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:

f(a, n__b, X) → f(X, X, X)
ca
cb
bn__b
activate(n__b) → b
activate(X) → X

The set Q consists of the following terms:

activate(x0)
b
c
f(a, n__b, x0)



QTRS
  ↳ DependencyPairsProof

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

f(a, n__b, X) → f(X, X, X)
ca
cb
bn__b
activate(n__b) → b
activate(X) → X

The set Q consists of the following terms:

activate(x0)
b
c
f(a, n__b, x0)


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

CB
ACTIVATE(n__b) → B
F(a, n__b, X) → F(X, X, X)

The TRS R consists of the following rules:

f(a, n__b, X) → f(X, X, X)
ca
cb
bn__b
activate(n__b) → b
activate(X) → X

The set Q consists of the following terms:

activate(x0)
b
c
f(a, n__b, x0)

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:

CB
ACTIVATE(n__b) → B
F(a, n__b, X) → F(X, X, X)

The TRS R consists of the following rules:

f(a, n__b, X) → f(X, X, X)
ca
cb
bn__b
activate(n__b) → b
activate(X) → X

The set Q consists of the following terms:

activate(x0)
b
c
f(a, n__b, x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 3 less nodes.