
sum₁(plus₂(cons₂(0, x), cons₂(y, l))) -> pred₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
pred₁(cons₂(s₁(x), nil)) -> cons₂(x, 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 strictly oriented and are deleted.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PLUS₂(x1, x2) = PLUS₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[PLUS₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
sum₁(plus₂(cons₂(0, x), cons₂(y, l))) -> pred₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
pred₁(cons₂(s₁(x), nil)) -> cons₂(x, 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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

APP₂(cons₂(x, l), k) -> APP₂(l, k)

The TRS R consists of the following rules:

app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
sum₁(plus₂(cons₂(0, x), cons₂(y, l))) -> pred₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
pred₁(cons₂(s₁(x), nil)) -> cons₂(x, 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 strictly oriented and are deleted.

APP₂(cons₂(x, l), k) -> APP₂(l, k)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP₂(x1, x2) = APP₁(x1)
cons₂(x1, x2) = cons₁(x2)

Lexicographic Path Order [19].
Precedence:

[APP₁, cons_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
sum₁(plus₂(cons₂(0, x), cons₂(y, l))) -> pred₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
pred₁(cons₂(s₁(x), nil)) -> cons₂(x, 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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

SUM₁(cons₂(x, cons₂(y, l))) -> SUM₁(cons₂(plus₂(x, y), l))

The TRS R consists of the following rules:

app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
sum₁(plus₂(cons₂(0, x), cons₂(y, l))) -> pred₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
pred₁(cons₂(s₁(x), nil)) -> cons₂(x, 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 strictly oriented and are deleted.

SUM₁(cons₂(x, cons₂(y, l))) -> SUM₁(cons₂(plus₂(x, y), l))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SUM₁(x1) = x1
cons₂(x1, x2) = cons₁(x2)
plus₂(x1, x2) = x2
0 = 0
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
sum₁(plus₂(cons₂(0, x), cons₂(y, l))) -> pred₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
pred₁(cons₂(s₁(x), nil)) -> cons₂(x, 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

          ↳ QDP

          ↳ QDP

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

SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> SUM₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

The TRS R consists of the following rules:

app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
sum₁(plus₂(cons₂(0, x), cons₂(y, l))) -> pred₁(sum₁(cons₂(s₁(x), cons₂(y, l))))
pred₁(cons₂(s₁(x), nil)) -> cons₂(x, nil)

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