Termination w.r.t. Q proof of TRS/TRCSREx4_4_Luc96b

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f₂(g₁(X), Y) -> f₂(X, n__f₂(n__g₁(X), activate₁(Y)))
f₂(X1, X2) -> n__f₂(X1, X2)
g₁(X) -> n__g₁(X)
activate₁(n__f₂(X1, X2)) -> f₂(activate₁(X1), X2)
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:

f₂(g₁(X), Y) -> f₂(X, n__f₂(n__g₁(X), activate₁(Y)))
f₂(X1, X2) -> n__f₂(X1, X2)
g₁(X) -> n__g₁(X)
activate₁(n__f₂(X1, X2)) -> f₂(activate₁(X1), X2)
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:

ACTIVATE₁(n__f₂(X1, X2)) -> F₂(activate₁(X1), X2)
F₂(g₁(X), Y) -> F₂(X, n__f₂(n__g₁(X), activate₁(Y)))
ACTIVATE₁(n__g₁(X)) -> G₁(activate₁(X))
ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__f₂(X1, X2)) -> ACTIVATE₁(X1)
F₂(g₁(X), Y) -> ACTIVATE₁(Y)

The TRS R consists of the following rules:

f₂(g₁(X), Y) -> f₂(X, n__f₂(n__g₁(X), activate₁(Y)))
f₂(X1, X2) -> n__f₂(X1, X2)
g₁(X) -> n__g₁(X)
activate₁(n__f₂(X1, X2)) -> f₂(activate₁(X1), X2)
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:

ACTIVATE₁(n__f₂(X1, X2)) -> F₂(activate₁(X1), X2)
F₂(g₁(X), Y) -> F₂(X, n__f₂(n__g₁(X), activate₁(Y)))
ACTIVATE₁(n__g₁(X)) -> G₁(activate₁(X))
ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__f₂(X1, X2)) -> ACTIVATE₁(X1)
F₂(g₁(X), Y) -> ACTIVATE₁(Y)

The TRS R consists of the following rules:

f₂(g₁(X), Y) -> f₂(X, n__f₂(n__g₁(X), activate₁(Y)))
f₂(X1, X2) -> n__f₂(X1, X2)
g₁(X) -> n__g₁(X)
activate₁(n__f₂(X1, X2)) -> f₂(activate₁(X1), X2)
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 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

F₂(g₁(X), Y) -> F₂(X, n__f₂(n__g₁(X), activate₁(Y)))
ACTIVATE₁(n__f₂(X1, X2)) -> F₂(activate₁(X1), X2)
ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__f₂(X1, X2)) -> ACTIVATE₁(X1)
F₂(g₁(X), Y) -> ACTIVATE₁(Y)

The TRS R consists of the following rules:

f₂(g₁(X), Y) -> f₂(X, n__f₂(n__g₁(X), activate₁(Y)))
f₂(X1, X2) -> n__f₂(X1, X2)
g₁(X) -> n__g₁(X)
activate₁(n__f₂(X1, X2)) -> f₂(activate₁(X1), X2)
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> X

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