Termination w.r.t. Q proof of TRS/TRCSREx6_Luc98

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__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

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(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

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(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

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(s₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
A__FROM₁(X) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.

MARK₁(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

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

POL( from₁(x₁) ) = x₁ + 1

POL( mark₁(x₁) ) = x₁

POL( first₂(x₁, x₂) ) = x₁ + x₂

POL( a__first₂(x₁, x₂) ) = x₁ + x₂

POL( a__from₁(x₁) ) = x₁ + 1

POL( MARK₁(x₁) ) = x₁

POL( A__FIRST₂(x₁, x₂) ) = x₂

POL( A__FROM₁(x₁) ) = x₁ + 1

POL( 0 ) = max{0, -1}

POL( s₁(x₁) ) = x₁ + 1

POL( nil ) = max{0, -1}

POL( cons₂(x₁, x₂) ) = x₁

The following usable rules [14] were oriented:

mark₁(0) -> 0
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(nil) -> nil
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
a__from₁(X) -> from₁(X)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__first₂(X1, X2) -> first₂(X1, X2)
a__first₂(0, X) -> nil
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

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:

MARK₁(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

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(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

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

POL( from₁(x₁) ) = x₁ + 1

POL( mark₁(x₁) ) = x₁

POL( first₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( a__first₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( a__from₁(x₁) ) = x₁ + 1

POL( MARK₁(x₁) ) = x₁ + 1

POL( A__FIRST₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( 0 ) = 1

POL( s₁(x₁) ) = 1

POL( nil ) = 0

POL( cons₂(x₁, x₂) ) = x₁ + 1

The following usable rules [14] were oriented:

a__from₁(X) -> from₁(X)
mark₁(0) -> 0
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(s₁(X)) -> s₁(mark₁(X))
a__first₂(X1, X2) -> first₂(X1, X2)
a__first₂(0, X) -> nil
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))

↳ 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__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(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.