Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx6_Luc98

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:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.

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

MARK₁(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

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

MARK₁(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.
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.

MARK₁(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(first₂(X1, X2)) -> MARK₁(X2)

Used argument filtering: MARK₁(x1) = x1
first₂(x1, x2) = first₂(x1, x2)
A__FIRST₂(x1, x2) = x2
mark₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
A__FROM₁(x1) = x1
cons₂(x1, x2) = x1
a__first₂(x1, x2) = a__first₂(x1, x2)
a__from₁(x1) = x1
0 = 0
nil = nil
Used ordering: Quasi Precedence: [first_2, a__first_2] > nil

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)
A__FROM₁(X) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.
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.

MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))

Used argument filtering: MARK₁(x1) = x1
s₁(x1) = x1
from₁(x1) = from₁(x1)
A__FROM₁(x1) = x1
mark₁(x1) = x1
cons₂(x1, x2) = x1
first₂(x1, x2) = x2
a__first₂(x1, x2) = x2
a__from₁(x1) = a__from₁(x1)
0 = 0
nil = nil
Used ordering: Quasi Precedence: [from_1, a__from_1] > nil 0 > nil

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

MARK₁(s₁(X)) -> MARK₁(X)
A__FROM₁(X) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPAfsSolverProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.
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.

MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

Used argument filtering: MARK₁(x1) = x1
s₁(x1) = x1
cons₂(x1, x2) = cons₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPAfsSolverProof

                        ↳ QDP

                          ↳ QDPAfsSolverProof

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

MARK₁(s₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.
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.

MARK₁(s₁(X)) -> MARK₁(X)

Used argument filtering: MARK₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPAfsSolverProof

                        ↳ QDP

                          ↳ QDPAfsSolverProof

                            ↳ QDP

                              ↳ PisEmptyProof

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

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

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.