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.

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 remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(A__F₂(x₁, x₂)) = 1 + x₁
POL(MARK₁(x₁)) = 1 + x₁
POL(a__f₂(x₁, x₂)) = 1 + 2·x₁
POL(f₂(x₁, x₂)) = 1 + 2·x₁
POL(g₁(x₁)) = 2 + 2·x₁
POL(mark₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

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 0 SCCs with 1 less node.