Termination w.r.t. Q proof of TRS/TRCSREx1_2_Luc02c_FR.trs

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:

2nd₁(cons₂(X, n__cons₂(Y, Z))) -> activate₁(Y)
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
cons₂(X1, X2) -> n__cons₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

2nd₁(cons₂(X, n__cons₂(Y, Z))) -> activate₁(Y)
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
cons₂(X1, X2) -> n__cons₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(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:

ACTIVATE₁(n__from₁(X)) -> FROM₁(activate₁(X))
ACTIVATE₁(n__cons₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
2ND₁(cons₂(X, n__cons₂(Y, Z))) -> ACTIVATE₁(Y)
FROM₁(X) -> CONS₂(X, n__from₁(n__s₁(X)))
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__cons₂(X1, X2)) -> CONS₂(activate₁(X1), X2)

The TRS R consists of the following rules:

2nd₁(cons₂(X, n__cons₂(Y, Z))) -> activate₁(Y)
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
cons₂(X1, X2) -> n__cons₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(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:

ACTIVATE₁(n__from₁(X)) -> FROM₁(activate₁(X))
ACTIVATE₁(n__cons₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
2ND₁(cons₂(X, n__cons₂(Y, Z))) -> ACTIVATE₁(Y)
FROM₁(X) -> CONS₂(X, n__from₁(n__s₁(X)))
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__cons₂(X1, X2)) -> CONS₂(activate₁(X1), X2)

The TRS R consists of the following rules:

2nd₁(cons₂(X, n__cons₂(Y, Z))) -> activate₁(Y)
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
cons₂(X1, X2) -> n__cons₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(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 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

ACTIVATE₁(n__cons₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

2nd₁(cons₂(X, n__cons₂(Y, Z))) -> activate₁(Y)
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
cons₂(X1, X2) -> n__cons₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(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__cons₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = x₁
POL(n__cons₂(x₁, x₂)) = 1 + x₁
POL(n__from₁(x₁)) = 1 + x₁
POL(n__s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

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

2nd₁(cons₂(X, n__cons₂(Y, Z))) -> activate₁(Y)
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
cons₂(X1, X2) -> n__cons₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(activate₁(X1), X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(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.