Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/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))

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 contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: INSERT₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
CHOOSE₄(x1, x2, x3, x4) = x2
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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 contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

              ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: CHOOSE₄(x1, x2, x3, x4) = x4
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ 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

                ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SORT₁(cons₂(x, y)) -> SORT₁(y)

Used argument filtering: SORT₁(x1) = x1
cons₂(x1, x2) = cons₁(x2)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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.