Termination w.r.t. Q proof of TRS/AG01#3.41.trs

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:

p₁(s₁(x)) -> x
fac₁(0) -> s₁(0)
fac₁(s₁(x)) -> times₂(s₁(x), fac₁(p₁(s₁(x))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p₁(s₁(x)) -> x
fac₁(0) -> s₁(0)
fac₁(s₁(x)) -> times₂(s₁(x), fac₁(p₁(s₁(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:

FAC₁(s₁(x)) -> P₁(s₁(x))
FAC₁(s₁(x)) -> FAC₁(p₁(s₁(x)))

The TRS R consists of the following rules:

p₁(s₁(x)) -> x
fac₁(0) -> s₁(0)
fac₁(s₁(x)) -> times₂(s₁(x), fac₁(p₁(s₁(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:

FAC₁(s₁(x)) -> P₁(s₁(x))
FAC₁(s₁(x)) -> FAC₁(p₁(s₁(x)))

The TRS R consists of the following rules:

p₁(s₁(x)) -> x
fac₁(0) -> s₁(0)
fac₁(s₁(x)) -> times₂(s₁(x), fac₁(p₁(s₁(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

      ↳ DependencyGraphProof

        ↳ QDP

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

FAC₁(s₁(x)) -> FAC₁(p₁(s₁(x)))

The TRS R consists of the following rules:

p₁(s₁(x)) -> x
fac₁(0) -> s₁(0)
fac₁(s₁(x)) -> times₂(s₁(x), fac₁(p₁(s₁(x))))

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