Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/TRCSRSLASHEx4_7_15_Bor03

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__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(X)

Q is empty.

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

A__F₁(s₁(0)) -> A__P₁(s₁(0))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> A__F₁(mark₁(X))
A__F₁(s₁(0)) -> A__F₁(a__p₁(s₁(0)))
MARK₁(p₁(X)) -> A__P₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(p₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(X)

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:

A__F₁(s₁(0)) -> A__P₁(s₁(0))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> A__F₁(mark₁(X))
A__F₁(s₁(0)) -> A__F₁(a__p₁(s₁(0)))
MARK₁(p₁(X)) -> A__P₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(p₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

A__F₁(s₁(0)) -> A__F₁(a__p₁(s₁(0)))

The TRS R consists of the following rules:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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.

A__F₁(s₁(0)) -> A__F₁(a__p₁(s₁(0)))

Used argument filtering: A__F₁(x1) = x1
s₁(x1) = s
0 = 0
a__p₁(x1) = a__p
p₁(x1) = p
Used ordering: Quasi Precedence: s > [a__p, p] > 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(f₁(X)) -> MARK₁(X)
MARK₁(p₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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(p₁(X)) -> MARK₁(X)

Used argument filtering: MARK₁(x1) = x1
s₁(x1) = x1
cons₂(x1, x2) = x1
f₁(x1) = x1
p₁(x1) = p₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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(f₁(X)) -> MARK₁(X)

Used argument filtering: MARK₁(x1) = x1
s₁(x1) = x1
cons₂(x1, x2) = x1
f₁(x1) = f₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ PisEmptyProof

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

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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.