Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/SK90SLASH4.17.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:

fac₁(s₁(x)) -> *₂(fac₁(p₁(s₁(x))), s₁(x))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

fac₁(s₁(x)) -> *₂(fac₁(p₁(s₁(x))), s₁(x))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

fac₁(s₁(x)) -> *₂(fac₁(p₁(s₁(x))), s₁(x))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))

The set Q consists of the following terms:

fac₁(s₁(x0))
p₁(s₁(0))
p₁(s₁(s₁(x0)))

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

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

The TRS R consists of the following rules:

fac₁(s₁(x)) -> *₂(fac₁(p₁(s₁(x))), s₁(x))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))

The set Q consists of the following terms:

fac₁(s₁(x0))
p₁(s₁(0))
p₁(s₁(s₁(x0)))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

fac₁(s₁(x)) -> *₂(fac₁(p₁(s₁(x))), s₁(x))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))

The set Q consists of the following terms:

fac₁(s₁(x0))
p₁(s₁(0))
p₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

P₁(s₁(s₁(x))) -> P₁(s₁(x))

The TRS R consists of the following rules:

fac₁(s₁(x)) -> *₂(fac₁(p₁(s₁(x))), s₁(x))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))

The set Q consists of the following terms:

fac₁(s₁(x0))
p₁(s₁(0))
p₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

P₁(s₁(s₁(x))) -> P₁(s₁(x))

Used argument filtering: P₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

fac₁(s₁(x)) -> *₂(fac₁(p₁(s₁(x))), s₁(x))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))

The set Q consists of the following terms:

fac₁(s₁(x0))
p₁(s₁(0))
p₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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:

fac₁(s₁(x)) -> *₂(fac₁(p₁(s₁(x))), s₁(x))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))

The set Q consists of the following terms:

fac₁(s₁(x0))
p₁(s₁(0))
p₁(s₁(s₁(x0)))

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