Termination w.r.t. Q proof of TRS/TRCSREx4_4_Luc96b

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__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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:

MARK(f(X1, X2)) → A__F(mark(X1), X2)
A__F(g(X), Y) → MARK(X)
A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
MARK(g(X)) → MARK(X)
MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

MARK(f(X1, X2)) → A__F(mark(X1), X2)
A__F(g(X), Y) → MARK(X)
A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
MARK(g(X)) → MARK(X)
MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

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

MARK(f(X1, X2)) → A__F(mark(X1), X2)
A__F(g(X), Y) → MARK(X)
A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
MARK(g(X)) → MARK(X)
MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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 oriented strictly and are deleted.

MARK(f(X1, X2)) → A__F(mark(X1), X2)
A__F(g(X), Y) → MARK(X)
MARK(f(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.

A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
MARK(g(X)) → MARK(X)

Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
f(x1, x2) = f(x1)
A__F(x1, x2) = A__F(x1)
mark(x1) = x1
g(x1) = x1
a__f(x1, x2) = a__f(x1)

Recursive path order with status [2].
Quasi-Precedence:

[f₁, a_{_{f_1]}} > A_{_F₁} > MARK₁

Status:

a_{_f₁}: [1]
f₁: [1]
MARK₁: multiset
A_{_F₁}: [1]

The following usable rules [14] were oriented:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
a__f(X1, X2) → f(X1, X2)
mark(g(X)) → g(mark(X))
mark(f(X1, X2)) → a__f(mark(X1), X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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 oriented strictly and are deleted.

MARK(g(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
g(x1) = g(x1)

Recursive path order with status [2].
Quasi-Precedence:

[MARK₁, g_1]

Status:

MARK₁: multiset
g₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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 oriented strictly and are deleted.

A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A__F(x1, x2) = A__F(x1)
g(x1) = g(x1)
mark(x1) = x1
f(x1, x2) = x2
a__f(x1, x2) = x2

Recursive path order with status [2].
Quasi-Precedence:

[A_{_F₁}, g_1]

Status:

A_{_F₁}: [1]
g₁: [1]

The following usable rules [14] were oriented:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
a__f(X1, X2) → f(X1, X2)
mark(g(X)) → g(mark(X))
mark(f(X1, X2)) → a__f(mark(X1), X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

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.