Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/AProVESLASHimproved

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:

f₂(a, x) -> f₂(g₁(x), x)
h₁(g₁(x)) -> h₁(a)
g₁(h₁(x)) -> g₁(x)
h₁(h₁(x)) -> x

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(a, x) -> f₂(g₁(x), x)
h₁(g₁(x)) -> h₁(a)
g₁(h₁(x)) -> g₁(x)
h₁(h₁(x)) -> x

Q is empty.

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

F₂(a, x) -> G₁(x)
F₂(a, x) -> F₂(g₁(x), x)
G₁(h₁(x)) -> G₁(x)
H₁(g₁(x)) -> H₁(a)

The TRS R consists of the following rules:

f₂(a, x) -> f₂(g₁(x), x)
h₁(g₁(x)) -> h₁(a)
g₁(h₁(x)) -> g₁(x)
h₁(h₁(x)) -> 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:

F₂(a, x) -> G₁(x)
F₂(a, x) -> F₂(g₁(x), x)
G₁(h₁(x)) -> G₁(x)
H₁(g₁(x)) -> H₁(a)

The TRS R consists of the following rules:

f₂(a, x) -> f₂(g₁(x), x)
h₁(g₁(x)) -> h₁(a)
g₁(h₁(x)) -> g₁(x)
h₁(h₁(x)) -> x

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

G₁(h₁(x)) -> G₁(x)

The TRS R consists of the following rules:

f₂(a, x) -> f₂(g₁(x), x)
h₁(g₁(x)) -> h₁(a)
g₁(h₁(x)) -> g₁(x)
h₁(h₁(x)) -> x

Q is empty.
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(h₁(x)) -> G₁(x)

Used argument filtering: G₁(x1) = x1
h₁(x1) = h₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

f₂(a, x) -> f₂(g₁(x), x)
h₁(g₁(x)) -> h₁(a)
g₁(h₁(x)) -> g₁(x)
h₁(h₁(x)) -> x

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

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

F₂(a, x) -> F₂(g₁(x), x)

The TRS R consists of the following rules:

f₂(a, x) -> f₂(g₁(x), x)
h₁(g₁(x)) -> h₁(a)
g₁(h₁(x)) -> g₁(x)
h₁(h₁(x)) -> x

Q is empty.
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.

F₂(a, x) -> F₂(g₁(x), x)

Used argument filtering: F₂(x1, x2) = x1
a = a
g₁(x1) = g
Used ordering: Precedence:

a > g

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

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

f₂(a, x) -> f₂(g₁(x), x)
h₁(g₁(x)) -> h₁(a)
g₁(h₁(x)) -> g₁(x)
h₁(h₁(x)) -> x

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