Termination w.r.t. Q proof of TRS/TRCSREx4_Zan97

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:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> 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:

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__g₁(X)) -> G₁(activate₁(X))
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)
G₁(s₁(X)) -> G₁(X)
ACTIVATE₁(n__f₁(X)) -> F₁(activate₁(X))
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__g₁(X)) -> G₁(activate₁(X))
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)
G₁(s₁(X)) -> G₁(X)
ACTIVATE₁(n__f₁(X)) -> F₁(activate₁(X))
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

G₁(s₁(X)) -> G₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> 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.

G₁(s₁(X)) -> G₁(X)

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

POL( G₁(x₁) ) = max{0, x₁ - 1}

POL( s₁(x₁) ) = x₁ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> 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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> 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.

ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)

The remaining pairs can at least be oriented weakly.

ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

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

POL( ACTIVATE₁(x₁) ) = max{0, x₁ - 1}

POL( n__g₁(x₁) ) = x₁ + 2

POL( n__f₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> 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.

ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

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

POL( ACTIVATE₁(x₁) ) = max{0, x₁ - 1}

POL( n__f₁(x₁) ) = x₁ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> 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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> 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.

SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))

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

POL( SEL₂(x₁, x₂) ) = max{0, x₁ - 1}

POL( s₁(x₁) ) = x₁ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> 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.