Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/TRCSRSLASHEx4_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 strictly oriented and are deleted.

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)

The remaining pairs can at least by weakly be oriented.

MARK₁(f₂(X1, X2)) -> MARK₁(X1)

Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = MARK₁(x1)
f₂(x1, x2) = x1
A__F₂(x1, x2) = x1
mark₁(x1) = x1
g₁(x1) = g₁(x1)
a__f₂(x1, x2) = x1

Lexicographic Path Order [19].
Precedence:

g₁ > MARK₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

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 strictly oriented and are deleted.

MARK₁(f₂(X1, X2)) -> MARK₁(X1)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = MARK₁(x1)
f₂(x1, x2) = f₁(x1)

Lexicographic Path Order [19].
Precedence:

f₁ > MARK₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ 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.