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__f₂(g₁(X), Y) -> a__f₂(mark₁(X), f₂(g₁(X), Y))
mark₁(f₂(X1, X2)) -> a__f₂(mark₁(X1), X2)
mark₁(g₁(X)) -> g₁(mark₁(X))
a__f₂(X1, X2) -> f₂(X1, X2)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__f₂(g₁(X), Y) -> a__f₂(mark₁(X), f₂(g₁(X), Y))
mark₁(f₂(X1, X2)) -> a__f₂(mark₁(X1), X2)
mark₁(g₁(X)) -> g₁(mark₁(X))
a__f₂(X1, X2) -> f₂(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:

MARK₁(f₂(X1, X2)) -> A__F₂(mark₁(X1), X2)
A__F₂(g₁(X), Y) -> MARK₁(X)
A__F₂(g₁(X), Y) -> A__F₂(mark₁(X), f₂(g₁(X), Y))
MARK₁(g₁(X)) -> MARK₁(X)
MARK₁(f₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__f₂(g₁(X), Y) -> a__f₂(mark₁(X), f₂(g₁(X), Y))
mark₁(f₂(X1, X2)) -> a__f₂(mark₁(X1), X2)
mark₁(g₁(X)) -> g₁(mark₁(X))
a__f₂(X1, X2) -> f₂(X1, X2)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

MARK₁(f₂(X1, X2)) -> A__F₂(mark₁(X1), X2)
A__F₂(g₁(X), Y) -> MARK₁(X)
A__F₂(g₁(X), Y) -> A__F₂(mark₁(X), f₂(g₁(X), Y))
MARK₁(g₁(X)) -> MARK₁(X)
MARK₁(f₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__f₂(g₁(X), Y) -> a__f₂(mark₁(X), f₂(g₁(X), Y))
mark₁(f₂(X1, X2)) -> a__f₂(mark₁(X1), X2)
mark₁(g₁(X)) -> g₁(mark₁(X))
a__f₂(X1, X2) -> f₂(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.

MARK₁(f₂(X1, X2)) -> A__F₂(mark₁(X1), X2)
MARK₁(g₁(X)) -> MARK₁(X)
MARK₁(f₂(X1, X2)) -> MARK₁(X1)

The remaining pairs can at least be oriented weakly.

A__F₂(g₁(X), Y) -> MARK₁(X)
A__F₂(g₁(X), Y) -> A__F₂(mark₁(X), f₂(g₁(X), Y))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MARK₁(x₁) ) = max{0, x₁ - 1}

POL( f₂(x₁, x₂) ) = x₁ + 2

POL( A__F₂(x₁, x₂) ) = max{0, x₁ - 2}

POL( mark₁(x₁) ) = x₁

POL( g₁(x₁) ) = x₁ + 2

POL( a__f₂(x₁, x₂) ) = x₁ + 2

The following usable rules [14] were oriented:

a__f₂(g₁(X), Y) -> a__f₂(mark₁(X), f₂(g₁(X), Y))
a__f₂(X1, X2) -> f₂(X1, X2)
mark₁(g₁(X)) -> g₁(mark₁(X))
mark₁(f₂(X1, X2)) -> a__f₂(mark₁(X1), X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

A__F₂(g₁(X), Y) -> MARK₁(X)
A__F₂(g₁(X), Y) -> A__F₂(mark₁(X), f₂(g₁(X), Y))

The TRS R consists of the following rules:

a__f₂(g₁(X), Y) -> a__f₂(mark₁(X), f₂(g₁(X), Y))
mark₁(f₂(X1, X2)) -> a__f₂(mark₁(X1), X2)
mark₁(g₁(X)) -> g₁(mark₁(X))
a__f₂(X1, X2) -> f₂(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 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

A__F₂(g₁(X), Y) -> A__F₂(mark₁(X), f₂(g₁(X), Y))

The TRS R consists of the following rules:

a__f₂(g₁(X), Y) -> a__f₂(mark₁(X), f₂(g₁(X), Y))
mark₁(f₂(X1, X2)) -> a__f₂(mark₁(X1), X2)
mark₁(g₁(X)) -> g₁(mark₁(X))
a__f₂(X1, X2) -> f₂(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__F₂(g₁(X), Y) -> A__F₂(mark₁(X), f₂(g₁(X), Y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( A__F₂(x₁, x₂) ) = max{0, x₁ - 1}

POL( g₁(x₁) ) = x₁ + 2

POL( mark₁(x₁) ) = x₁

POL( f₂(x₁, x₂) ) = 0

POL( a__f₂(x₁, x₂) ) = 0

The following usable rules [14] were oriented:

a__f₂(g₁(X), Y) -> a__f₂(mark₁(X), f₂(g₁(X), Y))
a__f₂(X1, X2) -> f₂(X1, X2)
mark₁(g₁(X)) -> g₁(mark₁(X))
mark₁(f₂(X1, X2)) -> a__f₂(mark₁(X1), X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

a__f₂(g₁(X), Y) -> a__f₂(mark₁(X), f₂(g₁(X), Y))
mark₁(f₂(X1, X2)) -> a__f₂(mark₁(X1), X2)
mark₁(g₁(X)) -> g₁(mark₁(X))
a__f₂(X1, X2) -> f₂(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.