Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/LJB01SLASHjones2.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:

f₂(empty, l) -> l
f₂(cons₂(x, k), l) -> g₃(k, l, cons₂(x, k))
g₃(a, b, c) -> f₂(a, cons₂(b, c))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

f₂(empty, l) -> l
f₂(cons₂(x, k), l) -> g₃(k, l, cons₂(x, k))
g₃(a, b, c) -> f₂(a, cons₂(b, c))

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:

f₂(empty, l) -> l
f₂(cons₂(x, k), l) -> g₃(k, l, cons₂(x, k))
g₃(a, b, c) -> f₂(a, cons₂(b, c))

The set Q consists of the following terms:

f₂(empty, x0)
f₂(cons₂(x0, x1), x2)
g₃(x0, x1, x2)

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

F₂(cons₂(x, k), l) -> G₃(k, l, cons₂(x, k))
G₃(a, b, c) -> F₂(a, cons₂(b, c))

The TRS R consists of the following rules:

f₂(empty, l) -> l
f₂(cons₂(x, k), l) -> g₃(k, l, cons₂(x, k))
g₃(a, b, c) -> f₂(a, cons₂(b, c))

The set Q consists of the following terms:

f₂(empty, x0)
f₂(cons₂(x0, x1), x2)
g₃(x0, x1, x2)

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:

F₂(cons₂(x, k), l) -> G₃(k, l, cons₂(x, k))
G₃(a, b, c) -> F₂(a, cons₂(b, c))

The TRS R consists of the following rules:

f₂(empty, l) -> l
f₂(cons₂(x, k), l) -> g₃(k, l, cons₂(x, k))
g₃(a, b, c) -> f₂(a, cons₂(b, c))

The set Q consists of the following terms:

f₂(empty, x0)
f₂(cons₂(x0, x1), x2)
g₃(x0, x1, x2)

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(cons₂(x, k), l) -> G₃(k, l, cons₂(x, k))

Used argument filtering: F₂(x1, x2) = x1
cons₂(x1, x2) = cons₁(x2)
G₃(x1, x2, x3) = x1
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

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

G₃(a, b, c) -> F₂(a, cons₂(b, c))

The TRS R consists of the following rules:

f₂(empty, l) -> l
f₂(cons₂(x, k), l) -> g₃(k, l, cons₂(x, k))
g₃(a, b, c) -> f₂(a, cons₂(b, c))

The set Q consists of the following terms:

f₂(empty, x0)
f₂(cons₂(x0, x1), x2)
g₃(x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.