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:

f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(activate(X))
activate(X) → X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(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__f(X1, X2)) → F(activate(X1), X2)
F(g(X), Y) → F(X, n__f(n__g(X), activate(Y)))
ACTIVATE(n__g(X)) → G(activate(X))
ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(X1, X2)) → ACTIVATE(X1)
F(g(X), Y) → ACTIVATE(Y)

The TRS R consists of the following rules:

f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(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__f(X1, X2)) → F(activate(X1), X2)
F(g(X), Y) → F(X, n__f(n__g(X), activate(Y)))
ACTIVATE(n__g(X)) → G(activate(X))
ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(X1, X2)) → ACTIVATE(X1)
F(g(X), Y) → ACTIVATE(Y)

The TRS R consists of the following rules:

f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(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:

F(g(X), Y) → F(X, n__f(n__g(X), activate(Y)))
ACTIVATE(n__f(X1, X2)) → F(activate(X1), X2)
ACTIVATE(n__g(X)) → G(activate(X))
ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(X1, X2)) → ACTIVATE(X1)
F(g(X), Y) → ACTIVATE(Y)

The TRS R consists of the following rules:

f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(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 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP

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

F(g(X), Y) → F(X, n__f(n__g(X), activate(Y)))
ACTIVATE(n__f(X1, X2)) → F(activate(X1), X2)
ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(X1, X2)) → ACTIVATE(X1)
F(g(X), Y) → ACTIVATE(Y)

The TRS R consists of the following rules:

f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(activate(X))
activate(X) → X

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