Termination w.r.t. Q proof of TRS/TRCSREx6_9_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:

2nd₁(cons1₂(X, cons₂(Y, Z))) -> Y
2nd₁(cons₂(X, X1)) -> 2nd₁(cons1₂(X, activate₁(X1)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

2nd₁(cons1₂(X, cons₂(Y, Z))) -> Y
2nd₁(cons₂(X, X1)) -> 2nd₁(cons1₂(X, activate₁(X1)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(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:

2ND₁(cons₂(X, X1)) -> 2ND₁(cons1₂(X, activate₁(X1)))
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
2ND₁(cons₂(X, X1)) -> ACTIVATE₁(X1)

The TRS R consists of the following rules:

2nd₁(cons1₂(X, cons₂(Y, Z))) -> Y
2nd₁(cons₂(X, X1)) -> 2nd₁(cons1₂(X, activate₁(X1)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(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:

2ND₁(cons₂(X, X1)) -> 2ND₁(cons1₂(X, activate₁(X1)))
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
2ND₁(cons₂(X, X1)) -> ACTIVATE₁(X1)

The TRS R consists of the following rules:

2nd₁(cons1₂(X, cons₂(Y, Z))) -> Y
2nd₁(cons₂(X, X1)) -> 2nd₁(cons1₂(X, activate₁(X1)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(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 0 SCCs with 3 less nodes.