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:

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

The set Q consists of the following terms:

h(x0)
activate(x0)
f(x0, x0)
a



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)
ab
activate(X) → X

The set Q consists of the following terms:

h(x0)
activate(x0)
f(x0, x0)
a


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)
ab
activate(X) → X

The set Q consists of the following terms:

h(x0)
activate(x0)
f(x0, x0)
a

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)
ab
activate(X) → X

The set Q consists of the following terms:

h(x0)
activate(x0)
f(x0, x0)
a

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 5 less nodes.