Termination w.r.t. Q proof of TRS/TRCSREx1_2_Luc02c

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:

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(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:

ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → S(activate(X))
2ND(cons(X, n__cons(Y, Z))) → ACTIVATE(Y)
FROM(X) → CONS(X, n__from(n__s(X)))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)

The TRS R consists of the following rules:

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → S(activate(X))
2ND(cons(X, n__cons(Y, Z))) → ACTIVATE(Y)
FROM(X) → CONS(X, n__from(n__s(X)))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)

The TRS R consists of the following rules:

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__s(X)) → S(activate(X))
2ND(cons(X, n__cons(Y, Z))) → ACTIVATE(Y)
FROM(X) → CONS(X, n__from(n__s(X)))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(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 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(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.

ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = ACTIVATE(x1)
n__cons(x1, x2) = n__cons(x1, x2)
n__from(x1) = x1
n__s(x1) = x1

Lexicographic path order with status [19].
Quasi-Precedence:

[ACTIVATE₁, n_{_{cons_2]}}

Status:

ACTIVATE₁: [1]
n_{_cons₂}: [2,1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

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

ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(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.

ACTIVATE(n__from(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__s(X)) → ACTIVATE(X)

Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = ACTIVATE(x1)
n__from(x1) = n__from(x1)
n__s(x1) = x1

Lexicographic path order with status [19].
Quasi-Precedence:

[ACTIVATE₁, n_{_{from_1]}}

Status:

ACTIVATE₁: [1]
n_{_from₁}: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

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

ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(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.

ACTIVATE(n__s(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = ACTIVATE(x1)
n__s(x1) = n__s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

[ACTIVATE₁, n_{_{s_1]}}

Status:

ACTIVATE₁: [1]
n_{_s₁}: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ PisEmptyProof

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

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(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.