Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/AG01SLASHHASH3.35.trs

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:

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

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:

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

The set Q consists of the following terms:

g₁(s₁(x0))
f₁(0)
f₁(s₁(x0))
g₁(0)

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

G₁(s₁(x)) -> F₁(x)
F₁(s₁(x)) -> G₁(x)

The TRS R consists of the following rules:

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

The set Q consists of the following terms:

g₁(s₁(x0))
f₁(0)
f₁(s₁(x0))
g₁(0)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

G₁(s₁(x)) -> F₁(x)
F₁(s₁(x)) -> G₁(x)

The TRS R consists of the following rules:

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

The set Q consists of the following terms:

g₁(s₁(x0))
f₁(0)
f₁(s₁(x0))
g₁(0)

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.

G₁(s₁(x)) -> F₁(x)
F₁(s₁(x)) -> G₁(x)

Used argument filtering: G₁(x1) = x1
s₁(x1) = s₁(x1)
F₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ PisEmptyProof

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

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

The set Q consists of the following terms:

g₁(s₁(x0))
f₁(0)
f₁(s₁(x0))
g₁(0)

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