Termination w.r.t. Q proof of /tmp/tmpPZwdbe/Ex14_Luc06

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:

h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
activate(X) → X

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G(a, X) → F(b, activate(X))
F(X, X) → H(a)
F(X, X) → A
G(a, X) → ACTIVATE(X)
H(X) → G(X, X)

The TRS R consists of the following rules:

h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
activate(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:

G(a, X) → F(b, activate(X))
F(X, X) → H(a)
F(X, X) → A
G(a, X) → ACTIVATE(X)
H(X) → G(X, X)

The TRS R consists of the following rules:

h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

G(a, X) → F(b, activate(X))
F(X, X) → H(a)
H(X) → G(X, X)

The TRS R consists of the following rules:

h(X) → g(X, X)
g(a, X) → f(b, activate(X))
f(X, X) → h(a)
a → b
activate(X) → X

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