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.

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 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.