Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/D33SLASH32.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

sort₁(nil) -> nil
sort₁(cons₂(x, y)) -> insert₂(x, sort₁(y))
insert₂(x, nil) -> cons₂(x, nil)
insert₂(x, cons₂(v, w)) -> choose₄(x, cons₂(v, w), x, v)
choose₄(x, cons₂(v, w), y, 0) -> cons₂(x, cons₂(v, w))
choose₄(x, cons₂(v, w), 0, s₁(z)) -> cons₂(v, insert₂(x, w))
choose₄(x, cons₂(v, w), s₁(y), s₁(z)) -> choose₄(x, cons₂(v, w), y, z)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

sort₁(nil) -> nil
sort₁(cons₂(x, y)) -> insert₂(x, sort₁(y))
insert₂(x, nil) -> cons₂(x, nil)
insert₂(x, cons₂(v, w)) -> choose₄(x, cons₂(v, w), x, v)
choose₄(x, cons₂(v, w), y, 0) -> cons₂(x, cons₂(v, w))
choose₄(x, cons₂(v, w), 0, s₁(z)) -> cons₂(v, insert₂(x, w))
choose₄(x, cons₂(v, w), s₁(y), s₁(z)) -> choose₄(x, cons₂(v, w), y, z)

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

sort₁(nil) -> nil
sort₁(cons₂(x, y)) -> insert₂(x, sort₁(y))
insert₂(x, nil) -> cons₂(x, nil)
insert₂(x, cons₂(v, w)) -> choose₄(x, cons₂(v, w), x, v)
choose₄(x, cons₂(v, w), y, 0) -> cons₂(x, cons₂(v, w))
choose₄(x, cons₂(v, w), 0, s₁(z)) -> cons₂(v, insert₂(x, w))
choose₄(x, cons₂(v, w), s₁(y), s₁(z)) -> choose₄(x, cons₂(v, w), y, z)

The set Q consists of the following terms:

sort₁(nil)
sort₁(cons₂(x0, x1))
insert₂(x0, nil)
insert₂(x0, cons₂(x1, x2))
choose₄(x0, cons₂(x1, x2), x3, 0)
choose₄(x0, cons₂(x1, x2), 0, s₁(x3))
choose₄(x0, cons₂(x1, x2), s₁(x3), s₁(x4))

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

SORT₁(cons₂(x, y)) -> INSERT₂(x, sort₁(y))
INSERT₂(x, cons₂(v, w)) -> CHOOSE₄(x, cons₂(v, w), x, v)
CHOOSE₄(x, cons₂(v, w), s₁(y), s₁(z)) -> CHOOSE₄(x, cons₂(v, w), y, z)
CHOOSE₄(x, cons₂(v, w), 0, s₁(z)) -> INSERT₂(x, w)
SORT₁(cons₂(x, y)) -> SORT₁(y)

The TRS R consists of the following rules:

sort₁(nil) -> nil
sort₁(cons₂(x, y)) -> insert₂(x, sort₁(y))
insert₂(x, nil) -> cons₂(x, nil)
insert₂(x, cons₂(v, w)) -> choose₄(x, cons₂(v, w), x, v)
choose₄(x, cons₂(v, w), y, 0) -> cons₂(x, cons₂(v, w))
choose₄(x, cons₂(v, w), 0, s₁(z)) -> cons₂(v, insert₂(x, w))
choose₄(x, cons₂(v, w), s₁(y), s₁(z)) -> choose₄(x, cons₂(v, w), y, z)

The set Q consists of the following terms:

sort₁(nil)
sort₁(cons₂(x0, x1))
insert₂(x0, nil)
insert₂(x0, cons₂(x1, x2))
choose₄(x0, cons₂(x1, x2), x3, 0)
choose₄(x0, cons₂(x1, x2), 0, s₁(x3))
choose₄(x0, cons₂(x1, x2), s₁(x3), s₁(x4))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

SORT₁(cons₂(x, y)) -> INSERT₂(x, sort₁(y))
INSERT₂(x, cons₂(v, w)) -> CHOOSE₄(x, cons₂(v, w), x, v)
CHOOSE₄(x, cons₂(v, w), s₁(y), s₁(z)) -> CHOOSE₄(x, cons₂(v, w), y, z)
CHOOSE₄(x, cons₂(v, w), 0, s₁(z)) -> INSERT₂(x, w)
SORT₁(cons₂(x, y)) -> SORT₁(y)

The TRS R consists of the following rules:

sort₁(nil) -> nil
sort₁(cons₂(x, y)) -> insert₂(x, sort₁(y))
insert₂(x, nil) -> cons₂(x, nil)
insert₂(x, cons₂(v, w)) -> choose₄(x, cons₂(v, w), x, v)
choose₄(x, cons₂(v, w), y, 0) -> cons₂(x, cons₂(v, w))
choose₄(x, cons₂(v, w), 0, s₁(z)) -> cons₂(v, insert₂(x, w))
choose₄(x, cons₂(v, w), s₁(y), s₁(z)) -> choose₄(x, cons₂(v, w), y, z)

The set Q consists of the following terms:

sort₁(nil)
sort₁(cons₂(x0, x1))
insert₂(x0, nil)
insert₂(x0, cons₂(x1, x2))
choose₄(x0, cons₂(x1, x2), x3, 0)
choose₄(x0, cons₂(x1, x2), 0, s₁(x3))
choose₄(x0, cons₂(x1, x2), s₁(x3), s₁(x4))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

INSERT₂(x, cons₂(v, w)) -> CHOOSE₄(x, cons₂(v, w), x, v)
CHOOSE₄(x, cons₂(v, w), s₁(y), s₁(z)) -> CHOOSE₄(x, cons₂(v, w), y, z)
CHOOSE₄(x, cons₂(v, w), 0, s₁(z)) -> INSERT₂(x, w)

The TRS R consists of the following rules:

sort₁(nil) -> nil
sort₁(cons₂(x, y)) -> insert₂(x, sort₁(y))
insert₂(x, nil) -> cons₂(x, nil)
insert₂(x, cons₂(v, w)) -> choose₄(x, cons₂(v, w), x, v)
choose₄(x, cons₂(v, w), y, 0) -> cons₂(x, cons₂(v, w))
choose₄(x, cons₂(v, w), 0, s₁(z)) -> cons₂(v, insert₂(x, w))
choose₄(x, cons₂(v, w), s₁(y), s₁(z)) -> choose₄(x, cons₂(v, w), y, z)

The set Q consists of the following terms:

sort₁(nil)
sort₁(cons₂(x0, x1))
insert₂(x0, nil)
insert₂(x0, cons₂(x1, x2))
choose₄(x0, cons₂(x1, x2), x3, 0)
choose₄(x0, cons₂(x1, x2), 0, s₁(x3))
choose₄(x0, cons₂(x1, x2), s₁(x3), s₁(x4))

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.

CHOOSE₄(x, cons₂(v, w), 0, s₁(z)) -> INSERT₂(x, w)

The remaining pairs can at least by weakly be oriented.

INSERT₂(x, cons₂(v, w)) -> CHOOSE₄(x, cons₂(v, w), x, v)
CHOOSE₄(x, cons₂(v, w), s₁(y), s₁(z)) -> CHOOSE₄(x, cons₂(v, w), y, z)

Used ordering: Combined order from the following AFS and order.
INSERT₂(x1, x2) = x2
cons₂(x1, x2) = cons₂(x1, x2)
CHOOSE₄(x1, x2, x3, x4) = x2
s₁(x1) = x1
0 = 0

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

              ↳ QDP

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

INSERT₂(x, cons₂(v, w)) -> CHOOSE₄(x, cons₂(v, w), x, v)
CHOOSE₄(x, cons₂(v, w), s₁(y), s₁(z)) -> CHOOSE₄(x, cons₂(v, w), y, z)

The TRS R consists of the following rules:

sort₁(nil) -> nil
sort₁(cons₂(x, y)) -> insert₂(x, sort₁(y))
insert₂(x, nil) -> cons₂(x, nil)
insert₂(x, cons₂(v, w)) -> choose₄(x, cons₂(v, w), x, v)
choose₄(x, cons₂(v, w), y, 0) -> cons₂(x, cons₂(v, w))
choose₄(x, cons₂(v, w), 0, s₁(z)) -> cons₂(v, insert₂(x, w))
choose₄(x, cons₂(v, w), s₁(y), s₁(z)) -> choose₄(x, cons₂(v, w), y, z)

The set Q consists of the following terms:

sort₁(nil)
sort₁(cons₂(x0, x1))
insert₂(x0, nil)
insert₂(x0, cons₂(x1, x2))
choose₄(x0, cons₂(x1, x2), x3, 0)
choose₄(x0, cons₂(x1, x2), 0, s₁(x3))
choose₄(x0, cons₂(x1, x2), s₁(x3), s₁(x4))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPOrderProof

              ↳ QDP

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

CHOOSE₄(x, cons₂(v, w), s₁(y), s₁(z)) -> CHOOSE₄(x, cons₂(v, w), y, z)

The TRS R consists of the following rules:

sort₁(nil) -> nil
sort₁(cons₂(x, y)) -> insert₂(x, sort₁(y))
insert₂(x, nil) -> cons₂(x, nil)
insert₂(x, cons₂(v, w)) -> choose₄(x, cons₂(v, w), x, v)
choose₄(x, cons₂(v, w), y, 0) -> cons₂(x, cons₂(v, w))
choose₄(x, cons₂(v, w), 0, s₁(z)) -> cons₂(v, insert₂(x, w))
choose₄(x, cons₂(v, w), s₁(y), s₁(z)) -> choose₄(x, cons₂(v, w), y, z)

The set Q consists of the following terms:

sort₁(nil)
sort₁(cons₂(x0, x1))
insert₂(x0, nil)
insert₂(x0, cons₂(x1, x2))
choose₄(x0, cons₂(x1, x2), x3, 0)
choose₄(x0, cons₂(x1, x2), 0, s₁(x3))
choose₄(x0, cons₂(x1, x2), s₁(x3), s₁(x4))

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.

CHOOSE₄(x, cons₂(v, w), s₁(y), s₁(z)) -> CHOOSE₄(x, cons₂(v, w), y, z)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
CHOOSE₄(x1, x2, x3, x4) = x4
cons₂(x1, x2) = cons₂(x1, x2)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

              ↳ QDP

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

sort₁(nil) -> nil
sort₁(cons₂(x, y)) -> insert₂(x, sort₁(y))
insert₂(x, nil) -> cons₂(x, nil)
insert₂(x, cons₂(v, w)) -> choose₄(x, cons₂(v, w), x, v)
choose₄(x, cons₂(v, w), y, 0) -> cons₂(x, cons₂(v, w))
choose₄(x, cons₂(v, w), 0, s₁(z)) -> cons₂(v, insert₂(x, w))
choose₄(x, cons₂(v, w), s₁(y), s₁(z)) -> choose₄(x, cons₂(v, w), y, z)

The set Q consists of the following terms:

sort₁(nil)
sort₁(cons₂(x0, x1))
insert₂(x0, nil)
insert₂(x0, cons₂(x1, x2))
choose₄(x0, cons₂(x1, x2), x3, 0)
choose₄(x0, cons₂(x1, x2), 0, s₁(x3))
choose₄(x0, cons₂(x1, x2), s₁(x3), s₁(x4))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

SORT₁(cons₂(x, y)) -> SORT₁(y)

The TRS R consists of the following rules:

sort₁(nil) -> nil
sort₁(cons₂(x, y)) -> insert₂(x, sort₁(y))
insert₂(x, nil) -> cons₂(x, nil)
insert₂(x, cons₂(v, w)) -> choose₄(x, cons₂(v, w), x, v)
choose₄(x, cons₂(v, w), y, 0) -> cons₂(x, cons₂(v, w))
choose₄(x, cons₂(v, w), 0, s₁(z)) -> cons₂(v, insert₂(x, w))
choose₄(x, cons₂(v, w), s₁(y), s₁(z)) -> choose₄(x, cons₂(v, w), y, z)

The set Q consists of the following terms:

sort₁(nil)
sort₁(cons₂(x0, x1))
insert₂(x0, nil)
insert₂(x0, cons₂(x1, x2))
choose₄(x0, cons₂(x1, x2), x3, 0)
choose₄(x0, cons₂(x1, x2), 0, s₁(x3))
choose₄(x0, cons₂(x1, x2), s₁(x3), s₁(x4))

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.

SORT₁(cons₂(x, y)) -> SORT₁(y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SORT₁(x1) = SORT₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

cons₂ > SORT₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

sort₁(nil) -> nil
sort₁(cons₂(x, y)) -> insert₂(x, sort₁(y))
insert₂(x, nil) -> cons₂(x, nil)
insert₂(x, cons₂(v, w)) -> choose₄(x, cons₂(v, w), x, v)
choose₄(x, cons₂(v, w), y, 0) -> cons₂(x, cons₂(v, w))
choose₄(x, cons₂(v, w), 0, s₁(z)) -> cons₂(v, insert₂(x, w))
choose₄(x, cons₂(v, w), s₁(y), s₁(z)) -> choose₄(x, cons₂(v, w), y, z)

The set Q consists of the following terms:

sort₁(nil)
sort₁(cons₂(x0, x1))
insert₂(x0, nil)
insert₂(x0, cons₂(x1, x2))
choose₄(x0, cons₂(x1, x2), x3, 0)
choose₄(x0, cons₂(x1, x2), 0, s₁(x3))
choose₄(x0, cons₂(x1, x2), s₁(x3), s₁(x4))

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